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AP Physics 2 : Circuits

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

Example Questions

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:

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 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 #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:

3.98V

2.57V

1.43V

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

Example Question #11 : 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_4 .

Possible Answers:

7.71V

4.20V

1.28V

3.30V

Correct answer:

7.71V

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:

$V_{R3}=V_{R4}$

Use Ohm's law:

I_3R_3=I_4R_4

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

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

Solve 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, we get 2.57A

Use:

V=IR

$V=2.57A*3\Omega$

7.71V=V

Report an Error

Example Question #32 : Circuit Properties

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_5 .

Possible Answers:

3.0V

4.12V

1.4V

3.95V

Correct answer:

4.12V

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, we get 2.06A

Use:

V=IR

$V=2.06A*2\Omega$

4.12V=V

Report an Error

Example Question #851 : Ap Physics 2

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_6 .

Possible Answers:

3.53V

3.0V

4.11V

1.90V

Correct answer:

4.11V

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, we get 1.37A

Use

V=IR

$V=1.37a*3\Omega$

4.11 V=V

Report an Error

Example Question #852 : Ap Physics 2

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=3 V

Determine the voltage drop across R_1

Possible Answers:

2.29 volts

None of these

3.98 volts

1.61 volts

4.48 volts

Correct answer:

2.29 volts

Explanation:

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combining R_1 with R_2 , R_3 with R_4 , R_5 with R_6 .

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

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

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

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

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

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

V=IR

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

Combing all voltage sources for the total voltage.

V_1+V_2+V_3=9 Volts

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

I=3.43 amps

The voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Using ohms law:

I_1R_1=I_2R_2

It is also true that:

$I_1+I_2=I_{Total}$

Using Subsitution

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

Solving for I_1 :

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

$\frac{3.43}{(1+\frac{1}{2})}=I_1$

2.29 amps=I_1

Plugging back into ohms law in order to find the voltage drop.

V=2.29*1

2.29 volts=V .

Report an Error

Example Question #31 : Circuits

identical resistors are placed in parallel. They are placed in a circuit with a battery. If the current through the battery is , determine the current through each resistor.

Possible Answers:

I_3=I_1=2A

I_2=6A

I_1=I_2=I_3=1A

I_1=I_2=I_3=3A

None of these

I_1=I_2=1A

I_3=3A

Correct answer:

I_1=I_2=I_3=1A

Explanation:

Since each resistor is in parallel, the voltage drop across each will be .

Since each resistor is identical, they all have the same resistance.

Using

$I_1+I_2+I_3=I_{total}=3A$

and

I_1=I_2=I_3

It is determined that

I_1=I_2=I_3=1A

Report an Error

Example Question #32 : Circuits

identical resistors are placed in parallel. They are placed in a circuit with a battery. If the current through the battery is , determine the resistance of each resistor.

Possible Answers:

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

None of these

$R_1=R_2=R_3=1\Omega$

$R_1=R_2=R_3=3\Omega$

$R_3=3\Omega$

$R_1=R_2=1\Omega$

Correct answer:

$R_1=R_2=R_3=3\Omega$

Explanation:

Since each resistor is in parallel, the voltage drop across each will be .

Since each resistor is identical, they all have the same resistance.

Using

$I_1+I_2+I_3=I_{total}=3A$

and

I_1=I_2=I_3

It is determined that

I_1=I_2=I_3=1A

Using

V=IR

3=1*R

$R=3\Omega$ for all three resistors

Report an Error

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AP Physics 2 : Circuits

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #4 : Current And Voltage

Example Question #4 : Current And Voltage

Example Question #31 : Circuits

Example Question #201 : Electricity And Magnetism

Example Question #11 : Current And Voltage

Example Question #32 : Circuit Properties

Example Question #851 : Ap Physics 2

Example Question #852 : Ap Physics 2

Example Question #31 : Circuits

Example Question #32 : Circuits

All AP Physics 2 Resources

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