Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : Current and Voltage

Study concepts, example questions & explanations for AP Physics 2

Create An Account

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Current And Voltage

A battery produces a 1.5A current in a piece of wire. What is the resistance of the wire?

Possible Answers:

$2\Omega$

$6\Omega$

$8\Omega$

There's not enough information to determine the resistance of the wire

$4\Omega$

Correct answer:

$6\Omega$

Explanation:

The relevant equation for this problem is Ohm's law.

V=IR

We're given the voltage and the current, so we need to rearrange to get the resistance.

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

$R=\frac{9}{1.5}$

$R= 6\Omega$

Report an Error

Example Question #2 : Current And Voltage

Consider a circuit connected to a 180V battery that has a resistance of $5 \Omega$ . How many moles of charge pass through this circuit within a 30s period?

$F=9.65*10^{4}\frac{C}{mol}$

Possible Answers:

$1.24*10^{-5}mol$

$1.24*10^{5}mol$

$1.12*10^{2}mol$

$1.12*10^{-2}mol$

Correct answer:

$1.12*10^{-2}mol$

Explanation:

To begin with, we'll need to calculate the current that is flowing through this circuit. We can do this by using Ohm's law.

V=IR

$I=\frac{V}{R}=\frac{180V}{5 \Omega }=36\frac{C}{s}$

Now that we have an expression that gives us the current, which is in units of charge per second, we'll need to calculate the moles of charge per second.

$I=\frac{Q}{t}$

$Q=It=36 \frac{C}{s}(30s)=1080C$

This expression directly above gives the total amount of charge that passes within a given time frame. But to find the moles of charge, we'll need to use Faraday's constant.

$moles\ of\ charge=\frac{Q}{F}$

$moles\:of\:charge=1080C\left ( \frac{1mol}{9.65* 10^{4} C} \right )$

$moles\:of\:charge=1.12*10^{-2}\: mol$

Report an Error

Example Question #2 : Current And Voltage

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the current through R_1 .

Possible Answers:

3.34A

2.29 A

4.49A

9.51A

Correct answer:

2.29 A

Explanation:

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohms law to determine the total current of the circuit

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plug in our values:

$\frac{9V}{2.62 \Omega }=3.43 A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

I_1R_1=I_2R_2

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

$I_1(1+\frac{R_1}{R_2})=I_{Total}$

Solve for I_1 :

$\frac{I_{Total}}{(1+\frac{R_1}{R_2})}=I_1$

$\frac{3.43A}{(1+\frac{1\Omega }{2\Omega })}=2.29A$

Report an Error

Example Question #1 : Current And Voltage

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the current through R_2 .

Possible Answers:

2.28A

39.85A

3.35A

1.14A

Correct answer:

1.14A

Explanation:

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit:

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plug in our values:

I=3.43A

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

I_1R_1=I_2R_2

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solving for I_2 :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

We plug in our values:

$\frac{3.43A}{(1+\frac{2\Omega }{1\Omega })}=I_2$

1.14A=I_2

Report an Error

Example Question #5 : Current And Voltage

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the current through R_3 .

Possible Answers:

.451A

.811A

.768A

.856A

Correct answer:

.856A

Explanation:

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohms law to determine the total current of the circuit.

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

I=3.43A

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Use Ohm's law:

I_3R_3=I_4R_4

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_3(1+\frac{R_3}{R_4})=I_{Total}$

Solve for I_3 :

$\frac{I_{Total}}{(1+\frac{R_3}{R_4})}=I_3$

Plug in our values:

$\frac{3.43 A}{(1+\frac{3 \Omega}{1\Omega})}=I_3$

.856A

Report an Error

Example Question #1 : Current And Voltage

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the current through R_4 .

Possible Answers:

2.57A

3.91A

3.48A

1.48A

Correct answer:

2.57A

Explanation:

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plugin our values:

$\frac{9V}{2.62 \Omega}=3.43A$

I=3.43A

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Use Ohm's law:

I_3R_3=I_4R_4

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_4(1+\frac{R_4}{R_3})=I_{Total}$

Solving for I_4 :

$\frac{I_{Total}}{(1+\frac{R_4}{R_3})}=I_4$

$\frac{3.43amps}{(1+\frac{1\Omega}{3\Omega})}=I_4$

Plug in our values:

I_4=2.57A

Report an Error

Example Question #4 : Current And Voltage

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the current through R_5 .

