We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics C Electricity : Electric Circuits

Study concepts, example questions & explanations for AP Physics C Electricity

Create An Account

All AP Physics C Electricity Resources

1 Diagnostic Test 46 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Understanding Resistors

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

Why was the potential of the battery measured differently when the wire was connected?

Possible Answers:

The wire has a slight internal resistance and caused a potential drop

Once the current is distributed across the wire, there is less charge per unit length of the circuit and so the voltage is lower

The wire connected to the battery must be connected to something else where this potential is being lost

The wire is causing electrons to be lost to the air, which lowers the potential measured

The difference in potential measured is small enough to be disregarded

Correct answer:

The wire has a slight internal resistance and caused a potential drop

Explanation:

All wires have at least some internal resistance. The most likely explanation for this is that the wire is displaying slight resistance, and therefore caused the measured potential to be less than it was before.

Report an Error

Example Question #1 : Calculating Resistance

Three resistors and a battery form the following circuit.

Circuit_1

Calculate the equivalent resistance if the resistors have the following values.

$R_1=40\Omega$

$R_2=25\Omega$

$R_3=38\Omega$

Possible Answers:

$R_{123}=2.5\Omega$

$R_{123}=92.3\Omega$

$R_{123}=24.5\Omega$

$R_{123}=134.5\Omega$

$R_{123}=45.2\Omega$

Correct answer:

$R_{123}=24.5\Omega$

Explanation:

First, calculate the equivalent resistance of R_2 and R_3 . Since these two resistors are arranged in series, we just take the sum of their values.

$R_{23}=R_{2}+R_{3}$

$R_{23}=25\Omega+38\Omega$

$R_{23}=63\Omega$

With resistors 2 and 3 combined together in a single value, the following circuit is formed.

Circuit_2

Notice that and $R_{23}$ are arranged in parallel. To calculate the equivalent resistance of this parallel pair, we use the following equation.

$\frac{1}{R_{123}}=\frac{1}{R_1}+\frac{1}{R_{23}}$

Plug in the values, and solve for $R_{123}$ .

$\frac{1}{R_{123}}=\frac{1}{40\Omega}+\frac{1}{63\Omega}$

$R_{123}=24.5\Omega$

This is just like the circuit shown below.

Circuit_3

Report an Error

Example Question #12 : Electric Circuits

A lamp has a 60W bulb. If the house wiring provides 110V to light up that bulb, how much current does the bulb draw?

Possible Answers:

$0.03\ \text{A}$

$5.83\ \text{A}$

$0.72\ \text{A}$

$0.55\ \text{A}$

$1.37\ \text{A}$

Correct answer:

$0.55\ \text{A}$

Explanation:

The formula for power is , P=IV and we are given the following values.

$P=60\ \text{W}$

$V=110\ \text{V}$

Solve for the current, .

$I=\frac{P}{V}=\frac{60\ \text{W}}{110\ \text{V}}=0.55\ \text{A}$

Report an Error

Example Question #11 : Electric Circuits

A particle accelerator with a radius of 500 meters can have up to $10^{15}$ protons circulating within it at once.

$q_{proton}=1.60*10^{-19}C$

How fast must the protons in the accelerator move in order to produce a current of 1A?

Possible Answers:

$3.00*10^8\frac{m}{s}$

$9.21*10^{7}\frac{m}{s}$

$1.96*10^7\frac{m}{s}$

$1.95*10^6\frac{m}{s}$

$2.04*10^8\frac{m}{s}$

Correct answer:

$1.96*10^7\frac{m}{s}$

Explanation:

The current produced is the total charge that circulates the particle accelerator per unit time.

$I=\frac{dQ}{dt}$

We calculate this by the equation:

$I = \frac{nqv}{2\pi R}$

is the number of protons, is the charge per proton, is the velocity of each proton, and is the radius of the particle accelerator.

Using the given current, we then solve for the velocity:

$1A = \frac{(10^{15})(1.60*10^{-19}C)v}{2\pi (500m)}$

$v=\frac{(1A)2\pi(500m)}{(10^{15})(1.60*10^{-19}C)}= 1.96*10^7\frac{m}{s}$

Report an Error

Example Question #1 : Circuit Relationships

Three resistors R₁, R₂, and R₃ and the capacitor, C, are connected to an ideal battery V to complete the circuit as shown.

Ps0_capacitorcircuit

After the circuit has been connected for a very long time, the currents in each branch of the circuit are measured to be some values I_1 , , and , and the charge on the capacitor some value .

Which of the following equations is a correct expression regarding the voltage of the circuit after a long time?

Possible Answers:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{2}I_{2}$

$0=-V-R_{2}I_{2} -R_{1}I_{1}$

$0=-\frac{Q}{}C-R_{3}I_{3} -R_{2}I_{2}$

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

Correct answer:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

Explanation:

Identifying the sum of the voltage drops and rises (Kirchoff's Loop Law) around the three possible loops of this circuit is the key to answering this question correctly. The following signs can be assigned to each of the circuit elements based on the direction of the currents given.

Ps0_capacitorcircuit_2

Use Ohm's law and the equation for capacitance to derive terms for the voltage across each element of the circuit.

V=IR

$C=\frac{Q}{V}\rightarrow V=\frac{Q}{C}$

There are three possible paths through the circuit, resulting in three correct equations that could be derived:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

$0=-\frac{Q}{}C-R_{3}I_{3} +R_{2}I_{2}$

$0=+V-R_{2}I_{2} -R_{1}I_{1}$

Only one of the given answer options matches up correctly to these.

Report an Error

Example Question #1 : Circuit Relationships

A simple circuit consists of a $2\ \text{M}\Omega$ resistor that's connected to a battery. How much power is dissipated by the resistor?

Possible Answers:

$8*10^{-4}W$

$2*10^{-2}W$

$5.5*10^{-6}W$

$7.2*10^{-2}W$

$4*10^{-5}W$

Correct answer:

$4*10^{-5}W$

Explanation:

One of the formulas for power is $P=\frac{V^2}{R}$ . We are given the values of the voltage and resistance.

$V=9\ \text{V}$

$R=2\ \text{M}\Omega=2*10^6\ \Omega$

Using these values, we can solve for power.

$P=\frac{(9\ \text{V})^2}{2*10^6\ \Omega}$

$P=4*10^{-5}\ \text{W}$

Report an Error

Example Question #2 : Circuit Relationships

A cicuit with a voltage source of 40V has a $15\Omega$ resistor and a $5\Omega$ resistor in series, followed by two $5\Omega$ resistors in parallel. After the parallel branches rejoin, there is a $10\Omega$ resistor before the circuit is closed at the voltage source.

What is the current supplied by the source?

Possible Answers:

0.82A

1.5A

1.2A

1.0A

Correct answer:

1.2A

Explanation:

The current supplied by the source can be calculated using a derivation of Ohm's law:

$I=\frac{V}{R}$

Start by finding the equivalent resistance of the circuit.

Sum the first two resistors in series:

$R_{12}=15\Omega+5\Omega=20\Omega$

Calculate the equivalent resistance of the two resistors in parallel:

$\frac{1}{R_{34}}=\frac{1}{5\Omega}+\frac{1}{5\Omega}=\frac{2}{5\Omega}$

$R_{34}=\frac{5}{2}\Omega=2.5\Omega$

Now, all of the reistances can be viewed as being in series.

$R_{eq}=R_{12}+R_{34}+R_{5}$

$R_{eq}=20\Omega+2.5\Omega+10\Omega=32.5\Omega$

Returning to our current calulation, we can find our final answer:

$I=\frac{V}{R}=\frac{40V}{32.5\Omega}=1.2A$

Report an Error

Example Question #1 : Using Ohm's Law

Ps0_resistorcircuit

$R_1=15\Omega,\ R_2=20\Omega,\ R_3=20\Omega$

V=50V

What is the current delivered to the curcuit by the battery?

Possible Answers:

4.6A

1.1A

0.91A

2.0A

Correct answer:

2.0A

Explanation:

Using Ohm's law to solve for the value of the current from the battery requires calculation of the equivalent resistance of the circuit.

$V=IR_{eq}$

The resistors R₂ and R₃ are in parallel with one another. Once combined, their requivalent resistor (R₂₃) is in series with R₁.

