• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
  • EDIT Edit this Article
  • EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
  • Browse Articles
  • Learn Something New
  • Quizzes Hot
  • This Or That Game New
  • Train Your Brain
  • Explore More
  • Support wikiHow
  • About wikiHow
  • Log in / Sign up
  • Education and Communications
  • Engineering

How to Solve Circuit Problems

Last Updated: December 24, 2022

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 14 people, some anonymous, worked to edit and improve it over time. This article has been viewed 35,120 times. Learn more...

Solving circuits is one of the most challenging tasks for the undergraduate student as it involves numerous theorems, concepts, and processes for solving the circuits. But following a planned problem solving strategy simplifies the problem, making it easier to solve.

Step 1 Develop confidence in your problem solving skills.

Community Q&A

Donagan

  • Use all the tools available: transforms, linear algebra, dimensional analysis, nonlinear modelling, chief among them. Thanks Helpful 0 Not Helpful 1

solving simple circuit problems

You Might Also Like

Read a Psychrometric Chart

  • ↑ https://en.wikipedia.org/wiki/Mesh_analysis
  • ↑ https://en.wikipedia.org/wiki/Nodal_analysis

About This Article

  • Send fan mail to authors

Did this article help you?

Am I a Narcissist or an Empath Quiz

Featured Articles

Study Efficiently

Trending Articles

Everything You Need to Know to Rock the Corporate Goth Aesthetic

Watch Articles

Cook Fresh Cauliflower

  • Terms of Use
  • Privacy Policy
  • Do Not Sell or Share My Info
  • Not Selling Info

Get all the best how-tos!

Sign up for wikiHow's weekly email newsletter

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Electrical engineering

Course: electrical engineering   >   unit 2, circuit analysis overview.

  • Kirchhoff's current law
  • Kirchhoff's voltage law
  • Kirchhoff's laws
  • Labeling voltages
  • Application of the fundamental laws (setup)
  • Application of the fundamental laws (solve)
  • Application of the fundamental laws
  • Node voltage method (steps 1 to 4)
  • Node voltage method (step 5)
  • Node voltage method
  • Mesh current method (steps 1 to 3)
  • Mesh current method (step 4)
  • Mesh current method
  • Loop current method
  • Number of required equations
  • Superposition
  • Element equations (Ohm's Law, etc.)
  • Schematics (wires, nodes, branches, loops, and meshes)
  • Simplifying series and parallel resistors
  • Kirchhoff's Laws for current and voltage

The general strategy

  • Create a set of independent equations based on the elements and circuit connections.
  • Solve the system of simultaneous equations for the independent variables (voltages or currents). This often involves using techniques from linear algebra.
  • Solve the remaining individual element voltages and currents.

The methods

  • Direct application of the fundamental laws (Ohm's Law and Kirchhoff's Laws)
  • Node Voltage Method
  • Mesh Current Method and its close relative, the Loop Current Method

A final note: decomposing problems

Epilogue - computer simulation of circuits, if i can solve a circuit with a simulation, why do i need to learn these circuit analysis methods, want to join the conversation.

  • Upvote Button navigates to signup page
  • Downvote Button navigates to signup page
  • Flag Button navigates to signup page

Incredible Answer

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Engineering LibreTexts

4.5: Simple Resistive Circuits

  • Last updated
  • Save as PDF
  • Page ID 81492

  • Donald E. Richards
  • Rose-Hulman Institute of Technology via Rose-Hulman Scholar

The only circuits we will study in this course are simple resistive circuits that satisfy the lumped circuit model. Subsequent courses, especially ES 203 - Electrical Systems, will introduce you to the more complicated circuits and more complete solution techniques of circuit theory.

Simple resistive circuits are constructed of resistors and current and voltage sources. A current source is a circuit element that maintains a specified current independent of the voltage change across its terminals. A voltage source is a circuit element that maintains a specified voltage difference across its terminals independent of the current flowing through the element. The junction where two or more circuit elements meet is called a node . The voltage at a node is called a node voltage .

The brute-force approach in solving a simple resistive circuit is to identify each node as a system and apply the conservation of charge to each system. In the circuit theory language, we would say "apply Kirchhoff's current law to each node." For \(N\) nodes, this should provide \(N\) independent equations to solve for \(N\) unknowns. If additional information is required, you can introduce node voltages using Ohm's Law to relate the branch current through a resistor to the resistance and the differences in node voltages across the resistor. Sometimes this process can be shortened by applying conservation of charge to larger systems that include several nodes. However, there is not a single node voltage for the system.

A simple algorithm is given here for solving simple resistive circuits that satisfy the lumped circuit model. Although it is not always the quickest approach, it will almost always lead you to an answer by brute force. In fact, you will learn other methods later. (In later courses on circuit analysis you will learn that this approach is called nodal analysis.) We will restrict our attention here to simple resistive circuits composed of current sources, voltage sources, and resistors. The following algorithm makes use of Kirchhoff's current law (conservation of charge for lumped circuits) and Ohm's law.

Solving Simple Resistive Circuits using Conservation of Charge and Ohm's law

  • Identify and label all nodes and branch currents. Nodes have a single voltage and at least one current input and one current output. Be sure to indicate the direction of the current in each branch of the circuit. Branches connect nodes and for our purposes a branch will consist of a resistor, a voltage source, or a current source.
  • Apply conservation of charge (Kirchhoff's current law) to each node to relate the branch currents entering and leaving the node. (Recall that by definition, a node cannot accumulate charge.) Applying Kirchhoff's current law to a circuit with \(N\) nodes results in at most \(N\) independent equations relating the branch currents in the circuit. If the entire circuit can be included in a system with no charge flowing across the system boundary, only \(N-1\) independent equations can be obtained from applying KCL to the \(N\) nodes.
  • Resistors - Relate the voltage drop across each resistance to the current flowing through the resistance using Ohm's resistance model. It is critical that your voltage drop correspond with the assumed direction of the current in a resistor. By definition, \(i = \left( V_{in} - V_{out} \right) / R\) where the current flows into the resistor at a voltage \(V_{in}\) and leaves the resistor at a voltage \(V_{out}\).
  • Voltage Sources - Specified voltage difference with no constraint on the direction or magnitude of the branch current

(Note that even though we call these sources, they are not generating charge within our system. It would be more correct to think of current and voltage sources as being charge movers with no charge storage.)

  • Select one node as the zero point and arbitrarily set its voltage value to zero. This is typically the grounded node. (For our purposes, we will assume that there is no current flowing into or out of ground. The existence of ground loops with non-zero current flowing often occurs in real systems; however, we will assume that our circuits are correctly grounded.) All node voltages are then defined with respect to this zero node.
  • Check to see if you have sufficient equations to handle the number of unknowns. If you have followed this procedure a system with \(N\) nodes and \(B\) branches will have at most \(N+B\) unknowns: \(N\) node voltages and \(B\) branch currents. The actual number of unknowns is reduces by the known voltages (including ground) and currents specified for the circuit.
  • Solve for the unknowns. (Matrix algebra can be a real time saver for large problems especially where there is a simple pattern for setting up the equations. For small systems of equations, it may be quicker to just set up the equations and let MAPLE or Mathematica solve them directly. The preferred method usually depends upon your familiarity with the software and how easily you can set up the matrices.)
  • Check your answers. One approach is to examine a system that contains several nodes and then check to see that conservation of charge is satisfied.

The following example shows how to apply this technique to a simple resistive circuit that includes both a voltage source and a current source.

Example — Simple Circuit #1

Find the unknown branch currents and node voltages in the circuit shown in the figure.

A circuit with two sub-loops, containing a total of 5 resistors and one 50-V battery. A value and direction of current along one of the outer boundaries of the circuit is given.

Figure \(\PageIndex{1}\): Circuit diagram for a circuit with 6 nodes, 1 battery, 5 resistors, and one known branch current.

Strategy \(\rightarrow\) This is clearly a circuit problem and will require application of conservation of charge along with suitable constitutive relationships.

First step is to label and locate all the nodes and branches. This gives

6 nodes: \(V_{1}, V_{2}, V_{3}, V_{4}, V_{5,} V_{6}\)

7 branches: \(i_{a}, i_{b}, i_{c}, i_{d}, i_{e}, i_{f}, i_{g}\)

Circuit diagram from above with all nodes and branch currents marked and labeled. Nodes are numbered in increasing order, clockwise from the top left corner. Currents are assumed to flow clockwise around the left loop of the circuit and counterclockwise around the right loop.

Figure \(\PageIndex{2}\): Circuit diagram from above with all nodes and branch currents marked and labeled.

Writing conservation of charge for each node assuming no charge accumulation (lumped circuit assumption), we get six equations relating the currents:

\[ \begin{align*} &\text{Node 1:} \quad 0 =i_{g} - i_{a} \quad\quad &\text{Node 4:} \quad 0=i_{d}-i_{c} \\ &\text{Node 2:} \quad 0=i_{a}+i_{b}-i_{e} \quad\quad\quad\quad &\text{Node 5:} \quad 0=i_{e}-i_{f}-i_{e} \\ &\text{Node 3:} \quad 0=i_{c}-i_{d} \quad \quad &\text{Node 6:} \quad 0=i_{f}-i_{g} \end{align*} \nonumber \]

Note that we have assumed no current in the ground connection. Also note that only five of these six equations are independent equations.

Now we can write the branch equations by using the appropriate constitutive relations:

\[ \begin{align*} & \text {Branch a: } \quad \left(V_{1}-V_{2}\right)=(3 \ \Omega) i_{a} \quad\quad \text { Branch e: } \quad\left(V_{2}-V_{5}\right)=(5 \ \Omega) i_{e} \\ & \text {Branch b: } \quad \left(V_{3}-V_{2}\right)=(2 \ \Omega) i_{b} \quad\quad \text { Branch f: } \quad \left(V_{5}-V_{6}\right)=(1 \ \Omega) i_{f} \\ & \text {Branch c: } \quad \left(V_{4}-V_{3}\right)=(1 \ \Omega) i_{c} \quad\quad \text { Branch g: } \quad i_{g}=4 \mathrm{~A} \\ & \text {Branch d: } \quad V_{5}-V_{4}=50 \text { volts } \end{align*} \nonumber \]

All seven of the branch equations are independent.

We now have 13 unknowns and 12 equations (5 node and 7 branch). The remaining equation comes from assigning the ground voltage to node 6.

\[ V_6 = 0 \nonumber \]

We should now have sufficient information to solve for the remaining voltages and currents. This is most easily done use a software package, e.g. MAPLE™ or EES.

MAPLE Solution:

There are 6 nodes in this problem (numbered 1,2,3,4,5,6). If we write conservation of charge for each node we will get 6 equations that relate the 6 unknown currents in the circuit:

Code to set up equations that relate charge for each node to the various currents in the circuit, using variables.

Figure \(\PageIndex{3}\): Code in MAPLE to set up equations relating conservation of charge for each node to the circuit currents.

Now writing the 7 branch equations:

Code to describe branch voltage for each branch of the circuit using constitutive relations, in terms of variables for all unknown values.

Figure \(\PageIndex{4}\): Code in MAPLE to set up equations for voltage drop in each branch of the circuit, based on constitutive relations.

There are 7 branch currents and 6 node voltages. Unfortunately only 5 of the 6 conservation of charge relations are independent, so we need an additional constraint. The remaining constraint is supplied by establishing a ground node with zero potential. Based on the circuit schematic this node should probably be Node 6.

Code assigning the value of 0 to the variable for voltage at node 6.

Figure \(\PageIndex{5}\): Code in MAPLE to assign a value of 0 to the ground node (node 6).

Now we can collect 12 independent equations to solve for the 6 currents and the 6 voltages.

Code to solve for all node voltages and branch currents from the equations set up above, using the equations for nodes 1-5, and the numerical values returned by the program for each of these variables.

Figure \(\PageIndex{6}\): Code to solve for the problem variables, using 5 of the conservation of charge equations, and the numerical solutions returned by the program.

To check our solutions, let's try some variations and see what happens. If any 5 of the conservation of charge equations work then we ought to be able to solve with a different set. Let's try for a couple of cases. First, let's try using Node 6 instead of Node 1:

Code to solve for all node voltages and branch currents from the equations set up above, using the equations for nodes 2-6, and the numerical values returned by the program for each of these variables.

Figure \(\PageIndex{7}\): Code to solve for the problem variables, using another set of 5 of the conservation of charge equations, and the numerical solutions returned by the program.

Now let's try using all of the nodes except Node 5:

Code to solve for all node voltages and branch currents from the equations set up above, using the equations for nodes 1, 2, 3, 4, and 6, and the numerical values returned by the program for each of these variables.

Figure \(\PageIndex{8}\): Code to solve for the problem variables, using another set of 5 of the conservation of charge equations, and the numerical solutions returned by the program.

Looks OK. Now let's see what happens if we don't include the GROUND condition and don't put \(v_6\) in the variable list:

Code from the previous figure is repeated, except for the omission of v6 from the variable list and the omission of GROUND from the conditions list. Solutions returned by the program are numerical for branch currents but are in terms of v6 for all node voltages.

Figure \(\PageIndex{9}\): Code from Figure \(\PageIndex{8}\) with \(v_6\) omitted from the variable list and the GROUND condition omitted from the conditions list, and the solutions returned by the program. All voltage solutions are in terms of \(v_6\).

Now let's try and see what happens if we just accidentally don't include the GROUND condition but we do include v6 in the variable list:

Code from Figure 8 above with the GROUND condition omitted from the conditions list, and the solutions returned by the program. All branch currents solutions are numerical, and all node voltages are in terms of v6.

Figure \(\PageIndex{10}\): Code from Figure \(\PageIndex{8}\) with the GROUND condition omitted from the conditions list, and the solutions returned by the program. All voltage solutions are in terms of \(v_6\).

Notice how all the voltages include a \(v_6\) component. What does this mean?

Finally, what happens if we just leave out the variables list?

Code from Figure 8 above with the variable list removed, and the numerical solutions returned by the program.

Figure \(\PageIndex{11}\): Code from Figure \(\PageIndex{8}\) with the variables list removed, and the numerical solutions returned by the program.

Same answers, just in a different order.

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Physics LibreTexts

3.4: General circuits and solution methods

  • Last updated
  • Save as PDF
  • Page ID 24997

  • David H. Staelin
  • Massachusetts Institute of Technology via MIT OpenCourseWare

Kirchoff’s Laws

Circuits are generally composed of lumped elements or “branches” connected at nodes to form two- or three-dimensional structures, as suggested in Figure 3.4.1. They can be characterized by the voltages vi at each node or across each branch, or by the currents ij flowing in each branch or in a set of current loops. To determine the behavior of such circuits we develop simultaneous linear equations that must be satisfied by the unknown voltages and currents. Kirchoff’s laws generally provide these equations.

Although circuit analysis is often based in part on Kirchoff’s laws, these laws are imperfect due to electromagnetic effects. For example, Kirchoff’s voltage law (KVL) says that the voltage drops vi associated with each lumped element around any loop must sum to zero, i.e.:

\[\sum_{i} v_{i}=0 \qquad\qquad\qquad \text{(Kirchoff’s voltage law [KVL])} \]

Figure 3.4.1.PNG

which can be derived from the integral form of Faraday’s law:

\[ \oint_{\mathrm{C}} \overline{\mathrm{E}} \bullet \mathrm{d} \overline{\mathrm{s}}=-(\partial / \partial \mathrm{t}) \oiint_{\mathrm{A}} \overline{\mathrm{B}} \bullet \mathrm{d} \overline{\mathrm{a}}\]

This integral of \(\overline{\mathrm{E}} \bullet \mathrm{d} \overline{\mathrm{s}} \) across any branch yields the voltage across that branch. Therefore the sum of branch voltages around any closed contour is zero if the net magnetic flux through that contour is constant; this is the basic assumption of KVL.

KVL is clearly valid for any static circuit. However, any branch carrying time varying current will contribute time varying magnetic flux and therefore voltage to all adjacent loops plus others nearby. These voltage contributions are typically negligible because the currents and loop areas are small relative to the wavelengths of interest (λ = c/f) and the KVL approximation then applies. A standard approach to analyzing circuits that violate KVL is to determine the magnetic energy or inductance associated with any extraneous magnetic fields, and to model their effects in the circuit with a lumped parasitic inductance in each affected current loop.

The companion relation to KVL is Kirchoff’s current law (KCL), which says that the sum of the currents i j flowing into any node is zero:

\[\sum_{j} \mathrm{i}_{j}=0 \qquad \qquad \qquad \text{(Kirchoff’s current law)} \]

This follows from conservation of charge (2.4.19) when no charge storage on the nodes is allowed:

\[(\partial / \partial \mathrm{t}) \int \int \int_{\mathrm{V}} \rho \mathrm{d} \mathrm{v}=-\oiint_{\mathrm{A}} \overline{\mathrm{J}} \bullet \mathrm{d} \overline{\mathrm{a}} \qquad\qquad\qquad \text{(conservation of charge) } \]

If no charge can be stored on the volume V of a node, then \((\partial / \partial \mathrm{t}) \int \int \int_{\mathrm{V}} \rho \mathrm{d} \mathrm{v}=0 \), and there can be V no net current into that node.

For static problems, KCL is exact. However, the physical nodes and the wires connecting those nodes to lumped elements typically exhibit varying voltages and \(\vec D\), and therefore have capacitance and the ability to store charge, violating KCL. If the frequency is sufficiently high that such parasitic capacitance at any node becomes important, that parasitic capacitance can be modeled as an additional lumped element attached to that node.

Solving circuit problems

To determine the behavior of any given linear lumped element circuit a set of simultaneous equations must be solved, where the number of equations must equal or exceed the number of unknowns. The unknowns are generally the voltages and currents on each branch; if there are b branches there are 2b unknowns.

Figure 3.4.2(a) illustrates a simple circuit with b = 12 branches, p = 6 loops, and n = 7 nodes. A set of loop currents uniquely characterizes all currents if each loop circles only one “hole” in the topology and if no additional loops are added once every branch in the circuit is incorporated in at least one loop. Although other definitions for the loop currents can adequately characterize all branch currents , they are not explored here. Figure 3.4.2(b) illustrates a bridge circuit with b = 6, p = 3, and n = 4.

Figure 3.4.2.PNG

The simplest possible circuit has one node and one branch, as illustrated in Figure 3.4.3(a).

Figure 3.4.3.PNG

It is easy to see from the figure that the number b of branches in a circuit is:

\[b=n+p-1 \]

As we add either nodes or branches to the illustrated circuit in any sequence and with any placement, Equation (3.4.5) is always obeyed. If we add voltage or current sources to the circuit, they too become branches.

The voltage and current for each branch are initially unknown and therefore any circuit has 2b unknowns. The number of equations is also b + (n – 1) + p = 2b, where the first b in this expression corresponds to the equations relating voltage to current in each branch, n-1 is the number of independent KCL equations, and p is the number of loops and KVL equations; (3.4.5) says (n – 1) + p = b. Therefore, since the numbers of unknowns and linear equations match, we may solve them. The equations are linear because Maxwell’s equations are linear for RLC circuits.

Often circuits are so complex that it is convenient for purposes of analysis to replace large sections of them with either a two-terminal Thevenin equivalent circuit or Norton equivalent circuit . This can be done only when that circuit is incrementally linear with respect to voltages imposed at its terminals. Thevenin equivalent circuits consist of a voltage source V Th (t) in series with a passive linear circuit characterized by its frequency-dependent impedance \(\underline{Z}(\omega)=\mathrm{R}+\mathrm{jX}\), while Norton equivalent circuits consist of a current source I No (t) in parallel with an impedance \(\underline{Z}(\omega)\).

An important example of the utility of equivalent circuits is the problem of designing a matched load \(\underline{Z}_{L}(\omega)=R_{L}(\omega)+j X_{L}(\omega) \) that accepts the maximum amount of power available from a linear source circuit, and reflects none. The solution is simply to design the load so its impedance \(\underline{Z}_{L}(\omega)\) is the complex conjugate of the source impedance: \(\underline{Z}_{\mathrm{L}}(\omega)=\underline{Z}^{*}(\omega) \). For both Thevenin and Norton equivalent sources the reactance of the matched load cancels that of the source [X L (ω) = - X(ω)] and the two resistive parts are set equal, R = R L .

One proof that a matched load maximizes power transfer consists of computing the timeaverage power P d dissipated in the load as a function of its impedance, equating to zero its derivative dP d /dω, and solving the resulting complex equation for R L and X L . We exclude the possibility of negative resistances here unless those of the load and source have the same sign; otherwise the transferred power can be infinite if R L = -R.

Example \(\PageIndex{A}\)

The bridge circuit of Figure 3.4.2(b) has five branches connecting four nodes in every possible way except one. Assume both parallel branches have 0.1-ohm and 0.2-ohm resistors in series, but in reverse order so that R 1 = R 4 = 0.1, and R 2 = R 3 = 0.2. What is the resistance R of the bridge circuit between nodes a and d if R 5 = 0? What is R if R 5 = ∞? What is R if R 5 is 0.5 ohms?

When R 5 = 0 then the node voltages v b = v c , so R 1 and R 3 are connected in parallel and have the equivalent resistance R 13// . Kirchoff’s current law “KCL” (3.4.3) says the current flowing into node “a” is I = (v a - v b )(R 1 -1 + R 3 -1 ). If V ab ≡ (v a - v b ), then V ab = IR 13// and R 13// = (R 1 -1 + R 3 -1 ) -1 = (10+5) -1 = 0.067Ω = R 24// . These two circuits are in series so their resistances add: R = R 13// + R 24// ≅ 0.133 ohms. When R 5 = ∞, R 1 and R 2 are in series with a total resistance R 12s of 0.1 + 0.2 = 0.3Ω = R 34s . These two resistances, R 12s and R 34s are in parallel, so R = (R 12s -1 + R 34s -1 ) -1 = 0.15Ω. When R 5 is finite, then simultaneous equations must be solved. For example, the currents flowing into each of nodes a, b, and c sum to zero, yielding three simultaneous equations that can be solved for the vector \(\overline{\mathrm{V}}=\left[\mathrm{v}_{\mathrm{a}}, \mathrm{v}_{\mathrm{b}}, \mathrm{v}_{\mathrm{c}}\right]\); we define v d = 0. Thus

\(\left(\mathrm{v}_{\mathrm{a}}-\mathrm{v}_{\mathrm{b}}\right) / \mathrm{R}_{1}+\left(\mathrm{v}_{\mathrm{a}}-\mathrm{v}_{\mathrm{c}}\right) / \mathrm{R}_{3}=\mathrm{I}=\mathrm{v}_{\mathrm{a}}\left(\mathrm{R}_{1}^{-1}+\mathrm{R}_{3}^{-1}\right)-\mathrm{v}_{\mathrm{b}} \mathrm{R}_{1}^{-1}-\mathrm{v}_{\mathrm{c}} \mathrm{R}_{3}^{-1}=15 \mathrm{v}_{\mathrm{a}}-10 \mathrm{v}_{\mathrm{b}}-5 \mathrm{v}_{\mathrm{c}}\).

KCL for nodes b and c similarly yield: -10v a + 17v b - 2v c = 0, and -5v a -2v b + 17v c = 0. If we define the current vector \(\overline{\mathrm{I}}=[\mathrm{I}, 0,0]\), then these three equations can be written as a matrix equation:

\[\overline{\overline{\mathrm{G}}} \overline{\mathrm{v}}=\overline{\mathrm{I}}, \text { where } \overline{\overline{\mathrm{G}}}=\left[\begin{array}{ccc} 15 & -10 & -5 \\ -10 & 17 & -2 \\ -5 & -2 & 17 \end{array}\right]. \nonumber\]

Since the desired circuit resistance between nodes a and d is R = v a /I, we need only solve for v a in terms of I, which follows from \(\overline{\mathrm{v}}=\overline{\overline{\mathrm{G}}}^{-1} \overline{\mathrm{I}}\), provided the conductance matrix \(\overline{\overline{\mathrm{G}}}\) is not singular (here it is not). Thus R = 0.146Ω, which is intermediate between the first two solutions, as it should be.

Resistors in Circuits

Practice problem 1.

  • the equivalent resistance
  • the current from the power supply
  • the current through each resistor
  • the voltage drop across each resistor
  • the power dissipated in each resistor

Follow the rules for series circuits.

Resistances in series add up.

Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.

Current is constant through resistors in series.

I T  =  I 1  =  I 2  =  I 3  =  1.25 A

The voltage drops can be found using Ohm's law.

Verify your calculations by adding the voltage drops. On a series circuit they should equal the voltage increase of the power supply.

We're good, so let's finish.

There are three equations for determining power. Since we have three resistors, let's apply a different equation to each as an exercise.

In a series circuit, the element with the greatest resistance consumes the most power.

Follow the rules for parallel circuits.

Resistances in parallel combine according to the sum-of-inverses rule.

(Note: we'll answer part iv before part iii.) On a parallel circuit, each branch experiences the same voltage drop.

V T  =  V 1  =  V 2  =  V 3  =  125 V

The current in each branch can be found using Ohm's law.

Verify your calculations by adding the currents. On a parallel circuit they should add up to the current from the power supply.

Good, it works.

Again as an exercise, use a different equation to determine the electric power of each resistor.

In a parallel circuit, the element with the least resistance consumes the most power.

practice problem 2

  • Draw a schematic diagram of this circuit.
  • Which of these appliances can be operated simultaneously without tripping the circuit breaker?

Outlets are wired in parallel so that the appliances on a circuit are independent of one another. Turning the coffee maker off will not result in the toaster turning off (assuming both were on at the same time). Each appliance will also get the same regulated voltage, which simplifies the design of electrical devices. The downside to this scheme is that the parallel currents can add up to dangerously high levels. A circuit breaker in series before the parallel branches can prevent overloads by automatically opening the circuit.

A 15 A circuit operating at 120 V consumes 1,800 W of total power.

P  =  VI  =  (120 V)(15 A) = 1,800 W

Total power in a parallel circuit is the sum of the power consumed on the individual branches.

On this circuit, only the coffee maker and toaster can be operated simultaneously. All other combinations will trigger the circuit breaker to open.

practice problem 3

  • the current through
  • the voltage drop across
  • the power dissipated by each resistor

The way to solve a complex problem is to break it down into a series of simpler problems. Be careful not to lose sight of your goal among all the bits and pieces, however. Before beginning plot your course. In this case we'll start by finding the effective resistance of the entire circuit and the current from the battery. This sets us up to get the current in all the different segments of the circuit. (The current divides and divides again in an effort to follow the path of least resistance.) After that, it's a simple matter to calculate the voltage drops in each resistor using V  =  IR and the power dissipated using P  =  VI . No part of this problem is difficult by itself, but since the circuit is so complex we'll be quite busy for a little while.

Let's begin the process by combining resistors. There are four series pairs in this circuit.

These pairs form two parallel circuits, one on the left and one on the right.

Each gang of four resistors is in series with another.

The left and right halves of the circuit are parallel to each other and to the battery.

Now that we have the effective resistance of the entire circuit, let's determine the current from the power supply using Ohm's law.

Now walk through the circuit (not literally of course). At each junction the current will divide with more taking the path with less resistance and less taking the path with more resistance. Since charge doesn't leak out anywhere on a complete circuit, the current will be the same for all those elements in series with one another.

The left and right halves of the circuit are identical in overall resistance, which means the current will divide evenly between them.

On each side the current divides again into two parallel branches.

Use V  =  IR over and over and over again to determine the voltage drops. (See the tables at the end of this solution.)

Use P  =  VI (or P  =  I 2 R or P  =  V 2 / R ) over and over again to determine the power dissipated. These last two tasks are so tedious you should use a spreadsheet application of some sort. Enter the resistance values given and the current values just calculated into columns and instruct your electronic device of choice to multiply appropriately. Something like this…

practice problem 4

  • Calculate the equivalent resistance of the circuit.
  • Calculate the current through the battery.
  • Graph voltage as a function of location on the circuit assuming that V a  = 0 V at the negative terminal of the battery.
  • Graph current as a function of location on the circuit.

Here are the solutions…

The total resistance in a series circuit is the sum of the individual resistances…

The total current can be found from Ohm's law…

The voltage in a circuit rises in a battery and drops in a resistor (when we follow the flow of conventional current). The rise in the battery is given as 12 V and the drops in each resistor can be found through repeated use of Ohm's law…

Starting at zero volts on the negative terminal of the battery, the voltage goes up 12 V then drops 2 V, 6 V, and 4 V, which brings us back to zero. (We are assuming that the battery and wires have negligible resistance.) Here's how it looks when graphed.

Here's how it looks when the graph is superimposed on the circuit.

Current is everywhere the same in a series circuit. We've already determined it's 0.667 A. All that remains is to draw a horizontal line at two-thirds of an amp.

How to Solve a Basic Circuit

license

Introduction: How to Solve a Basic Circuit

How to Solve a Basic Circuit

When it comes to electrical engineering, solving circuits is a fundamental skill. Before beginning this instructable it is important to note the three main aspects of a circuit, current (I) how much physical electricity is running through each wire, measured in Amps, potential difference, aka voltage (V) measured in Volts, essentially the potential a circuit has to push electricity through the wire(s), and total resistance (R) , measured in Ohms, how much the circuit is resisting the flow of electricity. An entire circuit is shown in Fig. 1 with each part labelled. Through this instructable you will learn how to solve for the total resistance of a circuit and then by using Ohm’s Law (I=V/R) to solve for the current (I) that a circuit will use.

Assumptions:

This instructable requires the basic knowledge of algebra along with the ability to use a calculator to solve algebraic equations. At Table 1 you can find tips for calculating the math values required in this instructable.

Step 1: Identify the Voltage (V) of the Circuit and Recognize the Type of Resistance

Identify the Voltage (V) of the Circuit and Recognize the Type of Resistance

The voltage of a circuit is displayed by the symbol found in Fig. 1 . You can simply transcribe this value and keep it until we are solving for current (I) in Step 3.

A Resistor is a small component of a circuit used to change how much resistance is within the circuit. It is illustrated as such in Fig. 2 .

These resistors can be organized in two basic ways, either in parallel or in series. Parallel resistors look like a “ladder” on a circuit, each one is stacked on top of each other so to speak. This is illustrated in Fig. 3.

Series resistors look like a “string” on a circuit, each one is placed end-to-end in a row, all traveling in the same direction. This can be seen in Fig. 4.

Look at the circuit you are given and identify which type of resistance your circuit uses, then you may proceed to step 2.

Step 2: Finding the Total Resistance

Finding the Total Resistance

Because Ohm’s Law is found to be “I=V/R”, then we need one single, total value for resistance (R).

To find the total resistance of a series configuration, you simply add them together. For example, if you have three resistors R1, R2, and R3, the total resistance is as such “R=R1+R2+R3”. This is illustrated in Fig. 5.

To find the total resistance of a parallel configuration, we must divide one by each resistor value separately, add them together, then divide one by this total. Such as (1/R1 + 1/R2 + 1/R3) = 1/R ==> R=___. This is illustrated by Fig. 6.

If you would like assistance with how to type these values into a calculator, please see Table 1.

Step 3: Solve for Current (I)

Solve for Current (I)

Once you have found the total resistance (R) and given voltage (V) we plug it into the Ohm’s Law equation (I=V/R). For example if our voltage was 4 Volts and our total resistance was 9 Ohm’s, then our current (I) would be 4/9 Amps, which is equal to 0.4444 Amps. Please see Fig. 7 for an illustration of this method.

Step 4: Try It for Yourself

Try It for Yourself

As seen in Figures 8-11 there are four circuit examples.

Feel free to try each example while going through this instructable. The correct values for each example are found in Table 2 below.

Step 5: Answers to Example Problems

Answers to Example Problems

Recommendations

Retro Internet Radio Using ESP32

Fix It Contest

Fix It Contest

Stay Warm Contest

Stay Warm Contest

Remake It - Autodesk Design & Make - Student Contest

Remake It - Autodesk Design & Make - Student Contest

Physics Problems with Solutions

Physics Problems with Solutions

  • Electric Circuits
  • Electrostatic
  • Calculators
  • Practice Tests
  • Simulations

solving simple circuit problems

Solve DC Circuits Problems

Kirchhoff's and Ohm's laws are used to solve DC circuits problems. There are 3 examples; solve in the order that they are presented; this will make it easier to fully understand them.

DC circuit example 1

More References and Links

Popular pages.

  • Privacy Policy

Youtube

  • TPC and eLearning
  • Read Watch Interact
  • What's NEW at TPC?
  • Practice Review Test
  • Teacher-Tools
  • Subscription Selection
  • Seat Calculator
  • Ad Free Account
  • Edit Profile Settings
  • Classes (Version 2)
  • Student Progress Edit
  • Task Properties
  • Export Student Progress
  • Task, Activities, and Scores
  • Metric Conversions Questions
  • Metric System Questions
  • Metric Estimation Questions
  • Significant Digits Questions
  • Proportional Reasoning
  • Acceleration
  • Distance-Displacement
  • Dots and Graphs
  • Graph That Motion
  • Match That Graph
  • Name That Motion
  • Motion Diagrams
  • Pos'n Time Graphs Numerical
  • Pos'n Time Graphs Conceptual
  • Up And Down - Questions
  • Balanced vs. Unbalanced Forces
  • Change of State
  • Force and Motion
  • Mass and Weight
  • Match That Free-Body Diagram
  • Net Force (and Acceleration) Ranking Tasks
  • Newton's Second Law
  • Normal Force Card Sort
  • Recognizing Forces
  • Air Resistance and Skydiving
  • Solve It! with Newton's Second Law
  • Which One Doesn't Belong?
  • Component Addition Questions
  • Head-to-Tail Vector Addition
  • Projectile Mathematics
  • Trajectory - Angle Launched Projectiles
  • Trajectory - Horizontally Launched Projectiles
  • Vector Addition
  • Vector Direction
  • Which One Doesn't Belong? Projectile Motion
  • Forces in 2-Dimensions
  • Being Impulsive About Momentum
  • Explosions - Law Breakers
  • Hit and Stick Collisions - Law Breakers
  • Case Studies: Impulse and Force
  • Impulse-Momentum Change Table
  • Keeping Track of Momentum - Hit and Stick
  • Keeping Track of Momentum - Hit and Bounce
  • What's Up (and Down) with KE and PE?
  • Energy Conservation Questions
  • Energy Dissipation Questions
  • Energy Ranking Tasks
  • LOL Charts (a.k.a., Energy Bar Charts)
  • Match That Bar Chart
  • Words and Charts Questions
  • Name That Energy
  • Stepping Up with PE and KE Questions
  • Case Studies - Circular Motion
  • Circular Logic
  • Forces and Free-Body Diagrams in Circular Motion
  • Gravitational Field Strength
  • Universal Gravitation
  • Angular Position and Displacement
  • Linear and Angular Velocity
  • Angular Acceleration
  • Rotational Inertia
  • Balanced vs. Unbalanced Torques
  • Getting a Handle on Torque
  • Torque-ing About Rotation
  • Properties of Matter
  • Fluid Pressure
  • Buoyant Force
  • Sinking, Floating, and Hanging
  • Pascal's Principle
  • Flow Velocity
  • Bernoulli's Principle
  • Balloon Interactions
  • Charge and Charging
  • Charge Interactions
  • Charging by Induction
  • Conductors and Insulators
  • Coulombs Law
  • Electric Field
  • Electric Field Intensity
  • Polarization
  • Case Studies: Electric Power
  • Know Your Potential
  • Light Bulb Anatomy
  • I = ∆V/R Equations as a Guide to Thinking
  • Parallel Circuits - ∆V = I•R Calculations
  • Resistance Ranking Tasks
  • Series Circuits - ∆V = I•R Calculations
  • Series vs. Parallel Circuits
  • Equivalent Resistance
  • Period and Frequency of a Pendulum
  • Pendulum Motion: Velocity and Force
  • Energy of a Pendulum
  • Period and Frequency of a Mass on a Spring
  • Horizontal Springs: Velocity and Force
  • Vertical Springs: Velocity and Force
  • Energy of a Mass on a Spring
  • Decibel Scale
  • Frequency and Period
  • Closed-End Air Columns
  • Name That Harmonic: Strings
  • Rocking the Boat
  • Wave Basics
  • Matching Pairs: Wave Characteristics
  • Wave Interference
  • Waves - Case Studies
  • Color Addition and Subtraction
  • Color Filters
  • If This, Then That: Color Subtraction
  • Light Intensity
  • Color Pigments
  • Converging Lenses
  • Curved Mirror Images
  • Law of Reflection
  • Refraction and Lenses
  • Total Internal Reflection
  • Who Can See Who?
  • Formulas and Atom Counting
  • Atomic Models
  • Bond Polarity
  • Entropy Questions
  • Cell Voltage Questions
  • Heat of Formation Questions
  • Reduction Potential Questions
  • Oxidation States Questions
  • Measuring the Quantity of Heat
  • Hess's Law
  • Oxidation-Reduction Questions
  • Galvanic Cells Questions
  • Thermal Stoichiometry
  • Molecular Polarity
  • Quantum Mechanics
  • Balancing Chemical Equations
  • Bronsted-Lowry Model of Acids and Bases
  • Classification of Matter
  • Collision Model of Reaction Rates
  • Density Ranking Tasks
  • Dissociation Reactions
  • Complete Electron Configurations
  • Enthalpy Change Questions
  • Equilibrium Concept
  • Equilibrium Constant Expression
  • Equilibrium Calculations - Questions
  • Equilibrium ICE Table
  • Ionic Bonding
  • Lewis Electron Dot Structures
  • Line Spectra Questions
  • Measurement and Numbers
  • Metals, Nonmetals, and Metalloids
  • Metric Estimations
  • Metric System
  • Molarity Ranking Tasks
  • Mole Conversions
  • Name That Element
  • Names to Formulas
  • Names to Formulas 2
  • Nuclear Decay
  • Particles, Words, and Formulas
  • Periodic Trends
  • Precipitation Reactions and Net Ionic Equations
  • Pressure Concepts
  • Pressure-Temperature Gas Law
  • Pressure-Volume Gas Law
  • Chemical Reaction Types
  • Significant Digits and Measurement
  • States Of Matter Exercise
  • Stoichiometry - Math Relationships
  • Subatomic Particles
  • Spontaneity and Driving Forces
  • Gibbs Free Energy
  • Volume-Temperature Gas Law
  • Acid-Base Properties
  • Energy and Chemical Reactions
  • Chemical and Physical Properties
  • Valence Shell Electron Pair Repulsion Theory
  • Writing Balanced Chemical Equations
  • Mission CG1
  • Mission CG10
  • Mission CG2
  • Mission CG3
  • Mission CG4
  • Mission CG5
  • Mission CG6
  • Mission CG7
  • Mission CG8
  • Mission CG9
  • Mission EC1
  • Mission EC10
  • Mission EC11
  • Mission EC12
  • Mission EC2
  • Mission EC3
  • Mission EC4
  • Mission EC5
  • Mission EC6
  • Mission EC7
  • Mission EC8
  • Mission EC9
  • Mission RL1
  • Mission RL2
  • Mission RL3
  • Mission RL4
  • Mission RL5
  • Mission RL6
  • Mission KG7
  • Mission RL8
  • Mission KG9
  • Mission RL10
  • Mission RL11
  • Mission RM1
  • Mission RM2
  • Mission RM3
  • Mission RM4
  • Mission RM5
  • Mission RM6
  • Mission RM8
  • Mission RM10
  • Mission LC1
  • Mission RM11
  • Mission LC2
  • Mission LC3
  • Mission LC4
  • Mission LC5
  • Mission LC6
  • Mission LC8
  • Mission SM1
  • Mission SM2
  • Mission SM3
  • Mission SM4
  • Mission SM5
  • Mission SM6
  • Mission SM8
  • Mission SM10
  • Mission KG10
  • Mission SM11
  • Mission KG2
  • Mission KG3
  • Mission KG4
  • Mission KG5
  • Mission KG6
  • Mission KG8
  • Mission KG11
  • Mission F2D1
  • Mission F2D2
  • Mission F2D3
  • Mission F2D4
  • Mission F2D5
  • Mission F2D6
  • Mission KC1
  • Mission KC2
  • Mission KC3
  • Mission KC4
  • Mission KC5
  • Mission KC6
  • Mission KC7
  • Mission KC8
  • Mission AAA
  • Mission SM9
  • Mission LC7
  • Mission LC9
  • Mission NL1
  • Mission NL2
  • Mission NL3
  • Mission NL4
  • Mission NL5
  • Mission NL6
  • Mission NL7
  • Mission NL8
  • Mission NL9
  • Mission NL10
  • Mission NL11
  • Mission NL12
  • Mission MC1
  • Mission MC10
  • Mission MC2
  • Mission MC3
  • Mission MC4
  • Mission MC5
  • Mission MC6
  • Mission MC7
  • Mission MC8
  • Mission MC9
  • Mission RM7
  • Mission RM9
  • Mission RL7
  • Mission RL9
  • Mission SM7
  • Mission SE1
  • Mission SE10
  • Mission SE11
  • Mission SE12
  • Mission SE2
  • Mission SE3
  • Mission SE4
  • Mission SE5
  • Mission SE6
  • Mission SE7
  • Mission SE8
  • Mission SE9
  • Mission VP1
  • Mission VP10
  • Mission VP2
  • Mission VP3
  • Mission VP4
  • Mission VP5
  • Mission VP6
  • Mission VP7
  • Mission VP8
  • Mission VP9
  • Mission WM1
  • Mission WM2
  • Mission WM3
  • Mission WM4
  • Mission WM5
  • Mission WM6
  • Mission WM7
  • Mission WM8
  • Mission WE1
  • Mission WE10
  • Mission WE2
  • Mission WE3
  • Mission WE4
  • Mission WE5
  • Mission WE6
  • Mission WE7
  • Mission WE8
  • Mission WE9
  • Vector Walk Interactive
  • Name That Motion Interactive
  • Kinematic Graphing 1 Concept Checker
  • Kinematic Graphing 2 Concept Checker
  • Graph That Motion Interactive
  • Rocket Sled Concept Checker
  • Force Concept Checker
  • Free-Body Diagrams Concept Checker
  • Free-Body Diagrams The Sequel Concept Checker
  • Skydiving Concept Checker
  • Elevator Ride Concept Checker
  • Vector Addition Concept Checker
  • Vector Walk in Two Dimensions Interactive
  • Name That Vector Interactive
  • River Boat Simulator Concept Checker
  • Projectile Simulator 2 Concept Checker
  • Projectile Simulator 3 Concept Checker
  • Turd the Target 1 Interactive
  • Turd the Target 2 Interactive
  • Balance It Interactive
  • Go For The Gold Interactive
  • Egg Drop Concept Checker
  • Fish Catch Concept Checker
  • Exploding Carts Concept Checker
  • Collision Carts - Inelastic Collisions Concept Checker
  • Its All Uphill Concept Checker
  • Stopping Distance Concept Checker
  • Chart That Motion Interactive
  • Roller Coaster Model Concept Checker
  • Uniform Circular Motion Concept Checker
  • Horizontal Circle Simulation Concept Checker
  • Vertical Circle Simulation Concept Checker
  • Race Track Concept Checker
  • Gravitational Fields Concept Checker
  • Orbital Motion Concept Checker
  • Balance Beam Concept Checker
  • Torque Balancer Concept Checker
  • Aluminum Can Polarization Concept Checker
  • Charging Concept Checker
  • Name That Charge Simulation
  • Coulomb's Law Concept Checker
  • Electric Field Lines Concept Checker
  • Put the Charge in the Goal Concept Checker
  • Circuit Builder Concept Checker (Series Circuits)
  • Circuit Builder Concept Checker (Parallel Circuits)
  • Circuit Builder Concept Checker (∆V-I-R)
  • Circuit Builder Concept Checker (Voltage Drop)
  • Equivalent Resistance Interactive
  • Pendulum Motion Simulation Concept Checker
  • Mass on a Spring Simulation Concept Checker
  • Particle Wave Simulation Concept Checker
  • Boundary Behavior Simulation Concept Checker
  • Slinky Wave Simulator Concept Checker
  • Simple Wave Simulator Concept Checker
  • Wave Addition Simulation Concept Checker
  • Standing Wave Maker Simulation Concept Checker
  • Color Addition Concept Checker
  • Painting With CMY Concept Checker
  • Stage Lighting Concept Checker
  • Filtering Away Concept Checker
  • InterferencePatterns Concept Checker
  • Young's Experiment Interactive
  • Plane Mirror Images Interactive
  • Who Can See Who Concept Checker
  • Optics Bench (Mirrors) Concept Checker
  • Name That Image (Mirrors) Interactive
  • Refraction Concept Checker
  • Total Internal Reflection Concept Checker
  • Optics Bench (Lenses) Concept Checker
  • Kinematics Preview
  • Velocity Time Graphs Preview
  • Moving Cart on an Inclined Plane Preview
  • Stopping Distance Preview
  • Cart, Bricks, and Bands Preview
  • Fan Cart Study Preview
  • Friction Preview
  • Coffee Filter Lab Preview
  • Friction, Speed, and Stopping Distance Preview
  • Up and Down Preview
  • Projectile Range Preview
  • Ballistics Preview
  • Juggling Preview
  • Marshmallow Launcher Preview
  • Air Bag Safety Preview
  • Colliding Carts Preview
  • Collisions Preview
  • Engineering Safer Helmets Preview
  • Push the Plow Preview
  • Its All Uphill Preview
  • Energy on an Incline Preview
  • Modeling Roller Coasters Preview
  • Hot Wheels Stopping Distance Preview
  • Ball Bat Collision Preview
  • Energy in Fields Preview
  • Weightlessness Training Preview
  • Roller Coaster Loops Preview
  • Universal Gravitation Preview
  • Keplers Laws Preview
  • Kepler's Third Law Preview
  • Charge Interactions Preview
  • Sticky Tape Experiments Preview
  • Wire Gauge Preview
  • Voltage, Current, and Resistance Preview
  • Light Bulb Resistance Preview
  • Series and Parallel Circuits Preview
  • Thermal Equilibrium Preview
  • Linear Expansion Preview
  • Heating Curves Preview
  • Electricity and Magnetism - Part 1 Preview
  • Electricity and Magnetism - Part 2 Preview
  • Vibrating Mass on a Spring Preview
  • Period of a Pendulum Preview
  • Wave Speed Preview
  • Slinky-Experiments Preview
  • Standing Waves in a Rope Preview
  • Sound as a Pressure Wave Preview
  • DeciBel Scale Preview
  • DeciBels, Phons, and Sones Preview
  • Sound of Music Preview
  • Shedding Light on Light Bulbs Preview
  • Models of Light Preview
  • Electromagnetic Radiation Preview
  • Electromagnetic Spectrum Preview
  • EM Wave Communication Preview
  • Digitized Data Preview
  • Light Intensity Preview
  • Concave Mirrors Preview
  • Object Image Relations Preview
  • Snells Law Preview
  • Reflection vs. Transmission Preview
  • Magnification Lab Preview
  • Reactivity Preview
  • Ions and the Periodic Table Preview
  • Periodic Trends Preview
  • Gaining Teacher Access
  • Tasks and Classes
  • Tasks - Classic
  • Subscription
  • Subscription Locator
  • 1-D Kinematics
  • Newton's Laws
  • Vectors - Motion and Forces in Two Dimensions
  • Momentum and Its Conservation
  • Work and Energy
  • Circular Motion and Satellite Motion
  • Thermal Physics
  • Static Electricity
  • Electric Circuits
  • Vibrations and Waves
  • Sound Waves and Music
  • Light and Color
  • Reflection and Mirrors
  • About the Physics Interactives
  • Task Tracker
  • Usage Policy
  • Newtons Laws
  • Vectors and Projectiles
  • Forces in 2D
  • Momentum and Collisions
  • Circular and Satellite Motion
  • Balance and Rotation
  • Electromagnetism
  • Waves and Sound
  • Forces in Two Dimensions
  • Work, Energy, and Power
  • Circular Motion and Gravitation
  • Sound Waves
  • 1-Dimensional Kinematics
  • Circular, Satellite, and Rotational Motion
  • Einstein's Theory of Special Relativity
  • Waves, Sound and Light
  • QuickTime Movies
  • About the Concept Builders
  • Pricing For Schools
  • Directions for Version 2
  • Measurement and Units
  • Relationships and Graphs
  • Rotation and Balance
  • Vibrational Motion
  • Reflection and Refraction
  • Teacher Accounts
  • Task Tracker Directions
  • Kinematic Concepts
  • Kinematic Graphing
  • Wave Motion
  • Sound and Music
  • About CalcPad
  • 1D Kinematics
  • Vectors and Forces in 2D
  • Simple Harmonic Motion
  • Rotational Kinematics
  • Rotation and Torque
  • Rotational Dynamics
  • Electric Fields, Potential, and Capacitance
  • Transient RC Circuits
  • Light Waves
  • Units and Measurement
  • Stoichiometry
  • Molarity and Solutions
  • Thermal Chemistry
  • Acids and Bases
  • Kinetics and Equilibrium
  • Solution Equilibria
  • Oxidation-Reduction
  • Nuclear Chemistry
  • NGSS Alignments
  • 1D-Kinematics
  • Projectiles
  • Circular Motion
  • Magnetism and Electromagnetism
  • Graphing Practice
  • About the ACT
  • ACT Preparation
  • For Teachers
  • Other Resources
  • Newton's Laws of Motion
  • Work and Energy Packet
  • Static Electricity Review
  • Solutions Guide
  • Solutions Guide Digital Download
  • Motion in One Dimension
  • Work, Energy and Power
  • Frequently Asked Questions
  • Purchasing the Download
  • Purchasing the CD
  • Purchasing the Digital Download
  • About the NGSS Corner
  • NGSS Search
  • Force and Motion DCIs - High School
  • Energy DCIs - High School
  • Wave Applications DCIs - High School
  • Force and Motion PEs - High School
  • Energy PEs - High School
  • Wave Applications PEs - High School
  • Crosscutting Concepts
  • The Practices
  • Physics Topics
  • NGSS Corner: Activity List
  • NGSS Corner: Infographics
  • About the Toolkits
  • Position-Velocity-Acceleration
  • Position-Time Graphs
  • Velocity-Time Graphs
  • Newton's First Law
  • Newton's Second Law
  • Newton's Third Law
  • Terminal Velocity
  • Projectile Motion
  • Forces in 2 Dimensions
  • Impulse and Momentum Change
  • Momentum Conservation
  • Work-Energy Fundamentals
  • Work-Energy Relationship
  • Roller Coaster Physics
  • Satellite Motion
  • Electric Fields
  • Circuit Concepts
  • Series Circuits
  • Parallel Circuits
  • Describing-Waves
  • Wave Behavior Toolkit
  • Standing Wave Patterns
  • Resonating Air Columns
  • Wave Model of Light
  • Plane Mirrors
  • Curved Mirrors
  • Teacher Guide
  • Using Lab Notebooks
  • Current Electricity
  • Light Waves and Color
  • Reflection and Ray Model of Light
  • Refraction and Ray Model of Light
  • Classes (Legacy Version)
  • Teacher Resources
  • Subscriptions

solving simple circuit problems

  • Newton's Laws
  • Einstein's Theory of Special Relativity
  • About Concept Checkers
  • School Pricing
  • Newton's Laws of Motion
  • Newton's First Law
  • Newton's Third Law
  • Combination Circuits
  • Circuit Symbols and Circuit Diagrams
  • Two Types of Connections

VIDThumbNail.png

When analyzing combination circuits, it is critically important to have a solid understanding of the concepts that pertain to both series circuits and parallel circuits . Since both types of connections are used in combination circuits, the concepts associated with both types of circuits apply to the respective parts of the circuit. The main concepts associated with series and parallel circuits are organized in the table below.

Each of the above concepts has a mathematical expression. Combining the mathematical expressions of the above concepts with the Ohm's law equation (ΔV = I • R) allows one to conduct a complete analysis of a combination circuit.

Analysis of Combination Circuits

The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit. Once transformed into a series circuit, the analysis can be conducted in the usual manner. Previously in Lesson 4 , the method for determining the equivalent resistance of parallel are equal, then the total or equivalent resistance of those branches is equal to the resistance of one branch divided by the number of branches.

This method is consistent with the formula

where R 1 , R 2 , and R 3 are the resistance values of the individual resistors that are connected in parallel. If the two or more resistors found in the parallel branches do not have equal resistance, then the above formula must be used. An example of this method was presented in a previous section of Lesson 4 .

By applying one's understanding of the equivalent resistance of parallel branches to a combination circuit, the combination circuit can be transformed into a series circuit. Then an understanding of the equivalent resistance of a series circuit can be used to determine the total resistance of the circuit. Consider the following diagrams below. Diagram A represents a combination circuit with resistors R 2 and R 3 placed in parallel branches. Two 4-Ω resistors in parallel is equivalent to a resistance of 2 Ω. Thus, the two branches can be replaced by a single resistor with a resistance of 2 Ω. This is shown in Diagram B. Now that all resistors are in series, the formula for the total resistance of series resistors can be used to determine the total resistance of this circuit: The formula for series resistance is

So in Diagram B, the total resistance of the circuit is 10  Ω .

Once the total resistance of the circuit is determined, the analysis continues using Ohm's law and voltage and resistance values to determine current values at various locations. The entire method is illustrated below with two examples.

The first example is the easiest case - the resistors placed in parallel have the same resistance. The goal of the analysis is to determine the current in and the voltage drop across each resistor.

As discussed above, the first step is to simplify the circuit by replacing the two parallel resistors with a single resistor that has an equivalent resistance. Two 8 Ω resistors in series is equivalent to a single 4 Ω resistor. Thus, the two branch resistors (R 2 and R 3 ) can be replaced by a single resistor with a resistance of 4 Ω. This 4 Ω resistor is in series with R 1 and R 4 . Thus, the total resistance is

R tot = 15  Ω

Now the Ohm's law equation (ΔV = I • R) can be used to determine the total current in the circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be used.

I tot = 4 Amp

The 4 Amp current calculation represents the current at the battery location. Yet, resistors R 1 and R 4 are in series and the current in series-connected resistors is everywhere the same. Thus,

For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I 2 + I 3 must equal 4 Amp. There are an infinite number of possible values of I 2 and I 3 that satisfy this equation. Since the resistance values are equal, the current values in these two resistors are also equal. Therefore, the current in resistors 2 and 3 are both equal to 2 Amp.

Now that the current at each individual resistor location is known, the Ohm's law equation (ΔV = I • R) can be used to determine the voltage drop across each resistor. These calculations are shown below.

ΔV 1 = I 1 • R 1 = (4 Amp) • (5 Ω) ΔV 1 = 20 V

ΔV 2 = I 2 • R 2 = (2 Amp) • (8 Ω)

ΔV 2 = 16 V

ΔV 3 = I 3 • R 3 = (2 Amp) • (8 Ω)

ΔV 3 = 16 V

ΔV 4 = I 4 • R 4 = (4 Amp) • (6 Ω)

ΔV 4 = 24 V  

The analysis is now complete and the results are summarized in the diagram below.

The second example is the more difficult case - the resistors placed in parallel have a different resistance value. The goal of the analysis is the same - to determine the current in and the voltage drop across each resistor.

As discussed above, the first step is to simplify the circuit by replacing the two parallel resistors with a single resistor with an equivalent resistance. The equivalent resistance of a 4-Ω and 12-Ω resistor placed in parallel can be determined using the usual formula for equivalent resistance of parallel branches:

1 / R eq = 1 / (4 Ω) + 1 / (12 Ω)

1 / R eq = 0.333 Ω -1

R eq = 1 / (0.333 Ω -1 )

R eq = 3.00 Ω

Based on this calculation, it can be said that the two branch resistors (R 2 and R 3 ) can be replaced by a single resistor with a resistance of 3 Ω. This 3 Ω resistor is in series with R 1 and R 4 . Thus, the total resistance is

R tot = 16  Ω

I tot = 1.5 Amp

The 1.5 Amp current calculation represents the current at the battery location. Yet, resistors R 1 and R 4 are in series and the current in series-connected resistors is everywhere the same. Thus,

For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I 2 + I 3 must equal 1.5 Amp. There are an infinite possibilities of I 2 and I 3 values that satisfy this equation. In the previous example, the two resistors in parallel had the identical resistance; thus the current was distributed equally among the two branches. In this example, the unequal current in the two resistors complicates the analysis. The branch with the least resistance will have the greatest current. Determining the amount of current will demand that we use the Ohm's law equation. But to use it, the voltage drop across the branches must first be known. So the direction that the solution takes in this example will be slightly different than that of the simpler case illustrated in the previous example.

To determine the voltage drop across the parallel branches, the voltage drop across the two series-connected resistors (R 1 and R 4 ) must first be determined. The Ohm's law equation (ΔV = I • R) can be used to determine the voltage drop across each resistor. These calculations are shown below.

ΔV 1 = I 1 • R 1 = (1.5 Amp) • (5 Ω) ΔV 1 = 7.5 V

ΔV 4 = I 4 • R 4 = (1.5 Amp) • (8 Ω)

ΔV 4 = 12 V

This circuit is powered by a 24-volt source. Thus, the cumulative voltage drop of a charge traversing a loop about the circuit is 24 volts. There will be a 19.5 V drop (7.5 V + 12 V) resulting from passage through the two series-connected resistors (R 1 and R 4 ). The voltage drop across the branches must be 4.5 volts to make up the difference between the 24 volt total and the 19.5-volt drop across R 1 and R 4 . Thus,

Knowing the voltage drop across the parallel-connected resistors (R 1 and R 4 ) allows one to use the Ohm's law equation (ΔV = I • R) to determine the current in the two branches.

I 2 = ΔV 2 / R 2 = (4.5 V) / (4 Ω) I 2 = 1.125 A

I 3 = ΔV 3 / R 3 = (4.5 V) / (12 Ω)

I 3 = 0.375 A

Developing a Strategy

The two examples above illustrate an effective concept-centered strategy for analyzing combination circuits. The approach demanded a firm grasp of the series and parallel concepts discussed earlier . Such analyses are often conducted in order to solve a physics problem for a specified unknown. In such situations, the unknown typically varies from problem to problem. In one problem, the resistor values may be given and the current in all the branches are the unknown. In another problem, the current in the battery and a few resistor values may be stated and the unknown quantity becomes the resistance of one of the resistors. Different problem situations will obviously require slight alterations in the approaches. Nonetheless, every problem-solving approach will utilize the same principles utilized in approaching the two example problems above.

The following suggestions for approaching combination circuit problems are offered to the beginning student:

  • If a schematic diagram is not provided, take the time to construct one. Use schematic symbols such as those shown in the example above.
  • When approaching a problem involving a combination circuit, take the time to organize yourself, writing down known values and equating them with a symbol such as I tot , I 1 , R 3 , ΔV 2 , etc. The organization scheme used in the two examples above is an effective starting point.
  • Know and use the appropriate formulae for the equivalent resistance of series-connected and parallel-connected resistors. Use of the wrong formulae will guarantee failure.
  • Transform a combination circuit into a strictly series circuit by replacing (in your mind) the parallel section with a single resistor having a resistance value equal to the equivalent resistance of the parallel section.
  • Use the Ohm's law equation (ΔV = I • R) often and appropriately. Most answers will be determined using this equation. When using it, it is important to substitute the appropriate values into the equation. For instance, if calculating I 2 , it is important to substitute the ΔV 2 and the R 2 values into the equation.

For further practice analyzing combination circuits, consider analyzing the problems in the Check Your Understanding section below.

We Would Like to Suggest ...

solving simple circuit problems

Check Your Understanding

a. The current at location A is _____ (greater than, equal to, less than) the current at location B.

b. The current at location B is _____ (greater than, equal to, less than) the current at location E.

c. The current at location G is _____ (greater than, equal to, less than) the current at location F.

d. The current at location E is _____ (greater than, equal to, less than) the current at location G.

e. The current at location B is _____ (greater than, equal to, less than) the current at location F.

f. The current at location A is _____ (greater than, equal to, less than) the current at location L.

g. The current at location H is _____ (greater than, equal to, less than) the current at location I.

The current outside the branches of a combination circuit is everywhere the same. The current inside of the branches is always less than that outside of the branches. When comparing the current of two parallel-connected resistors, the resistor with the least resistance will have the greatest current. The current within a single branch will be the same above and below the resistor.

b. The current at location B is greater than the current at location E.

c. The current at location G is less than the current at location F.

d. The current at location E is greater than the current at location G.

e. The current at location B is greater than the current at location F.

f. The current at location A is equal to the current at location L.

g. The current at location H is less than the current at location I.

a. The electric potential difference (voltage drop) between points B and C is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points J and K.

b. The electric potential difference (voltage drop) between points B and K is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points D and I.

c. The electric potential difference (voltage drop) between points E and F is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points G and H.

d. The electric potential difference (voltage drop) between points E and F is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points D and I.

e. The electric potential difference (voltage drop) between points J and K is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points D and I.

f. The electric potential difference between points L and A is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points B and K.

The voltage drop across a resistor is dependent upon the current in the resistor and the resistance of the resistor. In situations in which the current is the same for both resistors (such as for series-connected resistors), the resistor with the greatest resistance will have the greatest voltage drop.

b. The electric potential difference (voltage drop) between points B and K is greater than the electric potential difference (voltage drop) between points D and I.

c. The electric potential difference (voltage drop) between points E and F is equal to the electric potential difference (voltage drop) between points G and H.

d. The electric potential difference (voltage drop) between points E and F is equal to the electric potential difference (voltage drop) between points D and I.

e. The electric potential difference (voltage drop) between points J and K is greater than the electric potential difference (voltage drop) between points D and I.

f. The electric potential difference between points L and A is equal to the electric potential difference (voltage drop) between points B and K.

3. Use the concept of equivalent resistance to determine the unknown resistance of the identified resistor that would make the circuits equivalent.

For parallel-connected resistors:

1/R eq = 1/R 1 + 1/R 2 = 1 / (6 ) + 1 / (6 ) = 2 / (6 ) R eq = 3

For series-connected resistors:

R eq = R 1 + R 2 + R 3 = 3 + 3 + 5 R eq = 11
1/R eq = 1/R 1 + 1/R 2 = 1 / (12 ) + 1 / (6 ) = 3 / (12 ) R eq = 4
R eq = R 1 + R 2 + R 3 = 9 + 4 + 5 R eq = 18
1/R eq = 1/R 1 + 1/R 2 = 1 / (12 Ω) + 1 / (6 Ω) = 3 / (12 Ω) R eq = 4 Ω
R eq = R 1 + R 2 + R 3 + R 4 = 3 Ω + 6 Ω + 4 Ω + 5 Ω R eq = 18 Ω

4. Analyze the following circuit and determine the values of the total resistance, total current, and the current at and voltage drops across each individual resistor.

The first step is to simplify the circuit by replacing the two parallel resistors with a single resistor with an equivalent resistance. The equivalent resistance of a 4  Ω  and 6  Ω  resistor placed in parallel can be determined using the usual formula for equivalent resistance of parallel branches:

1 / R eq = 1 / (3 Ω) + 1 / (6 Ω)

1 / R eq = 0.500 Ω -1

R eq = 1 / (0.500 Ω -1 )

R eq = 2.00 Ω

Based on this calculation, it can be said that the two branch resistors (R 2 and R 3 ) can be replaced by a single resistor with a resistance of 2  Ω . This 2  Ω  resistor is in series with R 1 and R 4 . Thus, the total resistance is

R tot = 8 Ω

I tot = 3.0 Amp

The 3.0 Amp current calculation represents the current at the battery location. Yet, resistors R 1 and R 4 are in series and the current in series-connected resistors is everywhere the same. Thus,

For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I 2 + I 3 must equal 3.0 Amp. There are an infinite possibilities of I 2 and I 3 values which satisfy this equation. Determining the amount of current in either branch will demand that we use the Ohm's law equation. But to use it, the voltage drop across the branches must first be known. To determine the voltage drop across the parallel branches, the voltage drop across the two series-connected resistors (R 1 and R 4 ) must first be determined. The Ohm's law equation (ΔV = I • R) can be used to determine the voltage drop across each resistor. These calculations are shown below.

ΔV 1 = I 1 • R 1 = (3.0 Amp) • (2 ) = 6.0 V Δ V 4 = I 4 • R 4 = (3.0 Amp) • (4 ) = 12 V

This circuit is powered by a 24-volt source. Thus, the cumulative voltage drop of a charge traversing a loop about the circuit is 24 volts. There will be a 18.0 V drop (6.0 V + 12.0 V) resulting from passage through the two series-connected resistors (R 1 and R 4 ). The voltage drop across the branches must be 6.0 volts to make up the difference between the 24 volt total and the 18.0 volt drop across R 1 and R 4 . Knowing the voltage drop across the parallel-connected resistors (R 1 and R 4 ) allows one to use the Ohm's law equation (ΔV = I • R) to determine the current in the two branches.

I 2 = ΔV 2 / R 2 = (6.0 V) / (3 ) = 2.0 A I 3 = ΔV 3 / R 3 = (6.0 V) / (6 ) = 1.0 A

5. Referring to the diagram in question #4, determine the ...

a. ... power rating of resistor 4. b. ... rate at which energy is consumed by resistor 3.

The power rating of a resistor can be determined from any one of the following formulae:

Since all three variables (I,  Δ V and R) are known, either one of the equations can be used to calculate power.

a. P 4 = I 4 •  Δ V 4 = (3 A) • (12 V) = 36.0 Watt b. P 3 = I 3 •  Δ V 3 = (1 A) • (6 V) = 6.0 Watt

logo-type-white

Ohm’s Law Practice Problems | Review and Examples

  • The Albert Team
  • Last Updated On: December 5, 2023

solving simple circuit problems

Welcome to the fascinating world of electrical principles, where understanding Ohm’s Law is not just a skill but a necessity. Often considered the backbone of electrical engineering and physics, Ohm’s Law is a fundamental concept that illuminates the relationship between voltage, current, and resistance in an electrical circuit. In this comprehensive guide, we will cover the core of Ohm’s Law, explore its fundamental formula, and work through a series of Ohm’s Law practice problems. Mastering Ohm’s Law is a step towards unraveling the complexities of electronics.

What Does Ohm’s Law State?

Ohm’s Law is a fundamental principle in electronics and physics, providing a simple yet powerful way to understand the relationship between voltage, current, and resistance in electrical circuits. At its core, Ohm’s Law states that the current through a conductor between two points is directly proportional to the voltage across these points and inversely proportional to the resistance between them. This relationship is elegantly captured in the formula V = IR , where V stands for voltage, I for current, and R for resistance.

Understanding Voltage, Current, and Resistance

In order to fully appreciate Ohm’s law, let’s review the three components.

  • Voltage (V) : Often described as the electrical force or pressure that drives the flow of electrons through a conductor. It’s the potential difference between two points in a circuit.
  • Current (I) : This is the flow of electrical charge, measured in amperes (A). It represents how many electrons are flowing through the circuit.
  • Resistance (R) : Resistance is the opposition to the current flow in a circuit. It’s measured in ohms (Ω) and depends on the material, size, and temperature of the conductor.

By manipulating the Ohm’s Law formula, you can solve for any one of these three variables if the other two are known. This makes it an invaluable tool for understanding and designing electrical circuits.

Practical Applications

Ohm’s Law isn’t just a theoretical concept; it has numerous practical applications in everyday life and various industries. Here are a few examples:

  • Electronics Design: Engineers use Ohm’s Law to design circuits, select appropriate components, and ensure electrical devices function safely and efficiently.
  • Troubleshooting Electrical Problems: Technicians often use Ohm’s Law to diagnose issues in electrical systems, such as finding short circuits or identifying components that are not functioning correctly.
  • Educational Purposes: Ohm’s Law is a fundamental concept taught in physics and electronics courses, helping students understand the basics of electrical circuits.
  • Power Management: In larger-scale applications like power distribution, Ohm’s Law helps calculate the load that can be safely put on electrical systems without causing damage or inefficiency.

solving simple circuit problems

Understanding Ohm’s Law opens up possibilities for creating, managing, and troubleshooting electrical systems, from the smallest electronic devices to large-scale power grids.

What is the Formula for Ohm’s Law?

Ohm’s Law is elegant in its formulation, providing a precise mathematical relationship between voltage, current, and resistance in an electrical circuit.

In the formula for Ohm’s Law, V represents voltage measured in volts (V), I is the current measured in amperes (A), and R is the resistance measured in ohms (Ω). This formula is the cornerstone for analyzing and understanding electrical circuits, requiring two variables to solve.

Ohm’s Law Triangle

The Ohm’s Law triangle is a helpful tool for remembering how to calculate voltage, current, and resistance. It visually represents the formula V=IR in a graphic format, with V at the top, I on the left, and R on the right. By covering the variable you want to calculate, the other two variables show how they relate. For example, covering V shows I\times R , covering I shows /frac{V}{R} , and covering R shows \frac{V}{I} . This tool is handy for beginners and a reference for quick calculations.

The Ohm's Law Triangle is a useful tool for solving Ohm's Law practice problems.

Strategies for Solving Ohm’s Law Practice Problems

When solving problems using Ohm’s Law, it’s important to follow a systematic approach:

  • Identify Known Quantities: Start by determining which of the three variables (voltage, current, resistance) are known.
  • Determine the Unknown: Figure out which variable you need to calculate.
  • Use the Ohm’s Law Circle: Utilize the Ohm’s Law circle to understand the relationship between the variables and to choose the correct formula.
  • Solve Step-by-Step: Apply the formula and solve for the unknown variable step-by-step, ensuring accuracy in your calculations.
  • Check Units: Always check that your units are consistent (volts for voltage, amperes for current, ohms for resistance) and convert if necessary.

By applying these strategies, you can effectively use Ohm’s Law to solve a wide range of electrical problems, enhancing your understanding and skills in electrical theory and practice.

Examples of Ohm’s Law

Calculating Current: If a light bulb has a resistance of 240\text{ ohms} and is connected to a 120\text{-volt} power source, the current flowing through it can be calculated as:

Determining Voltage: For a toaster that draws a current of 5\text{ amperes} and has a resistance of 10\text{ ohms} , the voltage across it is:

Finding Resistance: If a hairdryer operates at 220\text{ volts} and draws a current of 11\text{ amperes} , its resistance is:

These examples demonstrate how Ohm’s Law is applied in practical situations, providing a clear understanding of how electrical components function in various devices.

Ohm’s Law Practice Problems

Here are eight practice problems involving Ohm’s Law, arranged in order of increasing complexity. These problems will help you apply the concepts of voltage, current, and resistance in various scenarios. Work through these on your own, then scroll down for solutions.

1. Basic Current Calculation

A circuit with a 9\text{-volt} battery and a resistor of 3\text{ ohms} . What is the current flowing through the circuit?

2. Resistance Determination

Find the resistance of a bulb that draws 0.5\text{ amperes} from a 120\text{-volt} supply.

3. Voltage Calculation

What is the voltage across a resistor of 15\text{ ohms} through which a current of 2\text{ amperes} is flowing?

4. Multiple Resistors (Series) 

In a series circuit with a 12\text{-volt} battery, if there are two resistors of 4\text{ ohms} and 6\text{ ohms} , what is the current flowing through the circuit?

5. Multiple Resistors (Parallel)

Calculate the total resistance in a parallel circuit with two resistors of 5\text{ ohms} and 10\text{ ohms} . If a voltage of 12\text{-volts} is applied across the circuit, what is the total current flowing through the circuit?