Possible Answers:

2.06A

1.5A

3.95A

4.12A

Correct answer:

2.06A

Explanation:

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

I=3.43A

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law:

I_5R_5=I_6R_6

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_5(1+\frac{R_5}{R_6})=I_{Total}$

Solve for I_5 :

$\frac{3.43A}{(1+\frac{2\Omega}{3\Omega})}=I_5$

Plug in our values:

I_5=2.06A

Report an Error

Example Question #4 : Current And Voltage

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the current through R_6 .

Possible Answers:

3.9A

1.37A

.58A

.98A

Correct answer:

1.37A

Explanation:

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

I=3.43A

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law:

I_5R_5=I_6R_6

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_6(1+\frac{R_6}{R_5})=I_{Total}$

Solve for I_6 :

$\frac{I_{Total}}{(1+\frac{R_6}{R_5})}=I_6$

$\frac{3.43A}{(1+\frac{3\Omega}{2\Omega})}=I_6$

Plug in our values:

I_6=1.37A .

Report an Error

Example Question #31 : Circuits

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the voltage drop across R_2 .

Possible Answers:

2.01A

1.14A

2.28A

3.98A

Correct answer:

2.28A

Explanation:

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plug in our values:

I=3.43A

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

I_1R_1=I_2R_2

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solve for I_2 :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

We plug in our values.

$\frac{3.43A}{(1+\frac{2\Omega }{1\Omega })}=I_2$

1.14A=I_2

Use Ohm's law:

V=IR

$V=1.14A*2\Omega$

V=2.28V

Report an Error

Example Question #3 : Current And Voltage

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

V_1=V_2=V_3=3V

Determine the voltage drop across R_3 .

Possible Answers:

2.01V

1.43V

2.57V

3.98V

Correct answer:

2.57V

Explanation:

Combine R_1 with R_2 :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine R_3 with R_4 :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine R_5 with R_6 :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

V=IR

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

Combine all of our voltage sources:

V_1+V_2+V_3=9 V

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

I=3.43A

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Using Ohm's law:

I_3R_3=I_4R_4

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_3(1+\frac{R_3}{R_4})=I_{Total}$

Solving for I_3 :

$\frac{I_{Total}}{(1+\frac{R_3}{R_4})}=I_3$

Plug in our values:

$\frac{3.43 A}{(1+\frac{3 \Omega}{1\Omega})}=I_3$

.856A

Use Ohm's law:

V=IR

$V=.856A*3\Omega$

V=2.57V

Report an Error

View Tutors

Patricia
Certified Tutor

SUNY New Paltz, Bachelors, French. SUNY New Paltz, Masters, Secondary Education: French.

View Tutors

Anjali
Certified Tutor

Delhi University, Bachelor of Science, Physics. Kumuan University, Master of Science, Physics.

View Tutors

Pankaj
Certified Tutor

University of Delhi, Electrical Engineer, Electrical Engineering. Columbia University in the City of New York, Master of Arts...

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Miami, English Tutors in Atlanta, SAT Tutors in Houston, Reading Tutors in Dallas Fort Worth, LSAT Tutors in Chicago, Biology Tutors in Phoenix, Calculus Tutors in San Francisco-Bay Area, Math Tutors in Denver, Calculus Tutors in Phoenix, Chemistry Tutors in Philadelphia

Popular Courses & Classes

LSAT Courses & Classes in Atlanta, ISEE Courses & Classes in San Diego, GMAT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in New York City, SSAT Courses & Classes in Miami, ISEE Courses & Classes in Phoenix, ACT Courses & Classes in Phoenix, LSAT Courses & Classes in Phoenix, GMAT Courses & Classes in Boston, ACT Courses & Classes in Seattle

Popular Test Prep

GMAT Test Prep in Boston, ACT Test Prep in Denver, GMAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Chicago, ISEE Test Prep in Boston, MCAT Test Prep in Chicago, ISEE Test Prep in Phoenix, ACT Test Prep in San Diego, GRE Test Prep in Dallas Fort Worth, GRE Test Prep in Washington DC

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 2 : Current and Voltage

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Current And Voltage

Example Question #2 : Current And Voltage

Example Question #2 : Current And Voltage

Example Question #1 : Current And Voltage

Example Question #5 : Current And Voltage

Example Question #1 : Current And Voltage

Example Question #4 : Current And Voltage

Example Question #4 : Current And Voltage

Example Question #31 : Circuits

Example Question #3 : Current And Voltage

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: