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AP Physics 2 : Current and Voltage

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

Example Questions

Example Question #21 : Circuits

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

Possible Answers:

$8\Omega$

$4\Omega$

$6\Omega$

$2\Omega$

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

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 #21 : Circuits

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:

$1.24*10^{5}mol$

$1.12*10^{2}mol$

$1.12*10^{-2}mol$

$1.24*10^{-5}mol$

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

4.49A

2.29 A

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 #24 : 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 current through R_2 .

Possible Answers:

3.35A

1.14A

2.28A

39.85A

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

3.98A

1.14A

2.28A

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 #201 : Electricity And Magnetism

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:

2.57V

1.43V

3.98V

2.01V

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

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AP Physics 2 : Current and Voltage

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #21 : Circuits

Example Question #21 : Circuits

Example Question #2 : Current And Voltage

Example Question #24 : Circuits

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 #201 : Electricity And Magnetism

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