6. Combined Ohm’s Law and Power

A device using 18\text{ watts} of power is connected to a 9\text{-volt} battery. Calculate the current drawn by the device and determine the resistance of the device.

7. Variable Resistance

If the current in a circuit is 0.25\text{ amperes} and the voltage is 10\text{ volts} , what must be the resistance?

8. Complex Circuit Analysis

In a circuit, a 6\text{-ohm} resistor and a 12\text{-ohm} resistor are connected in series to a 9\text{-volt} battery. Calculate the current through each resistor.

solving simple circuit problems

Solutions to Ohm’s Law Practice Problems

Are you ready to see how you did? Review below to see the solutions for the Ohm’s Law practice problems.

We have a simple circuit with a 9\text{ V} battery and a 3\ \Omega resistor. In order to solve this, use Ohm’s Law, V=IR to find the current:

Therefore, the current flowing through this circuit is 3\text{ amperes} , typical for small electronic devices.

A bulb is connected to a 120\text{ V} supply and draws 0.5\text{ A} .To find the resistance, rearrange Ohm’s Law to R = V/I :

The bulb has a resistance of 240\ \Omega , indicating it’s suitable for moderate power applications.

A resistor of 15\ \Omega carries a current of 2\text{ A} . Apply V=IR to find the voltage across the resistor:

The voltage across this resistor is 30\text{ V} , typical for small household circuits.

We have a series circuit with a 12\text{ V} battery and two resistors ( 4\ \Omega and 6\ \Omega ). First, sum the resistances in series. Then, apply Ohm’s Law.

Summing the resistance:

Now, apply Ohm’s Law with the total resistance, rearranged for the current:

The current of 1.2\text{ A} flows uniformly through each component in this series circuit.

In this scenario, there is a parallel circuit with two resistors of 5\ \Omega and 10\ \Omega . First, calculate the total resistance in parallel using the reciprocal formula:

Then, apply Ohm’s Law with the total resistance, rearranged for the current:

The total current flowing through the circuit is approximately 3.6\text{ A} .

An 18\text{ W} device is connected to a 9\text{ V} battery. First, find the current using by rearranging the power formula P=VI :

The device draws a current of 2\text{ A} . Next, use Ohm’s Law rearranged for resistance:

The device’s resistance is 4.5\ \Omega .

For a circuit with a current of 0.25\text{ A} and a voltage of 10\text{ V} , apply Ohm’s Law to find the resistance:

The circuit has a resistance of 40\ \Omega , indicating a relatively high resistance for the given current and voltage.

First, calculate the total resistance:

For a series circuit, all elements receive the same current. Each resistor in this series circuit experiences a current of 0.5\text{ A} .

As we reach the end of our exploration into Ohm’s Law, it’s clear that this fundamental principle is more than just a formula; it’s a key to unlocking the mysteries of electrical circuits. Through this guide, we’ve journeyed from the basic understanding of voltage, current, and resistance to applying these concepts in various practical scenarios. The practice problems provided various challenges, from straightforward calculations to more complex circuit analyses, each designed to strengthen your grasp of Ohm’s Law.

Remember, the journey of mastering Ohm’s Law is as much about practice as it is about understanding the theory. Each problem you solve, and each circuit you analyze adds to your skill set, making you more adept at navigating the world of electronics.

Interested in a school license?​

Popular posts.

AP® Physics I score calculator

AP® Score Calculators

Simulate how different MCQ and FRQ scores translate into AP® scores

solving simple circuit problems

AP® Review Guides

The ultimate review guides for AP® subjects to help you plan and structure your prep.

solving simple circuit problems

Core Subject Review Guides

Review the most important topics in Physics and Algebra 1 .

solving simple circuit problems

SAT® Score Calculator

See how scores on each section impacts your overall SAT® score

solving simple circuit problems

ACT® Score Calculator

See how scores on each section impacts your overall ACT® score

solving simple circuit problems

Grammar Review Hub

Comprehensive review of grammar skills

solving simple circuit problems

AP® Posters

Download updated posters summarizing the main topics and structure for each AP® exam.

Interested in a school license?

solving simple circuit problems

Bring Albert to your school and empower all teachers with the world's best question bank for: ➜ SAT® & ACT® ➜ AP® ➜ ELA, Math, Science, & Social Studies aligned to state standards ➜ State assessments Options for teachers, schools, and districts.

Gurumuda Networks

Electric circuits – problems and solutions

1. R 1 , = 6 Ω, R 2 = R 3 = 2 Ω, and voltage = 14 volt, determine the electric current in circuit as shown in figure below.

Electric circuits – problems and solutions 1

Resistor 1 (R 1 ) = 6 Ω

Resistor 2 (R 2 ) = 2 Ω

Resistor 3 (R 3 ) = 2 Ω

Voltage (V) = 14 Volt

Wanted : Electric current (I)

Equivalent resistor (R) :

R 2 and R 3 are connected in parallel. The equivalent resistor :

1/R 23 = 1/R 2 + 1/R 3 = 1/2 + 1/2 = 2/2

R 23 = 2/2 = 1 Ω

R 1 and R 23 are connected in series. The equivalent resistor :

R = R 1 + R 23 = 6 Ω + 1 Ω

Electric current (I) :

I = V / R = 14 / 7 = 2 Ampere

2. Which one of the electric circuits as shown below has the bigger current.

Electric circuits – problems and solutions 2

The resistance of the resistor is R and the electric voltage is V.

R 1 , R 2 and R 3 are connected in series. The equivalent resistor :

R A = R 1 + R 2 + R 3 = R + R + R = 3R

Electric circuits – problems and solutions 3

R 1 , R 2 and R 3 are connected in parallel. The equivalent resistor :

1/R = 1/R 1 + 1/R 2 + 1/R 3 = 1/R + 1/R + 1/R = 3/R

Electric circuits – problems and solutions 4

1/R 23 = 1/R 2 + 1/R 3 = 1/R + 1/R = 2/R

R C = R 1 + R 23 = R + R/2 = 2R/2 + R/2 = 3R/2

Electric circuits – problems and solutions 5

R 1 and R 2 are connected in parallel. The equivalent resistor :

1/R 12 = 1/R 1 + 1/R 2 = 1/R + 1/R = 2/R

R 12 and R 3 are connected in series. The equivalent resistor :

R D = R 12 + R 3 = R/2 + R = R/2 + 2R/2 = 3R/2

Electric circuits – problems and solutions 6

3. R 1 = 4 ohm, R 2 = 6 ohm, R 3 = 2 ohm, and V = 24 volt. What is the electric current in circuit as shown in figure below.

Electric circuits – problems and solutions 7

Resistor 1 (R 1 ) = 4 Ohm

Resistor 2 (R 2 ) = 6 Ohm

Resistor 3 (R 3 ) = 2 Ohm

Voltage (V) = 24 Volt

Wanted : Electric current in circuit

R = R 1 + R 2 + R 3 = 4 + 6 + 2

Electric current :

I = V / R = 24 / 12 = 2 Ampere

4. Which one of the electric circuits as shown below has the bigger current.

Electric circuits – problems and solutions 8

Electric current in circuit A.

The equivalent resistor :

R 1 = 3 Ω, R 2 = 4 Ω, R 3 = 4 Ω, V = 12 Volt

1/R 23 = 1/R 2 + 1/R 3 = 1/4 + 1/4 = 2/4 = 1/2

R 23 = 2/1 = 2 Ω

R = R 1 + R 23 = 3 Ω + 2 Ω = 5 Ω

I = V / R = 12 / 5 = 2.4 Ampere

Electric current in circuit B.

R 1 = 8 Ω, R 2 = 2 Ω, R 3 = 2 Ω, V = 36 Volt

R = R 1 + R 2 + R 3 = 8 + 2 + 2 = 12 Ω

I = V / R = 36 / 12 = 3 Ampere

Electric current in circuit C.

R 1 = 4 Ω, R 2 = 4 Ω, R 3 = 6 Ω, V = 12 Volt

1/R 23 = 1/R 2 + 1/R 3 = 1/4 + 1/4 + 1/6 = 3/12 + 3/12 + 2/12 = 8/12

R 23 = 12/8 = 1.5 Ω

I = V / R = 12 / 1.5 = 8 Ampere

Electric current in circuit D.

R 1 = 3 Ω, R 2 = 3 Ω, R 3 = 3 Ω, R 4 = 3 Ω, R 5 = 6 Ω, V = 24 Volt

R 2 , R 3 and R 4 are connected in parallel. The equivalent resistor :

1/R 234 = 1/R 2 + 1/R 3 + 1/R 4 = 1/3 + 1/3 + 1/3 = 3/3

R 234 = 3/3 = 1 Ω

R 1 , R 234 and R 5 are connected in series The equivalent resistor :

R = R 1 + R 234 + R 5 = 3 + 1 + 6 = 9 Ω

I = V / R = 24 / 9 = 2.6 Ampere

5. According to figure as shown below, determine :

Electric circuits – problems and solutions 9

B. Electric current in circuit

C. Current I 1

D. Current I 2

Resistor 1 (R 1 ) = 4 Ω

Resistor 2 (R 2 ) = 4 Ω

Resistor 4 (R 4 ) = 3 Ω

Electric voltage (V) = 12 Volt

A. Total resistance (R)

Resistor R 2 and resistor R 3 are connected in series. The equivalent resistor :

R 23 = R 2 + R 3 = 4 Ω + 2 Ω = 6 Ω

Resistor R 23 and resistor R 4 are connected in parallel. The equivalent resistor :

1/R 234 = 1/R 23 + 1/R 4 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6

R 234 = 6/3 = 2 Ω

Resistor R 1 and resistor R 234 are connected series. The equivalent resistor :

R = R 1 + R 234 = 4 Ω + 2 Ω = 6 Ω

The total resistance is 6 Ohm.

B. Electric current in circuit (I)

V = electric voltage, I = electric current, R = electric resistance

I = V / R = 12 Volt / 6 Ohm = 2 Ampere

C. Electric current I 1

Electric current in resistor R 1 = electric current in circuit = 2 Ampere.

Electric circuits – problems and solutions 10

Electric current in resistor R 234 = electric current in resistor R 1 = 2 Ampere.

Voltage in resistor R 234 is:

V = I R 234 = (2 A)(2 Ohm) = 4 Volt

Voltage in resistor R 234 = voltage in resistor R 4 = voltage in resistor R 23 = 4 Volt.

The equivalent resistor R 23 is 6 Ohm.

Electric current in resistor R 23 is :

I = V / R = 4 Volt / 6 Ohm = 2/3 Ampere

Electric current in resistor R 23 = Electric current in resistor R 2 = electric current in resistor R 3 = 2/3 Ampere.

6. R 1 = R 2 = 10 Ω and R 3 = R 4 = 8 Ω. What is the electric current in circuit as shown in figure below ?

Electric circuits – problems and solutions 11

Resistor R 1 = Resistor R 2 = 10 Ω

Resistor R 3 = Resistor R 4 = 8 Ω

Wanted : electric current (I)

The equivalent resistor

Resistor R 3 and resistor R 4 are connected in parallel, the equivalent resistor :

1/R 34 = 1/R 3 + 1/R 4 = 1/8 + 1/8 = 2/8

R 34 = 8/2 = 4 Ω

Resistor R 1 , R 2 and R 34 are connected in series, the equivalent resistor :

R = R 1 + R 2 + R 34 = 10 Ω + 10 Ω + 4 Ω = 24 Ω

I = V / R = 12 Volt / 24 Ohm = 0,5 Volt/Ohm = 0.5 Ampere

7. If the internal resistance of battery ignored, what is the electric current in the circuit shown in figure below.

Electric circuits – problems and solutions 12

Resistor R 1 = 3 Ohm

Resistor R 2 = 3 Ohm

Resistor R 3 = 6 Ohm

Electric voltage (V) = 6 Volt

Equivalent resistor

Resistor R 1 and R 2 are connected in series. The equivalent resistor :

R 12 = R 1 + R 2 = 3 Ohm + 3 Ohm = 6 Ohm

Resistor R 12 and resistor 3 are connected in parallel. The equivalent resistor :

1/R = 1/R 12 + 1/R 3 = 1/6 + 1/6 = 2/6

R = 6/2 = 3 Ohm

I = V / R = 6 / 3 = 2 Ampere

8. What is the total electric current in circuit as shown in figure below.

Electric circuits – problems and solutions 13

Resistor R 1 = 6 Ohm

Resistor R 2 = 4 Ohm

Electric current (V) = 6 Volt

Internal resistance (r) = 0.6 Ohm

Wanted : Electric current

Resistor R 1 and resistor R 2 are connected in parallel. The equivalent resistor :

1/R P = 1/R 1 + 1/R 2 = 1/6 + 1/4 = 4/24 + 6/24 = 10/24

R P = 24/10 = 2.4 Ohm

Resistor R P and internal resistance (r) are connected in series. The equivalent resistor :

R = R P + r = 2.4 Ohm + 0.6 Ohm = 3.0 Ohm

Electric current in circuit :

I = V / R = 6 Volt / 3 Ohm = 2 Ampere

  • Answer : An electric circuit is a closed path or loop in which electric current can flow continuously. It typically consists of sources of voltage (like batteries), loads (like resistors, LEDs, motors), and conductors to connect them.
  • Answer : In a series circuit, components are connected end-to-end, so there’s a single path for current. In a parallel circuit, components are connected across common points or junctions, providing multiple paths for current.
  • Answer : Ohm’s Law states that the current ( I ) flowing through a conductor between two points is directly proportional to the voltage ( V ) across the two points and inversely proportional to the resistance ( R ). It’s represented as I = V/R ​ .
  • Answer : A switch controls the flow of current in a circuit. When closed, it allows current to flow; when open, it interrupts or stops the current flow.
  • Answer : In a short circuit, the resistance is very low, causing a very high current to flow. This can lead to overheating, fires, or damage to components and should be protected against with fuses or circuit breakers.
  • Answer : Both fuses and circuit breakers are protective devices designed to interrupt a circuit if the current exceeds a predetermined safe level. While fuses “blow” or “melt”, breaking the circuit, circuit breakers “trip”, and can be reset after they interrupt the circuit.
  • Answer : Kirchhoff’s Current Law states that the sum of currents entering a junction is equal to the sum of currents leaving that junction. This is essentially a statement of the conservation of electric charge.
  • Answer : DC refers to the unidirectional flow of electric charge, typically from a battery or a DC power supply. AC, on the other hand, is an electric charge that changes direction periodically, like what’s supplied from the power grid in many countries.
  • Answer : “Ground” refers to a reference point in an electrical circuit from which other voltages are measured, or a common return path for electric current, or a direct physical connection to the Earth.
  • Why are capacitors used in electric circuits?
  • Answer : Capacitors store and release electrical energy. They’re used in circuits for various purposes, such as filtering, smoothing voltage fluctuations, coupling and decoupling AC signals, and timing elements in oscillators.

Print Friendly, PDF & Email

Share this:

Discover more from physics.

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

You must be logged in to post a comment.

Physexams Logo

  • Exam Center
  • Ticket Center
  • Flash Cards
  • AP Physics Problems

AP Physics 2: Circuits Practice Problems with Answers

Here, a few sample AP Physics C problems on circuits are gathered and solved. These questions involve the various configuration of capacitors, resistances, and electric power in a DC circuit. 

Circuit Practice Problems

Problem (1): In the circuit below, find the equivalent resistance between points $A$ and $B$.

ap-circuits-problem-1

Solution : One of the typical questions in all circuit practice problems is finding the equivalent resistance of a given circuit. 

Recall that the equivalent of two resistors in series is $R_{eq}=R_1+R_2$ and in parallel is $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$. 

The two resistors $3-{\rm \Omega}$ are in series, so their equivalent resistance is \[R_{3,3}=R_3+R_3=3+3=6\,{\rm \Omega}\] It's obvious that resistors of $12\,{\rm \Omega}$ and $6\,{\rm \Omega}$ are in parallel, so their effective or equivalent resistance is \begin{align*}\frac{1}{R_{12,6}}&=\frac{1}{R_{12}}+\frac{1}{R_6}\\\\&=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}\\\\\Rightarrow R_{12,6}&=4\quad{\rm \Omega}\end{align*} Now, $R_{12,6}$ is in series with the resistor of $8\,{\rm \Omega}$ which both have a effective resistance of \[R_{12,6,8}=R_{12,6}+R_8=4+8=12\,{\rm \Omega}\] In the end, we are left two resistance of $12\,{\rm \Omega}$ and $6\,{\rm \Omega}$ which are placed in parallel in the circuit (Because each are positioned on one side of points $A$ and $B$). Thus, the overall resistance of the circuit is determined as below \begin{align*}\frac{1}{R_{eq}}&=\frac{1}{R_{3,3}}+\frac{1}{R_{12,6,8}}\\\\&=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}\\\\\Rightarrow R_{eq}&=4\quad{\rm \Omega}\end{align*}  Note: When two resistors $R_1$ and $R_2$ are in parallel, we can use the simple formula below to avoid using reciprocals \[R_{eq}=\frac{R_1 R_2}{R_1+R_2}\]

Problem (2): The equivalent resistance between points $a$ and $b$ is $40\,{\rm \Omega}$ in the following electric circuit. If resistors $R$ are equal, find their value.

ap-circuits-problem-2

Solution : The two resistances of $120\,{\rm \Omega}$ and $40\,{\rm \Omega}$ are in parallel, so we can remove these two resistors and replace them with their equivalent resistance which is \[R_{120,40}=\frac{R_1 R_2}{R_1+R_2}=\frac{120\times 40}{120+40}=30\,{\rm \Omega}\] Two resistors $5\,{\rm \Omega}$ and $R$ in the leftmost vertical branch are in series, so their equivalent resistance is $R_{R,5}=R+5$. These two latter equivalent resistances are also in parallel having an effective resistance of \[\frac{30\times (R+5)}{30+(R+5)}=\frac{30R+150}{R+35}\] In the end, this resistance is placed in series with the $R$ resistor. Thus, the overall resistance of the circuit is obtained as \[R_{eq}=R+\frac{30\times (R+5)}{30+(R+5)}\] Setting this $40\,{\rm \Omega}$ and Solving for $R$, we get the value of the unknown resistor \begin{gather*} R+\frac{30R+150}{R+35}=40 \\\\ \frac{R(R+35)+30R+150}{R+35}=40 \\\\ \Rightarrow 40R+1400=R^2+35R+30R+150 \\\\ \Rightarrow R^2+25R-1250=0 \end{gather*} To solve the quadratic equation $ax^2+bx^2+c=0$ where $a,b,c$ are some constants, use the following equation \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] In this case, $a=1,b=25,c=-1250$, substituting these numerical values into the above equation, we obtain two solutions as \[R=\frac{2(1)\pm\sqrt{25^2-4(1)(-1250)}}{2(1)}\] The plus sign gives $\boxed{R=25\,{\rm \Omega}}$ and the minus sign gives $R=-50\,{\rm \Omega}$ which is incorrect (since the resistance can not be a negative value). 

Problem (3): In the following circuit, how much must $R_3$ be so that the equivalent resistance between $A$ and $B$ is $R_1$?

ap-circuits-problem-3

Solution : Resistors $R_1$ and $R_2$ are in parallel, so their equivalent can be obtained by reciprocal-sum rule or the equation below \[R_{12}=\frac{R_1 \times R_2}{R_1+R_2}\] The result is a resistor which is in series with resistor $R_3$. Thus, the total equivalent resistance of the circuit is \[R_{eq}=R_{12}+R_3\] In the question it's told that the equivalent resistance of the circuit is $R_1$, so set $R_{eq}=R_1$ in the above equation, and solve for $R_1$. \begin{gather*} R_1 =\frac{R_1 \times R_2}{R_1+R_2}+R_3 \\\\ \Rightarrow R_3=R_1-\frac{R_1 \times R_2}{R_1+R_2}=\frac{R_1^2+R_1 R_2-R_1 R_2}{R_1+R_2} \\\\ \Longrightarrow \boxed{R_3=\frac{R_1^2}{R_1+R_2}}\end{gather*}

Problem (4): In the circuit below, how much current passes through the $1.5-{\rm \Omega}$ resistor?

ap-circuits-problem-4

Solution : In this kind of question, first of all, find the total equivalent resistance of the circuit, then using Ohm's law find the amount of total current delivered by the battery. 

Two resistors $3\,{\rm \Omega}$ and $1.5\,{\rm \Omega}$ are in parallel, so their equivalent resistance is \[R_{eq}=\frac{3\times 1.5}{3+1.5}=1\,{\rm \Omega}\] This effective resistor is placed in series with the remaining external resistance of the circuit, $3.5\,{\rm \Omega}$. With this simplification, the circuit involves the resistances of $1\,{\rm \Omega}$, $3.5\,{\rm \Omega}$, and the internal resistance of the battery. Thus, combine them to find the total equivalent resistance of the circuit \[R_t=1+3.5=4.5\,{\rm \Omega}\]  

ap-circuits-problem-4-solution

All these resistors are in series with each other, so a common current flows through each. This current flow is the same current delivered by the buttery into the circuit. But how much current?

When a circuit is connected to a source of emf (battery), the total current through the battery is found by the following formula \[I_t=\frac{\mathcal{E}}{r+R_{eq}}=\frac{6}{1.5+4.5}=1\,{\rm A}\] Thus, through each resistor a current of $1\,{\rm A}$ flows. The next step is to find the current through the $1.5\,{\rm \Omega}$ resistor which is part of $1\,{\rm \Omega}$ combined resistor. 

The two $1.5\,{\rm \Omega}$ and $3\,{\rm \Omega}$ resistors are in parallel with each other. Recall that the voltages across resistors in parallel are the same i.e. $V_3=V_{1.5}$. Using this fact and Ohm's law, we find a relation between the current through these two resistors as below \begin{gather*} V_3=V_{1.5} \\\\ I_3 R_3=I_{1.5}R_{1.5}\\\\ 3I_3=1.5I_{1.5} \\\\ \Rightarrow \boxed{I_3=\frac 12 I_{1.5}}\end{gather*} The $1-{\rm A}$ total current provided by the battery when reaches the junction $a$, splits into the two currents $I_3$ and $I_{1.5}$. So, their sum must be the same total current (i.e., $I_t=I_3+I_{1.5}$). 

In this case, we have $I_{1.5}+I_3=1\,{\rm A}$. Therefore, substituting the boxed relation above into this, we get \begin{gather*} I_{1.5}+\frac 12 I_{1.5}=1 \\\\ \Rightarrow (1.5)I_{1.5}=1 \\\\ \Rightarrow I_{1.5}=\frac{1}{1.5}=\frac 23 \,{\rm A}\end{gather*} Thus, the current through $1.5\,{\rm \Omega}$ is $1.5\,{\rm A}$.

Further reading:      Ohm's law Practice problems with solutions

Problem (5): The $33-{\rm \Omega}$ resistor dissipates $0.8\,{\rm W}$ in the circuit shown below. Find the voltage across the battery.

ap-circuits-problem-5

Solution : The battery voltage is unknown. By having the total current flowing from the battery and the equivalent resistance of the circuit, using Ohm's law, $V_t=I_t R_t$, we can find the voltage across the battery.

The power dissipated by resistance is found by the following formulas \[P=R i^2=\frac{V^2}{R}=Vi\] where $V$ is the voltage across the resistance and $i$ is the current through it. 

In this circuit the $33-{\rm \Omega}$ resistor is connected directly to the battery, so all the current delivered by the battery flows through it. In other words, the current passing through this resistor gives the total current of the circuit.

Using the power formula $P=R i^2$, and solving for $i$, we can find the current flowing from the $33-{\rm \Omega}$ resistor \[i=\sqrt{\frac{P}{R}}=\sqrt{\frac{0.8}{33}}=0.156\,{\rm A}\] This is the total current flowing from the battery $I_t=0.156\,{\rm A}$. Now we must find the equivalent resistance of the circuit. 

The two resistors of $68\,{\rm \Omega}$ and $85\,{\rm \Omega}$ are in parallel, so their equivalent resistance is found as below \[R_{eq}=\frac{68\times 85}{68+85}=37.7\,{\rm \Omega}\] This new resistor is placed in series with the $33-{\rm \Omega}$ resistor, so the total equivalent resistance of the circuit is \[R_t=33+37.7=70.7\quad {\rm \Omega}\] Given the total current and resistance of the circuit, one can find the voltage across the battery using Ohm's law as below \[V_t=I_t R_t=0.15\times 70.7=11\,{\rm V}\] 

Problem (6): In the circuit shown below, the $2-{\rm \Omega}$ resistor dissipates a power of $18\,{\rm W}$ in the circuit. What is the voltage between points $A$ and $B$? 

ap-circuits-problem-6

Solution : The voltage between points $A$ and $B$ is the sum of voltages across $4-{\rm \Omega}$ resistor and the loop. \[V_{AB}=V_{cd}+V_{ef}\] Our task is to find these two unknown voltages.

The resistors $2-{\rm \Omega}$ and $6-{\rm \Omega}$ are in series, so the currents through them are the same. Given the power dissipated by $2-{\rm \Omega}$ resistor and using the power formula $P=Ri^2$, we can find the current flowing from this resistor \[i=\sqrt{\frac{P}{R}}=\sqrt{\frac{18}{2}}=3\,{\rm A}\] Thus, $I_2=I_4=3\,{\rm A}$. The equivalent resistance of these two is \[R_{2,6}=R_2+R_6=2+6=8\,{\rm \Omega}\] The total current and equivalent resistance of the upper branch are known, so using Ohm's law, we can find the voltage across this branch as \[V_{up}=IR=3\times 8=24\,{\rm V}\] The lower branch also has the same voltage as the up branch according to this rule that the voltages across parallel branches are the same. Therefore, \[V_{upper}=V_{lower}=V_{ef}=24\,{\rm V}\]

ap-circuits-problem-6-solution

Therefore, using Ohm's law the voltage $V_{cd}$ is calculated as \[V_{cd}=i_{cd}R_4=5\times 4=20\,{\rm V}\] Hence, the voltage across the points $A$ and $B$ is \[V_{AB}=V_{cd}+V_{ef}=20+24=44\,{\rm V}\]

Further Reading: Kirchhoff's law solved problems for the AP Physics C exam

Problem (7): In the circuit shown below, the voltage drop across the internal resistance and the $2.5-{\rm \Omega}$ resistor are $0.25\,{\rm V}$ and $1.25\,{\rm V}$, respectively. Find the emf and internal resistance of the battery.

ap-circuits-problem-7

Solution : The potential drop across the internal resistance of a battery is defined as $V_r=Ir$, and across the resistor external to the battery is $V_R=IR$ where $I$ is the total current through the circuit. 

Using these two facts and their given numerical values, we can find their ratio \begin{align*} \frac{V_r}{V_R}&=\frac{Ir}{IR}\\\\ \frac{0.25}{1.25}&=\frac{r}{2.5}\\\\ \Rightarrow r&=0.5\,{\rm \Omega}\end{align*} Thus, the internal resistance of the battery is $\boxed{r=0.5\,{\rm \Omega}}$. 

ap-circuits-problem-7-solution

Problem (8): In the circuit below, when the switch is opened, the voltmeter shows $12\,{\rm V}$, and when it is closed, the voltmeter reading is $10\,{\rm V}$. What are the emf and the internal resistance of the battery? 

ap-circuits-problem-8

Solution : Remember that in a circuit including a source of emf with an internal resistance of $r$ and an equivalent resistance of $R_{eq}$ due to other resistances external to the battery, the total current flowing through the circuit is found by the formula below \[I=\frac{\mathcal{E}}{r+R_{eq}}\] Here, the voltmeter is connected across the terminals of the battery, so it shows the battery's terminal voltage. Always, the terminal voltage is given by $\Delta V=\mathcal{E}-Ir$, where $I$ is the total current through the circuit. 

ap-circuits-problem-8-solution

When the switch is opened, the middle branch is removed from the circuit, so the equivalent resistance of the circuit equals $R_{eq}=R$. Correspondingly, the current in this case is \[I_O=\frac{\mathcal{E}}{r+R_{eq}}=\frac{\mathcal{E}}{r+R}\] where $I_O$ is the current when the switch is open. By opening the switch, the voltmeter reading is also found as \begin{align*} V_O&=\mathcal{E}-I_O r\\\\ &=\mathcal{E}-\left(\frac{\mathcal{E}}{r+R}\right)r \\\\&=\frac{\mathcal{E}R}{R+r}\end{align*} Now we turn to the case of closing the switch $S$. In this case, the equivalent resistance is \[R_{eq}=\frac{R\times R}{R+R}=\frac R2\] Since the resistors are in parallel. The ammeter shows the total current flowing from the battery whose value is \[I_C=\frac{\mathcal{E}}{r+R_{eq}}=\frac{\mathcal{E}}{r+R/2}\] On the other side, as mentioned, the voltmeter indicates the voltage across the battery or terminal voltage with the following value \begin{align*} V_C&=\mathcal{E}-I_C r\\\\ &=\mathcal{E}-\left(\frac{\mathcal{E}}{r+R/2}\right)r \\\\&=\frac{\mathcal{E}R}{R+2r}\end{align*} Now we compare the currents and voltages before and after closing the switch $K$: \begin{gather*} \frac{\mathcal{E}}{r+R/2}> \frac{\mathcal{E}}{r+R} \\\\ \Rightarrow \boxed{I_C > I_O} \\\\ \frac{\mathcal{E}R}{R+r} > \frac{\mathcal{E}R}{R+2r} \\\\\Rightarrow \boxed{V_O > V_C} \end{gather*}

Problem (10): (a) Four $3.2-{\rm \mu F}$ capacitors are connected in series. What is the equivalent capacitance? (b) Assuming now they are connected in parallel, what is their equivalent capacitance? 

Solution : (a) When a number of capacitors are connected in series, say $C_1, C_2, C_3,\cdots$, their equivalent capacitance is given by the following formula \[\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\cdots\] Therefore, in this case, the equivalent capacitance is \[\frac{1}{C_{eq}}=\frac{1}{3.2}+\frac{1}{3.2}+\frac{1}{3.2}+\frac{1}{3.2}=\frac{4}{3.2}\] By inverting, we have $C_{eq}=0.8\,{\rm \mu F}$. 

(b) When those capacitors are in parallel, their equivalent capacitance becomes \[C_{eq}=C_1+C_2+C_3+\cdots\] Thus, in this case, the equivalent capacitance is \[C_{eq}=4(3.2)=12.8\,{\rm \mu F}\]

Problem (11): In the following circuit, what is the potential difference across the capacitor? a. 8V        b. 10V        c.    6V        d. 15V

ap-circuits-problem-11

Solution : As can be seen from the circuit, the two $4-{\rm \Omega}$ resistors are in parallel, so we compute their combined resistance and replace them with those. \[R_{eq}=\frac{R_1 R_2}{R_1+R_2}=\frac{4\times 4}{4+4}=2\,{\rm \Omega}\] Note that when a fully charged capacitor is present in an electric circuit, that branch acts as an open circuit, which means the current no longer flows through it. Because, in practice, the capacitor behaves like a resistor with infinite resistance. 

On the other side, the current delivered by a source of emf (battery) with internal resistance $r$ and external resistance $R_{eq}$ is found by \[I=\frac{\mathcal{E}}{r+R_{eq}}=\frac{12}{1+2}=4\,{\rm A}\] All this current flows through the equivalent resistance of $2\,{\rm \Omega}$. Given that, we can find the potential drop across this combined resistor, using Ohm's law, as  \[V=IR=4\times 2=8\,{\rm V}\] Now pay attention to this fact that the capacitor and $2-{\rm \Omega}$ equivalent resistance are in parallel which means the voltage across the capacitor is the same as the voltage across the resistor (i.e., $V_C=V_{2R}$). 

Hence, the potential (voltage) difference across the capacitor is $\boxed{8\,{\rm V}}$.

Problem (12): In the circuit diagram shown below, how much charge is on the plates of the capacitor stored? 

ap-circuits-problem-12

Solution : The charge stored on the plates of a capacitor is related to the voltage across it by $Q=C\Delta V$. The capacitance of the capacitor is known, the only thing that we must find is the potential drop (voltage) across it. 

On the other side, you can see that the capacitor is connected in parallel with the $4-{\rm \Omega}$ resistor, which means that the voltage difference across the capacitor is the same as the resistor. So, our task is to find the voltage across the $4-{\rm \Omega}$ resistor, $\Delta V_4=IR$, which is subsequently needed to find the current through that resistor. 

As a rule of thumb, ignore that branch of the circuit that includes the capacitor since the capacitors in an electric circuit (when it is fully charged) act as a resistor with an infinite resistor and do open the circuit.  

With this long introduction, let's move on to solving the problem. 

By ignoring the capacitor, the three $4-{\rm \Omega}$ resistors are in series with a equivalent resistance of  $R_{eq}=3(4)=12\,{\rm \Omega}$. This new resistor is in parallel with the other $4-{\rm \Omega}$ resistor in the middle branch. 

ap-circuits-problem-12-solution

Problem (13): In the circuit shown below, the capacitor does not have any charge on its plates. Immediately after closing the switch $S$, what are the currents $I_1$ and $I_2$, respectively? a. 1, 1        b. 1.5, 0        c. 0, 1.5        d. 1.5, 3

ap-circuits-problem-13

Conversely, after passing a long time, it behaves like a resistor with infinite resistance. 

Initially, the capacitor has no charge, so the potential difference across it, using the capacitance formula $\Delta V=\frac{Q}{C}$, is zero. On the other side, this capacitor is in parallel with the $5-{\rm \Omega}$ resistor. 

Recall that when two electronic components (e.g., a capacitor or resistor) are in parallel, the potential drop across them is the same. According to this rule, the potential difference across the $5-{\rm \Omega}$ resistor is also zero. Applying Ohm's law, $\Delta V=IR$, we find that the current flowing through this resistor must also be zero. 

Consequently, the current $I_2$ shown in the circuit, immediately after closing the switch, must be zero. In this situation, the $5-{\rm \Omega}$ resistor is removed and the circuit converts to the following 

ap-circuits-problem-13-solution

Note that in all AP Physics circuit multiple-choice problems, just after placing an initially uncharged capacitor in a circuit (like immediately after closing a switch in the circuit), it acts as a typical wire with zero resistance, and subsequently, all branches in parallel with it can be removed from the circuit.

Problem (14): In the circuit below, the capacitor is initially uncharged. A long time after closing the switch $S$, find the currents $I_1$ and $I_2$.  a. 1, 1        b. 1.5, 0        c. 0, 1.5        d. 1.5, 3

ap-circuits-problem-14

After a long time, when the switch is closed, the charge accumulation on the capacitor reaches its maximum value. At this moment, this charge, which is built on the plates of the capacitor, impedes completely the charge flowing by the battery (precisely, the current through this branch). 

So, the current flowing through this branch is stopped and all total current delivered by the battery (i.e., $I_1$) reaching the junction $A$, flows into the branch where the $5-{\rm \Omega}$ resistor is presented. In other words, $I_1=I_2$. 

ap-circuits-problem-14-solution

On the other side, the current flowing through the circuit is found as \[I_t=\frac{\mathcal{E}}{r+R_{eq}}=\frac{15}{0+15}=1\,{\rm A}\] Therefore, \[I_1=I_2=I_t=1\,{\rm A}\] The correct answer is a. 

In all AP Physics circuit multiple-choice questions, note that, after a long time of placing an initially uncharged capacitor in a circuit, it acts as an open circuit, and we can remove that branch from the circuit. 

Author : Dr. Ali Nemati      Page Published : 10/13/2021  

© 2015 All rights reserved. by Physexams.com

How to Solve Complicated Circuits with Kirchhoff's Voltage Law (KVL)?

Published Aug 04, 2020

We have gone over Kirchhoff’s Current Law (KCL) in a previous tutorial and Kirchhoff’s Voltage Law (KVL) is very similar but focused on the voltage in a circuit, not the current. Kirchhoff’s Voltage law states that the sum of the voltages in a closed loop will equal zero. In other words, if you look at any loop that goes completely all the way around, any increases in voltage throughout the loop will be offset with an equal amount of decreases in voltage. Visually, this can be seen in the image below.

Using this concept, much as how we can use nodal analysis with KCL, we can use mesh analysis because of KVL. While a mesh is basically any loop within a circuit, for mesh analysis, we will need to define meshes that don’t enclose any other meshes.

You can see that if we make a loop around the ‘outside’ of the entire circuit, technically it is a mesh because the loop can be completed. However, for purposes of analysis, we need to break it into three different meshes. So let’s go over the steps of how to solve a circuit using mesh analysis before jumping into a few examples.

There are 5 steps that we recommend, and as we did with the KCL/nodal analysis steps, two of the steps are to calm down and step back, making sure that everything makes sense intuitively.

  • Take your time, breathe, and assess the problem. Write down what info you’ve been given and any intuitive insights you have.
  • Assign mesh currents to all of the meshes. There should be one current assigned per mesh. You need to choose which direction your current is flowing - this is semi-arbitrary because as long as you do your math right, it doesn’t really matter. But in most cases, people assume a clockwise current direction.
  • Apply KVL to each of the meshes, using Ohm’s Law to show the voltages in terms of the current.
  • Solve the simultaneous equations (like we did with KCL) to find the actual values.
  • Sanity check. Take a moment to review what you’ve done and see if the numbers make sense and are internally consistent.

We’ll go over some examples now and frankly, after these examples, the only real additions and changes will be complications that make the math more difficult. The problems shouldn’t get much harder conceptually but the math can get significantly harder. Please, don’t get lost in the math. If the numbers start to lose their numbers, make sure to come up for air and remember what you’re doing and what you’re trying to do.

A simple example - 1 mesh.

Let’s start here! This is a simple circuit, so simple that we could solve this using tools we already know. But I want to start simple so that we can focus on the concepts and the steps. So, let’s do it.

Step 1: Let’s take stock of the circuit. It obviously only has one loop, and we’ve got a voltage source and two resistors. We’ve been given the value of the voltage source and both resistors, so all we need is to find out the current around the loop and the voltage drops over the resistors. And as soon as we find one, we can quickly use Ohm’s Law to get the other. This is going to be easy.

Step 2: We already noticed in step 1 that there will only be one mesh, so let’s draw in our mesh current, give it a direction, and give it a name. We’ll go clockwise and call it i 1 . Now, I’m usually sloppy and don’t distinguish between i 1 and I 1 , but in this case, we will do a lowercase ‘i’. This will be important in later examples. And, we know that since we have one mesh, there will only be one equation.

Step 3: Let’s create our equations based off of KVL. This is the first step that requires any math. So, with KVL, let’s figure out our equation.

There are two ways of looking this, which can cause untold confusion. I will explain the differences and, as long as you’re consistent in each equation (not even necessarily in each problem, but sheesh, why would you confuse yourself unnecessarily?) then everything will be fine.

In the first option, as we go around the loop, we see that we increase by 5V across the voltage source and then drop voltage across R 1 and R 2 , giving us our positive 5 volts and then our two negatives. To me, this is more intuitive because you’re going up in voltage across the voltage source in the way we defined the current flow, and you drop voltage across the resistors as the current flows through them. However, it is extremely common for people to learn it the second way.

In the second option, you just use the sign of the voltage on the side of your branch that the current enters into. With the voltage source, since we are going clockwise, the current sees the negative sign first, so it is a minus. As the voltage is dropping from positive to negative over the resistors, the current sees the positive sign on the resistors first, so you add them. If this is more intuitive for you - use it! Neither of these options are wrong, you see that you get the same equations (just multiply both sides by -1) but make sure you’re consistent with each equation. Please.

Step 4: Since there are no unknowns, we can simply plug in the values for R 1 and R 2 and find out what i 1 is.

And now we can find the voltages across R 1 and R 2 .

Step 5: Sanity check! Note that V 1 + V 2 basically equals 5V (rounding errors!) which means that the voltage that drops over the two resistors is the same as the voltage increase from the voltage source.

Let’s make things a little more complicated.

Step 1: What have we got here? It looks like we have two meshes that share a common resistor in the middle, R 3 . Again, we have all the values of voltage sources and resistors, so we should be able to get actual values for the current and voltages through/across those resistors. Even without any values, we could do the analysis and show relationships but it is a bit more satisfying to me to actually come up with a numerical answer. We do need to know how to treat R 3 , but we’ll take care of that in step 3.

Step 2: Let’s identify the meshes. We’ll make both current loops flow clockwise and we’ll name the left hand one i 1 and the right hand one i 2 . Note that these are still lower case. And it matters this time, because we also have the current through the resistor R 1 , which is I 1 (note the capitalized “I”), the current through resistor R 2 , which is I 2 , and then through R 3 , which is I 3 . The capitalization is how to distinguish between the mesh currents (i 1 and i 2 ) and the branch currents (I 1 , I 2 , and I 3 ).

Step 3: Create the equations for the meshes. This will be quite straightforward but we need to know what to do about the voltage across R 3 . Let’s actually do the equation for i 1 and then talk about it for a moment.

So, looking at that equation, you’re probably wondering why there is i 2 in our equation for mesh current i 1 . Remember that each section is in reference to voltage. We increase by 10V, which is straightforward. We drop a voltage across R 1 that is equal to i 1 *R 2 , still fairly straightforward. But the voltage drop across R 3 is the amount of current flowing downward as i 1 minus the amount of current flowing upward as i 2 multiplied by R 3 .

With our clockwise direction, we have stated that i 2 is flowing up through R 3 . Obviously, in reality, current is only flowing one way, but we don’t know which way right now, and mathematically, we have said that there is both current flowing down through R 3 as i 1 and flowing up through R 3 as i 2 . The trick here is that if we had defined i 2 in the opposite (counterclockwise) direction, we would have to add the current i 2 to i 1 to figure out the voltage drop across R 3 .

So with this, pause, take a second, make sure you understand why we created the equation we did for the first mesh current. Then see what you come up with for the second mesh current before checking to see what we come up with. You’re going to have to control your eyeballs, though, because the answer is right below this text.

Is this what you got? Remember that, with our definition that the current is flowing clockwise, the voltage is dropping as we go from ground across R 3 , and still dropping as we go across R 2 , before coming to the voltage source, which since we’ve defined this clockwise direction, gives us a negative 5 volts. This is where it’s incredibly important to understand what’s going on intuitively - if you get bogged down into the math too much without knowing what’s going on, you’re going to be setting up and solving the wrong equations! Trust me - I speak from much painful experience.

So now we have two equations and two unknowns. We can either solve this with substitution or by getting ready to do some linear algebra. Let’s do substitution.

Insert the values for the resistors.

Simplify the first equation.

Simplify the second equation a bit before replacing i 1 .

Then i 1 is equal to,

Example 3 (Supermeshes)

With KCL, if we had a voltage source that wasn’t connected directly to reference ground, we would create a supernode and then, as part of the process, we would need to do a bit of KVL to finish the analysis. With KVL, if we have a current source that is shared between two meshes, we need to treat it in a similar way. We get rid of the current source and anything that is connected in series with it. We then treat the remainder as a single, larger supermesh.

Once we make that mesh, we create the equation to describe it. In this case, we get:

Now we have the equation for the supermesh but we have two unknowns and only one equation. So let’s put the current source back in with any elements that were in series with it and do KCL at the node where they connect to the bigger circuit. Once it’s in place, we use KCL at that node to create a second equation.

Now we have two equations and two unknowns! Let’s put this in the format needed to do some linear algebra and see what we get.

So our two equations are:

Which we put into a linear equation solver to get:

As I’m prone to making math errors, I prefer the linear equation method as it is usually faster and less likely that I screw up it. With the supermesh, this isn’t a common concern as, unless you’re dealing with transistors or CMOS level circuit design, current sources aren’t very typical. However, it’s a good tool to have in your toolbag in case it crops up and helps us better understand the relationship between the physical circuits and the mathematical representations.

That is our brief overview of Kirchhoff’s Voltage Law and how that leads to mesh analysis. You’ll note that we sometimes used mesh analysis and KVL interchangeably. While technically not the same, it is very common to hear them used like that. Depending on where you are and who you’ve studied with, you may find some other differences in approach, naming conventions, and even direction assumptions. However, despite these superficial differences, all mesh analysis comes down to finding the voltage across the different elements in a mesh. As long as you’re consistent and have a good understanding of what you’re doing, you should be able to get the answer you’re looking for.

  • Circuit Theory (18)
  • Kirchhoff's Voltage Law (2)
  • Mesh Analysis (2)

Authored By

Josh bishop.

Interested in embedded systems, hiking, cooking, and reading, Josh got his bachelor's degree in Electrical Engineering from Boise State University. After a few years as a CEC Officer (Seabee) in the US Navy, Josh separated and eventually started working on CircuitBread with a bunch of awesome people. Josh currently lives in southern Idaho with his wife and four kids.

Related Tutorials

  • Electronic Basics

Explore CircuitBread

  • 216 Tutorials
  • 9 Textbooks
  • 12 Study Guides
  • 295 Equations Library
  • 201 Reference Materials
  • 91 Glossary of Terms

Friends of CircuitBread

Try OLC's Bill of Materials Tool

Check Out Our Free Ohm's Law Calculator

Free eBook & Resource Library + 1 Perk

Get the latest tools and tutorials, fresh from the toaster.

What are you looking for?

Message sent.

Thanks for the message, our team will review it shortly.

Log In or Sign Up

Username should have no spaces, underscores and only use lowercase letters.

Thanks for joining CircuitBread!

Please confirm your email address by clicking the link in the email we sent you.

Didn't receive anything? Resend now.

StickMan Physics

StickMan Physics

Animated Physics Lessons

Complex Circuit

Complex circuits.

Complex circuits have components that are in series and some that are in parallel .  Lets start by reviewing series and parallel circuits and then see how a complex circuit works as a combination.

Series Circuits and Rules (Click Here to Go to the Lesson)

A series circuit you have no branches , current has only one path to follow.  The same current remains together through all components from the positive terminal and back to the negative of the battery.

Ohm's Law   (V=IR):

You can use this at a single location (T, 1, 2, 3, and any more) as seen below

  • (V T ) = (I T )(R T )
  • (V 1 ) = (I 1 )(R 1 )
  • (V 2 ) = (I 2 )(R 2 )
  • (V 3 ) = (I 3 )(R 3 )

Overall Series Circuit Rules

  • V T = V 1 + V 2 + V 3 + …

I T = I 1 = I 2 = I 3 = …

  • R T = R 1 + R 2 + R 3 + …

Parallel Circuit Rules (Click Here to Go to That Lesson)

In a parallel circuit you have branches , multiple paths to follow.  So current splits up but later comes back together as it returns to the battery.

You can use Ohm's Law (V=IR) at a single location the same as above in series circuits.

Overall Parallel Circuit Rules

  • V T = V 1 = V 2 = V 3 = …
  • I T = I 1 + I 2 + I 3 + …
  • 1/R T = 1/R 1 + 1/R 2 + 1/R 3 + ...

Follow the current from the positive terminal of the battery in the animation.  Some parts of the circuit are in series and some in parallel.

  • The battery and resistor 1   are in series: 2A of current starts at the batter and flows through both.
  • The branch with resistor 2, the branch with Resistor 3   and 4, and the branch with Resistor 5   are in parallel :  Current splits between these branches.
  • The resistor 3   and resistor 4 are in series : Once current goes down the branch that same current must flow from both resistors.

Solving A Complex Circuit For Resistance

When solving for the resistance in a series circuit, the goal is to break down all the different parts making act like one single series circuit.  Follow the steps or our example as we do this.

Example Problems

1. Solve for the total resistance of the complex circuit below.

Complex Circuit Resistance

Example 1 Step 1

First add resistor 3 and resistor 4 which are in series down a branch

R T = R 1 + R 2

R T = 10Ω + 5Ω = 15Ω

1

Example 1 Step 2

Now take resistor 2, the equivalent from resistor 3 and 4, and resistor 5 which are all on branches and put them together following parallel rules.  In this case R T will be the equivalent of these three branches.

R T = ((1/R 1 ) + (1/R 2 ) + (1/R 3 )) -1

R T = ((1/10Ω) + (1/15Ω) + (1/20Ω)) -1 = 4.615Ω

2

Example 1 Step 3

Current must flow through the 5Ω resistor and 4.615Ω resistor equivalent so they act like they are in series together.  Use series rules to put these together.

R T = 5Ω + 4.615Ω = 9.615Ω

This is the resistance placed at the battery equivalent to the resistance of the total circuit.

3

2. Solve for the unknown components of the following complex circuit.

Complex Circuit Example 2

Example 2 Step 1

Combine Resistor 2 and resistor 3 in parallel

R T = ((1/R 1 ) + (1/R 2 )) -1

R T = ((1/30Ω) + (1/30Ω)) -1 = 15Ω

1

Example 2 Step 2

Combine the 15Ω equivalent from resistor 2 and 3 with resistor 1 which is in series

R T = 15Ω + 5Ω = 20Ω

This is the resistance of the circuit placed at the battery.  We used and no longer need the equivalent or resistor 2 and resistor 3 so you will see that equivalent dropped off the next step.

2

Example 2 Step 3

Now solve for current at the battery using Ohm's Law

I T =V T /R T

I T = 40V/20Ω = 2A

3

Example 2 Step 4

2A of current flows from the battery and does not branch before resistor 1.  This make resistor 1 have the same current

Series rule

solving simple circuit problems

Example 2 Step 5

Now solve for voltage at resistor 1 using Ohm's Law

V 1 = (I T )(R T )

V 1 = (2A)(5Ω) = 10V

5

Example 2 Step 6

The 10V is dropped crossing the first resistor leaving 30V for the rest of the circuit.  In parallel the voltage left will be dropped all and equally on both the branches which have resistor 2 and resistor 3

The parallel rule we followed was this:

V T = V 1 = V 2

6

Example 2 Step 7

Now we can use Ohm's Law to finish this problem off

At both resistor 2 and resistor 3 you would do the follwing.

I= 30V/30Ω = 1A

7

For our final check we can see that the 2A of current from resistor split mathematically correct and know we are right.

I T = I 1 + I 2

2A = 1A + 1A

3. Solve for the unknown components of the following complex circuit.

Complex Circuit Example 3

Example 3 Step 1

Use Ohm's Law to solve for the current at resistor 4

V 4 =( I 4 )(R 4 )

I 4 =(V 4 )/(R 4 )

I 4 =(16V)/(16Ω)= 1A

1

Example 3 Step 2

Resistor 2 and resistor 3 are in series.  In series current is equal

Series Circuit Rule

So the 4A of current at resistor also passed trough resistor 2 to get there.  Since I 3  = 4A, I 2 = 4A.

2

Example 3 Step 3

Now use Ohm's Law to solve for the voltage drop of resistor 3.

V 3 =I 3 R 3

V 3 =(4A)(1Ω) = 4V

3

Example 3 Step 4

  • The branch that has Resistor 2 and resistor 3 are in parallel with the branch that has resistor 4.
  • So the entire branch that has resistor 2 and resistor 3 must be equal to the 16V that drops on the other branch.
  • Now observe that resistor 2 and resistor 3 are in series and current must flow through both resistors.
  • they mus add up to 16V following the series rule.

two resistor on a branch in series must drop the same voltage as the other branch

Therefore the voltage drop at resistor 2 must be 12 V since V 2 and V 3 dropping an additional 4V must add to the 16V of the other branch.

4

Example 3 Step 5

Now you can use Ohm's Law to solve for the resistance of resistor 2.

V 2 =I 2 R 2

R 2 =V 2 /I 2

R 2 = 12V/4A = 3Ω

5

Example 3 Step 6

Now look at the branch that has resistor 2 and resistor 3.  4 amps of current flows down that branch.  It is the same 4 amps of current since these two are in series.  (Look at the current parallel circuit animation earlier in this lesson if you are confused)

An additional 1 amp of current flows down the separate branch with resistor 4

So the total current that split off coming from resistor 1 was 5 amps.

6

Example 3 Step 7

Now do Ohm's law to determine voltage drop at resistor 1.

V 1 =I 1 R 1

V 1 = (5A)(8Ω) = 40V 

and Ohm's Law for the voltage drop at resistor 5

V 5 =I 5 R 5

V 5 = (5A)(4Ω) = 20V 

8

Example 3 Step 8

Lastly do Ohm's Law to determine the resistance of the entire circuit at the battery.

V T =I T R T

R T =V T /I T

R T =76V/5A = 15.2Ω

9

You can spend a little time doing your final check calculating the resistance of the entire circuit as a last check and you will get a R T = 15.2Ω when solving for total resistance of the circuit correctly.

Visual of Current Flowing in Example 3

Click on the picture below to see the animation of current flowing.  This is for understanding of the final answers for current.

  • Back to the Main Current and Circuits Page
  • Back to the  Stickman Physics Home Page
  • Equation Sheet

Terms and Conditions - Privacy Policy

Solve AC Circuits Problems with Solutions

Table of Contents

Kirchhoff's and Ohm's law is extended and used to solve AC circuits problems using impedances in complex forms. All the quantities such as voltages, currents and impedances are represented by complex numbers in standard and polar forms.

Review of Complex Numbers

The imaginary unit is defined by \( j = \sqrt {-1} \) or \( j^2 = - 1 \) A complex number \( Z \) in the standard the form \( Z = a + j b \) may be written in polar form as \( Z = r \; \angle \; \theta \) where \( r \) and \( \theta \) are the modulus and argument , of \( Z \), respectively and are defined by \( r = |Z| = \sqrt {a^2 + b^2} \) and \( \theta = \arctan \left( \dfrac{b}{a} \right) \) within the range \( -\pi \lt \theta \le \pi \) It is easier to divide or multiply complex numbers in polar form Let \( Z_1 = r_1 \; \angle \; \theta_1 \) and \( Z_2 = r_2 \; \angle \; \theta_2 \) \( Z_1 \cdot Z_2 = r_1 \cdot r_2 \; \angle \; \theta_1 + \theta_2 \) \( \dfrac{Z_1}{Z_2} = \dfrac{r_1}{r_2} \; \angle \; \theta_1 - \theta_2 \)

Problems with Solutions

More problems with answers.

Answer to Problem 5 Answers Currents: \( I_1 = 0.054 \angle 10.55^{\circ} \; , \; I_2 = 0.048 \angle 23.01^{\circ} \; , \; I_3 = 0.013 \angle -42.22^{\circ} \) Voltages: \( V_{R_1} = 5.49 \angle 10.55^{\circ} \; , \; V_{R_2} = 4.71 \angle -12.32^{\circ} \; , \; V_{R_3} = 2.60 \angle -42.22^{\circ} \; , \; V_{R_4} = 2.77 \angle 15.63^{\circ} \)

More References and Links

Short your Time We open your answer

Fed up of solving complex circuits with tons of bridges and pillars or too lazy even to solve a simple circuit. fret not check out our super cool circuit solver that could solve any kind of resistive circuit. excited to check it out.

Get Started

Simulate any kind of circuit you have designed and get the results instantly on just a single click.

Design your own custom circuit using the components given in the palette in the web editor by simple drag and drop.

Electrically4U

Superposition Theorem with solved problems

solving simple circuit problems

Superposition theorem is used to solve the complex electric network, which consists of two or more sources and several resistances, by considering and analyzing all the sources individually.

Superposition theorem states that,

“In any linear, bilateral network having more than one source, the response across any element is the sum of the responses obtained from each source considered separately and all other sources are replaced by their internal resistance.”

It can also be stated as,

If a linear circuit consists of more than one independent source, then the current flowing through any part of the circuit is equal to the algebraic sum of the currents produced by each independent source, when it is considered separately.

Illustration of Superposition theorem

Let us understand this theorem with the following circuit, which consists of three independent sources(I 1 , V 1 , V 2 ), several resistances(R 1 , R 2 , R 3 ) and a load resistor R L .

Illustration of Superposition theorem 1

Now, to determine the current flowing through the load resistor R L , the circuit is analyzed by considering each source independently. While considering a single source, all other voltage sources in the circuit must be short-circuited and the current source, if any should be open-circuited.

By considering the current source I 1 alone acting in the circuit, as shown below, the current I L1 through the load is determined.

Illustration of Superposition theorem 2

Now, this current source is open-circuited and the second source(Here, it is voltage source V 1 ) is connected in the circuit, as shown below. Considering this voltage source V 1 alone acting in the circuit, the current through the load I L2 is determined.

Illustration of Superposition theorem 3

Similarly, the circuit is reconnected with the next voltage source V 2 , while the other voltage source is short-circuited and the current source is open-circuited. Considering this voltage source V 2 alone acting in the circuit, the current through the load I L3 is determined.

Illustration of Superposition theorem 4

Now, according to the superposition theorem, the current flowing through any part of the circuit is equal to the algebraic sum of the currents produced by each independent source.

Thus the current flowing through the load R L is given by,

I L = I L1 + I L2 + I L3

Steps to solve the circuits using Superposition Theorem

  • Identify the load resistor (R L ) in the given problem.
  • Considering a single source alone acting in the circuit, short-circuit the other voltage sources and open the current sources, if any.
  • Calculate the current flowing through the load resistor R L  due to a single source.
  • Repeat steps 2 and 3 for all other sources in the circuit.
  • To find the total current through the load resistor, perform an algebraic sum of the currents produced by each independent source.

Solved Problem 1

Find the current through 3 Ω resistor using superposition theorem.

Superposition Theorem - Solved Problem 1

Let I 1 and I 2 are the currents flowing through the 3 Ω resistor, due to the voltage sources 20 v and 40 v respectively.

(i) To find I 1 .

Consider 20 v voltage source alone . Hence, Short circuit the other voltage source and the circuit is redrawn as below,

Problem solution 1_1

Now, to find the current through 3 Ω resistor, it is necessary to determine the total current supplied by the source (I T ).

When you observe the circuit, 3 Ω and 6 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 5 Ω resistor. Hence the equivalent or total resistance is obtained as below,

\[R_T = 5 + \frac{3 * 6}{3 + 6} = 7 \Omega \]

By applying Ohm’s law ,

\[I_T = \frac{V}{R_T} = \frac{20}{7} = 2.857 A \]

Now, the current through 3 Ω resistor is determined by using current division rule . It is given by,

\[ I_1 = I_T * \frac{6}{6 + 3} = 2.857 * 0.667 = 1.904 A\]

(ii) To find I 2 .

Consider 40 v voltage source alone . Hence, Short circuit the other voltage source and the circuit is redrawn as below,

Problem solution 1_2

When you observe the circuit, 3 Ω and 5 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 6 Ω resistor. Hence the equivalent or total resistance is obtained as below,

\[R_T = 6 + \frac{3 * 5}{3 + 5} = 7.875 \Omega \]

The below figure shows the resultant circuit, which depicts the currents produced because of two voltage sources 20 v and 40 v acting individually.

Problem solution 1_3

By superposition theorem, the total current is determined by adding the individual currents produced by 20 v and 40 v.

Thus the current through 3 Ω resistor is = I 1 + I 2 = 1.904 + 3.174 = 5.078 A

Solved Problem 2

Find the voltage across through 15 Ω resistor using superposition theorem.

Superposition Theorem - Solved Problem 2

Let V 1 , V 2 , V 3 , V 4 be the voltages across the 15 Ω resistor when each source (20v, 10v, 10A, 5A sources) are considered separately. Hence the resultant voltage is given by,

V T = V 1 + V 2 + V 3 + V 4

(i) To find V 1

Consider 20v source alone. Hence the other two current sources are open-circuited and a voltage source is short-circuited. The circuit is redrawn as below,

Problem solution 2_1

By applying KVL , the current through the 15 Ω resistor is given by,

\[I_1 = \frac{20}{15 + 40} = 0.363 A \]

Thus,the voltage across the 15 Ω resistor is given by,

\[V_1 = I_1 * R = 0.363 * 15 = 5.445 V \]

(ii) To find V 2

Consider 10v source alone. Hence the other two current sources are open-circuited and a voltage source is short-circuited. The circuit is redrawn as below,

Problem solution 2_2

(iii) To find V 3

Consider 10A source alone. Hence the other two voltage sources are short-circuited and a current source is open-circuited. The circuit is redrawn as below,

Problem solution 2_3

Here, 15 Ω and 40 Ω resistors in parallel with each other, hence by using current division rule , current through the 15 Ω resistor is given by,

\[I_3 = 10 * \frac{40}{15 + 40} = 7.27 A \]

Thus, the voltage across the 15 Ω resistor is given by,

\[V_3 = I_3 * R = 7.27 * 15 = 109.05 V \]

(iv) To find V 4

Consider 5A source alone. Hence the other two voltage sources are short-circuited and a current source is open-circuited. The circuit is redrawn as below,

solving simple circuit problems

Here, since the circuit has a low resistance path, a negligible amount of current flows through 15 Ω resistor. Hence, current through the 15 Ω resistor is given by,

\[ I_4 = 0 \]

Therefore, by the superposition theorem, the resultant voltage is given,

\[V_T = V_1 + V_2 + V_3 + V_4 = 5.445 + 2.73 - 109.05 + 0 = -100.875 V\]

In the above expression, the negative sign in V 3 indicates that the direction of the current, in that case, is opposite to the assumed direction in the circuit.

Sharing is Caring

solving simple circuit problems

An Assistant Professor in the Department of Electrical and Electronics Engineering, Certified Energy Manager, Photoshop designer, a blogger and Founder of Electrically4u.

Related Posts

Norton’s-Theorem

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

19 Comments

Hope you will solve more solutions

Solve problems on all theorems ?

In step iii, Since we need current through 15 ohm resistor, shouldn’t it be: I3= 10*15/(15+40) ?

No. that’s the wrong way. Since it is a parallel circuit, the current division rule is applied. Kindly go through the rule to get a clear idea.

very good notes.Easily understandable.Thank you

Glad to know that the contents are understandable. Thank you.

Sir actually i want to ask u a numerical question related to superposition theorem how i can sen u the question

You can post the question here or you can send it through the mail: [email protected]

Sir, in the second problem to actually verify superpositon theorem v total should also be calculated when all the sources are active which you didn’t mention there will you please explain the process involved there in detail?

The superposition theorem is used not to verify but to solve complex electric circuits. To verify this theorem, you can either simulate the circuit in any simulation software or you can do it practically. In the superposition theorem, the current or voltage due to each individual source is determined and then they are added to find the resultant(total) current or voltage across the given load. The same is followed for both the problems. kindly check the above steps to solve .

Am so glad because my problems in this area are now fully solved. Receive m gratitude

for problem 2, step (iv) can you please explain how/where the low resistance path works and why the 15 ohm resistor has a negligible resistance. thanks

in step (iv), for the 5A source, there are two paths for the flow of current. One path is through 15Ω resistance and the other path is the short-circuited path, which is across the current source. In an electrical circuit, a short circuit will have very low resistance. Since there is very low resistance in this path, all the current from the source rushes through the short circuit path, making a negligible amount of current to flow through 15Ω resistance. That is the reason, I4 =0. And in your question, you have asked, “why the 15 ohm resistor has a negligible resistance.” 15Ω resistor will never have negligible resistance. It has its own resistance of 15Ω.

In part iv, to find for V4, you said the other current source is open-circuited but on the re-drawing of the circuit it is short-circuited.

Yeah, you are right. It was corrected now and thanks for showing the mistake.

I don’t also understand why v4 in problem 2 is completely zero. I try using Ohms law for it, it was 0.060v. Can you help explain to me more better.

Could you please tell me how you applied Ohm’s law?

I don’t know how I can easily understand part iv, the reason for completely ignoring 5A, does it mean it’s not contributing any voltage to resistor 15, if it does then why not considering it. Secondly, I am asking, in part iii, when we connect to 10A, resistors 15 and 40 are considered to be in parallel, what about when on IV, when connected to 5A, are they in parallel or series?

Subscribe to Electrically4u..!

Enter your Email Address to get all our updates about new articles to your inbox.

IMAGES

  1. How to Solve Any Series and Parallel Circuit Problem

    solving simple circuit problems

  2. How to Solve Circuit Problems: 9 Steps (with Pictures)

    solving simple circuit problems

  3. How to Solve a Basic Circuit : 5 Steps

    solving simple circuit problems

  4. Solving Circuit Problems using Kirchhoff's Rules

    solving simple circuit problems

  5. Circuit Diagram Problems Kirchoff's Rules

    solving simple circuit problems

  6. Solving Circuit Problems using Laplace

    solving simple circuit problems

VIDEO

  1. Circuits: Example 5

  2. Problems based on Equivalent Circuits Part 1

  3. Solving Simple Equation

  4. Circuits: Example 4

  5. What is a Circuit?

  6. The simplified circuit for the following circuit is

COMMENTS

  1. 5.6: Circuit Problem Solving

    Circuit problem solving procedure: 1) Calculate the equivalent resistance of the circuit. First combine all the series resistors and then calculate the parallel ones. Use the following equations: series: Req = n ∑ i Ri. parallel: 1 Req = n ∑ i 1 Ri. 2) Use your result of equivalent resistance to find the total current coming out of the battery:

  2. Calculations from circuit diagrams (practice)

    You might need: Calculator A student builds the circuit below. What is the resistance of R ? Ω Show Calculator Stuck? Use a hint. Report a problem Do 4 problems Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more.

  3. Electric Circuits Problem Sets

    Problem Set EC1: Electric Current Relate the amount of charge passing a point on a circuit to the current and the time. Some problems requiring multiple steps (for example, unit conversions or calculating the Coulombs of charge from the number of electrons). Includes 9 problems. Problem Set EC2: Electric Resistance

  4. How to Solve Circuit Problems: 9 Steps (with Pictures)

    1 Develop confidence in your problem solving skills. Confidence is the first and most prominent task to solve the circuit problems. Believe that you can solve the problems, then, ultimately, the formulas, information and ideas will come to your mind. 2

  5. Electric Circuits Problem Sets

    Problem 1: Over the course of an 8 hour day, 3.8x10 4 C of charge pass through a typical computer (presuming it is in use the entire time). Determine the current for such a computer. Audio Guided Solution Show Answer Problem 2: The large window air conditioner in Anita Breeze's room draws 11 amps of current.

  6. Circuit analysis overview (article)

    The general strategy After simplifying a circuit as much as possible, all circuit analysis methods are some version of the following strategy: Create a set of independent equations based on the elements and circuit connections. Solve the system of simultaneous equations for the independent variables (voltages or currents).

  7. How To Solve Any Resistors In Series and Parallel Combination Circuit

    14K 1M views 6 years ago Electronic Circuits This physics video tutorial explains how to solve any resistors in series and parallel combination circuit problems. The first thing you need to...

  8. Engineering Math for Basic Circuits

    Method 2: Linear Equations Calculator. Solving with a linear equations calculator can make your life a lot easier, especially as you get more than two equations and two unknowns. And while solving with a calculator like the TI-89 is possible, it's quite tedious, and there are easier-to-use calculators online, like ours: CircuitBread - Tools ...

  9. 4.5: Simple Resistive Circuits

    The junction where two or more circuit elements meet is called a node. The voltage at a node is called a node voltage. The brute-force approach in solving a simple resistive circuit is to identify each node as a system and apply the conservation of charge to each system. In the circuit theory language, we would say "apply Kirchhoff's current ...

  10. 3.4: General circuits and solution methods

    Solving circuit problems. To determine the behavior of any given linear lumped element circuit a set of simultaneous equations must be solved, where the number of equations must equal or exceed the number of unknowns. ... Figure 3.4.2(a) illustrates a simple circuit with b = 12 branches, p = 6 loops, and n = 7 nodes.

  11. Ohm's Law with Examples

    Use Ohm's Law to Solve Simple Circuits Problems Example 1 Find the current I through a resistor of resistance R = 2 Ω if the voltage across the resistor is 6 V. Solution to Example 1 Substitute R by 2 and V by 6 in Ohm's law V = R I. 6 = 2 I Solve for I I = 6 / 2 = 3 A Example 2

  12. Resistors in Circuits

    solution Follow the rules for series circuits. Resistances in series add up. Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit. IT = VT / RT IT = 125 V/100 Ω IT = 1.25 A Current is constant through resistors in series. IT = I1 = I2 = I3 = 1.25 A

  13. How to Solve a Basic Circuit : 5 Steps

    Step 1: Identify the Voltage (V) of the Circuit and Recognize the Type of Resistance The voltage of a circuit is displayed by the symbol found in Fig. 1. You can simply transcribe this value and keep it until we are solving for current (I) in Step 3.

  14. Solve DC Circuits Problems

    Kirchhoff's and Ohm's laws are used to solve DC circuits problems. There are 3 examples; solve in the order that they are presented; this will make it easier to fully understand them. Problem 1 Find current i, voltages VR1 and VR2 in the ciruit below given that the voltage source e = 20 Volts, the resistances R1 = 100 Ω and R2 = 300 Ω .

  15. Physics Tutorial: Combination Circuits

    Combination Circuits. Previously in Lesson 4, it was mentioned that there are two different ways to connect two or more electrical devices together in a circuit. They can be connected by means of series connections or by means of parallel connections. When all the devices in a circuit are connected by series connections, then the circuit is ...

  16. Ohm's Law Practice Problems

    By manipulating the Ohm's Law formula, you can solve for any one of these three variables if the other two are known. This makes it an invaluable tool for understanding and designing electrical circuits. Practical Applications Ohm's Law isn't just a theoretical concept; it has numerous practical applications in everyday life and various industries.

  17. Electric circuits

    Solution : The resistance of the resistor is R and the electric voltage is V. Answer A. R1, R2 and R3 are connected in series. The equivalent resistor : RA = R1 + R2 + R3 = R + R + R = 3R Electric current (I) : Answer B. R1, R2 and R3 are connected in parallel. The equivalent resistor : 1/R = 1/R1 + 1/R2 + 1/R3 = 1/R + 1/R + 1/R = 3/R RB = R/3

  18. AP Physics 2: Circuits Practice Problems with Answers

    Problem (1): In the circuit below, find the equivalent resistance between points A A and B B. Solution : One of the typical questions in all circuit practice problems is finding the equivalent resistance of a given circuit.

  19. How to Solve Complicated Circuits with Kirchhoff's Voltage Law (KVL)?

    Example 1 A simple example - 1 mesh. Let's start here! This is a simple circuit, so simple that we could solve this using tools we already know. But I want to start simple so that we can focus on the concepts and the steps. So, let's do it. Step 1: Let's take stock of the circuit.

  20. Complex Circuit

    Solving A Complex Circuit For Resistance. When solving for the resistance in a series circuit, the goal is to break down all the different parts making act like one single series circuit. Follow the steps or our example as we do this. Example Problems. 1. Solve for the total resistance of the complex circuit below.

  21. Solve AC Circuits Problems with Solutions

    Problem 1 Find the magnitude and phase shift (polar form) of the current through and voltages across the resistor R R and the capacitor C C in the circuit below given that the voltage source vi = 10cos(ωt) v i = 10 cos ( ω t) V, the resistances R = 100 Ω R = 100 Ω, C = 0.47 μF C = 0.47 μ F, frequency f = 1 f = 1 kHz and ω = 2πf ω = 2 π f .

  22. Resistive Circuit Solver

    Resistive Circuit Solver Simulate Simulate any kind of circuit you have designed and get the results instantly on just a single click. Design Design your own custom circuit using the components given in the palette in the web editor by simple drag and drop

  23. Superposition Theorem with solved problems

    Solved Problem 2. Find the voltage across through 15 Ω resistor using superposition theorem. Let V 1, V 2, V 3, V 4 be the voltages across the 15 Ω resistor when each source (20v, 10v, 10A, 5A sources) are considered separately. Hence the resultant voltage is given by, VT = V1 + V2 + V3 + V4. (i) To find V1.