Ps0_resistorcircuit_2

The equivalent resistance is thus given by:

$R_{eq}=\left ( \frac{1}{R_{2}} + \frac{1}{R_{3}} \right )^{-1} + R_{1}$

$R_{eq}=\left ( \frac{1}{20\Omega} + \frac{1}{20\Omega} \right )^{-1} + 15\Omega$

$R_{eq}=10\Omega + 15\Omega=25\Omega$

Use this value and the given voltage to solve for the current in the circuit:

$I = \frac{V}{R_{eq}}$

$I=\frac{50V}{25\Omega}=2.0A$

Report an Error

Example Question #1 : Circuit Relationships

A straight copper wire has a fixed voltage applied across its length. Which of the following changes would increase the power dissipated by this wire?

Possible Answers:

Increasing the temperature of the wire

Changing the material of the wire to a metal of greater resistivity

Decreasing the cross-sectional area of the wire

None of these options result in increased power dissipation

Decreasing the length of the wire

Correct answer:

Decreasing the length of the wire

Explanation:

Relevant equations:

V = IR

P = IV = I^2 R

$R = \frac{\rho L}{A}$

Current and resistance are inversely proportional to one another, assuming voltage is fixed. Since P = I^2 R , changes in current effect the power more than changes in resistance do. Thus, we need current to increase, meaning that resistance must decrease.

To decrease resistance, we could:

1. Change the material of the wire to one of lesser resistivity

2. Decrease the length of the wire

3. Increase the cross-sectional area of the wire

4. Decrease the temperature of the wire (very slight effect on resistance)

Report an Error

Example Question #1 : Circuit Relationships

A simple circuit contains two $12\Omega$ resistors in parallel, connected to a 20V source. What power is being provided by the source to the circuit?

Possible Answers:

56W

42W

20W

67W

Correct answer:

67W

Explanation:

The power supplied to the circuit can be calculated using the equation:

$P=IV=\frac{V^2}{R}$

To use this equation, we need to find the equivalent resistance of the circuit. Use the equation for equivalent resistance in parallel:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}=\frac{1}{12\Omega}+\frac{1}{12\Omega}=\frac{1}{6\Omega}$

$R_{eq}=6\Omega$

Now that we have the resistance and the voltage, we can solve for the power.

$P=\frac{V^2}{R}=\frac{(20V)^2}{6\Omega}\approx 67W$

Report an Error

View Tutors

Dilip
Certified Tutor

University of Calcutta, Bachelor of Science, Physics. Indiana Institute of Technology, Master of Science, Physics.

View Tutors

Tyler
Certified Tutor

University of Illinois at Urbana-Champaign, Bachelor of Science, Physics.

View Tutors

Robert
Certified Tutor

Rensselaer Polytechnic Institute, Bachelors, Electrical Engineering. Brooklyn Polytechnic, Masters, Computer Science.

All AP Physics C Electricity Resources

1 Diagnostic Test 46 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Chemistry Tutors in Dallas Fort Worth, LSAT Tutors in New York City, LSAT Tutors in Seattle, Spanish Tutors in Atlanta, Statistics Tutors in Seattle, ACT Tutors in Atlanta, Math Tutors in Dallas Fort Worth, Calculus Tutors in Houston, LSAT Tutors in Houston, French Tutors in Washington DC

Popular Courses & Classes

SAT Courses & Classes in San Diego, ACT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in San Diego, MCAT Courses & Classes in Miami, GRE Courses & Classes in Washington DC, ISEE Courses & Classes in Los Angeles, ACT Courses & Classes in Miami, GMAT Courses & Classes in Philadelphia, ACT Courses & Classes in Atlanta, ISEE Courses & Classes in Washington DC

Popular Test Prep

MCAT Test Prep in Washington DC, ACT Test Prep in Atlanta, GMAT Test Prep in San Diego, GMAT Test Prep in Philadelphia, ACT Test Prep in San Francisco-Bay Area, SAT Test Prep in San Diego, SAT Test Prep in New York City, GMAT Test Prep in Seattle, LSAT Test Prep in Los Angeles, LSAT Test Prep in Atlanta

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics C Electricity : Electric Circuits

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #1 : Understanding Resistors

Example Question #1 : Calculating Resistance

Example Question #12 : Electric Circuits

Example Question #11 : Electric Circuits

Example Question #1 : Circuit Relationships

Example Question #1 : Circuit Relationships

Example Question #2 : Circuit Relationships

Example Question #1 : Using Ohm's Law

Example Question #1 : Circuit Relationships

Example Question #1 : Circuit Relationships

All AP Physics C Electricity Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: