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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 #11 : Ohm's Law

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the current flowing through R_2 ?

Possible Answers:

1.9 amps

None of these

.672 amps

Impossible to determine

3.81 amps

Correct answer:

.672 amps

Explanation:

R_1 , R_2 , and R_3 are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , R_4 , and R_5 are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, we need to find the total current of the circuit, we simply use:

V=IR

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

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because R_1 , R_2 and R_3 are in parallel,

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

Also, the voltage drop must be the same across all three

V_1=V_2=V_3

Using

V=IR

I_1R_1=I_2R_2=I_3R_3

Using algebraic subsitution we get:

$I_2+I_2 \frac{R_2}{R_1}+I_2 \frac{R_2}{R_3}=I_{total}$

Solving for I_2

$\frac{I_{total}}{1+\frac{R_2}{R_1}+\frac{R_2}{R_3}}=I_2$

$\frac{195/67}{1+\frac{5}{2}+\frac{5}{6}}=I_2$

I_2=.672 amps

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Example Question #11 : Circuits

Three parallel resistors

R_1=R_4=R_5=2 ohms

R_2=5 ohms

R_3=6 ohms

What is the current flowing through R_5 ?

Possible Answers:

$\frac{195}{67} amps$

None of these

$\frac{67}{195} amps$

15 amps

3 amps

Correct answer:

$\frac{195}{67} amps$

Explanation:

R_1 , R_2 , and R_3 are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , R_4 , and R_5 are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

We will then find the total current of the circuit. This will also be the current of R_5 because this resistor is not in parallel with any others.

V=IR

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

$I=\frac{15V}{(67/13)\Omega}$

$I=\frac{195}{67} amps$

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Example Question #13 : Ohm's Law

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R2=5\Omega$

$R_3=6 \Omega$

What is the current flowing through R_1 ?

Possible Answers:

1.68 amps

2.57 amps

Impossible to determine

3.18 amps

None of these

Correct answer:

1.68 amps

Explanation:

R_1 , R_2 , and R_3 are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , R_4 , and R_5 are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, we need to find the total current of the circuit, we simply use:

V=IR

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

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because R_1 , R_2 and R_3 are in parallel,

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

Also, the voltage drop must be the same across all three

V_1=V_2=V_3

Using

V=IR

I_1R_1=I_2R_2=I_3R_3

Using algebraic subsitution we get:

$I_1+I_1 \frac{R_1}{R_2}+I_1 \frac{R_1}{R_3}=I_{total}$

Solving for I_1

$\frac{I_{Total}}{1+\frac{R_1}{R_2}+\frac{R_1}{R_3}}=I_1$

$\frac{195/67}{1+\frac{2}{5}+\frac{2}{6}}=I_1$

1.68 amps=I_1

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Example Question #11 : Circuits

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the voltage drop across R_5 ?

Possible Answers:

5.82 volts

1.12 volts

6.91 volts

2.34 volts

None of these

Correct answer:

5.82 volts

Explanation:

R_1 , R_2 , and R_3 are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , R_4 , and R_5 are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

We will then determine the total current of the circuit.

V=IR

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

$I=\frac{15V}{(67/13)\Omega}$

$I=\frac{195}{67} amps$

Once again, using

V=IR

Where is the resistance of the resistor in question, we get

$V=\frac{195}{67}\cdot 2\Omega$

5.82 volts=V

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Example Question #12 : Circuits

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the voltage drop across R_2 ?

Possible Answers:

1.98 volts

3.98 volts

None of these

3.36 volts

2.27 volts

Correct answer:

3.36 volts

Explanation:

R_1 , R_2 , and R_3 are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , R_4 , and R_5 are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, we need to find the total current of the circuit, we simply use:

V=IR

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

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because R_1 , R_2 and R_3 are in parallel,

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

Also, the voltage drop must be the same across all three

V_1=V_2=V_3

Using

V=IR

I_1R_1=I_2R_2=I_3R_3

Using algebraic subsitution we get:

$I_2+I_2 \frac{R_2}{R_1}+I_2 \frac{R_2}{R_3}=I_{total}$

Solving for I_2

$\frac{I_{total}}{1+\frac{R_2}{R_1}+\frac{R_2}{R_3}}=I_2$

$\frac{195/67}{1+\frac{5}{2}+\frac{5}{6}}=I_2$

I_2=.672 amps

Using

V=IR

$V=.672A\cdot5\Omega$

3.36 volts=V

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Example Question #13 : Circuits

Parallel circuit 1 jpeg

The current through R_1 is 10A , R_1 has a resistance of $1.5\Omega$ . Determine the voltage of the battery.

Possible Answers:

12V

28V

15V

Correct answer:

15V

Explanation:

Due to the "loop rule", the voltage increase in the battery will need to be equal to the voltage drop in the resistor.

$V_1=V_{R1}$

Use Ohm's law:

V=IR

Combine equations:

$V_1=I_{R1}R_{R1}$

Plug in values:

V_1=10*1.5

V_1=15V

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Example Question #11 : Circuit Properties

Parallel circuit 1 jpeg

If the battery has a voltage of , and R_2 has a resistance of $2\Omega$ , determine the current through R_2 .

Possible Answers:

10A

Correct answer:

Explanation:

Due to the "loop rule", the voltage increase in the battery will need to be equal to the voltage drop in the resistor.

$V_1=V_{R2}$

Use Ohm's law:

V=IR

Combine equations:

$V_1=I_{R2}R_{R2}$

Plug in values:

$6=2*I_{R2}$

$I_{R2}=3A$

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Example Question #11 : Circuits

A $9\textup{ V}$ battery is placed in series with five $9\ \Omega$ resistors. Find the total resistance.

Possible Answers:

$1.8\ \Omega$

$9\ \Omega$

$15\ \Omega$

$5\ \Omega$

$45\ \Omega$

Correct answer:

$45\ \Omega$

Explanation:

In series, resistance adds conventionally.

$5*9\Omega=45\Omega$

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Example Question #16 : Ohm's Law

A $9\textup{ V}$ battery is placed in series with five $9\ \Omega$ resistors. Find the voltage drop across each resistor.

Possible Answers:

$6\textup{ V}$

$10\textup{ V}$

$3.6\textup{ V}$

$1.8\textup{ V}$

$9\textup{ V}$

Correct answer:

$1.8\textup{ V}$

Explanation:

In series, resistance adds conventionally.

$5*9\Omega=45\Omega$

Using Ohm's law for total current:

V=IR

9=I*45

I=.2A

Now use Ohm's law for an individual resistor:

V=IR

V=.2*9

V=1.8V

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Example Question #17 : Circuits

A $9\textup{ V}$ battery is placed in series with five $9\ \Omega$ resistors. Find the current.

Possible Answers:

$9\textup{ A}$

$1\textup{ A}$

$45\textup{ A}$

$3\textup{ A}$

$.2\textup{ A}$

Correct answer:

$.2\textup{ A}$

Explanation:

In series, resistance adds conventionally.

$5*9\Omega=45\Omega$

Using V=IR

9=I*45

I=.2A

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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 #11 : Ohm's Law

Example Question #11 : Circuits

Example Question #13 : Ohm's Law

Example Question #11 : Circuits

Example Question #12 : Circuits

Example Question #13 : Circuits

Example Question #11 : Circuit Properties

Example Question #11 : Circuits

Example Question #16 : Ohm's Law

Example Question #17 : Circuits

All AP Physics 2 Resources

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