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AP Physics 1 : Circuits

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Example Questions

Example Question #23 : Circuit Power

A capacitor with a capacitance of 10 F $\displaystyle 10 F$ stores 500 J $\displaystyle 500 J$ of energy. What is the voltage across this capacitor?

Possible Answers:

100 V $\displaystyle 100 V$

20 V $\displaystyle 20 V$

10 V $\displaystyle 10 V$

50 V $\displaystyle 50 V$

500 V $\displaystyle 500 V$

Correct answer:

10 V $\displaystyle 10 V$

Explanation:

The correct answer is 10 V $\displaystyle 10 V$.

We can obtain this answer with the following equation regarding capacitors:

$U= \frac{1}{2}CV^{2}$ $\displaystyle U= \frac{1}{2}CV^{2}$

$\displaystyle U$ is the energy stored on the capacitor, $\displaystyle C$ is the capacitance, and $\displaystyle V$ is the voltage.

$(500J)=\frac{1}{2}(10F)(V^{2})$ $\displaystyle (500J)=\frac{1}{2}(10F)(V^{2})$

Solving for $\displaystyle V$ we obtain:

V= 10 V $\displaystyle V= 10 V$

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Example Question #31 : Circuit Power

What is the power of a circuit with a resistance of $12\Omega$ $\displaystyle 12\Omega$ and $\displaystyle 4A$ current?

Possible Answers:

406W $\displaystyle 406W$

12W $\displaystyle 12W$

192W $\displaystyle 192W$

92W $\displaystyle 92W$

Correct answer:

192W $\displaystyle 192W$

Explanation:

We calculate the power in a circuit using the following equation

P=I^2R $\displaystyle P=I^2R$

Given $R=12\Omega$ $\displaystyle R=12\Omega$ and I=4A $\displaystyle I=4A$

$\displaystyle P=(4)^2(12)=192W$

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Example Question #32 : Circuit Power

What is the power of a circuit with a resistance of $6\Omega$ $\displaystyle 6\Omega$ and 15A $\displaystyle 15A$ current?

Possible Answers:

4050W $\displaystyle 4050W$

560W $\displaystyle 560W$

1350W $\displaystyle 1350W$

192W $\displaystyle 192W$

Correct answer:

1350W $\displaystyle 1350W$

Explanation:

We calculate the power in a circuit using the following equation

P=I^2R $\displaystyle P=I^2R$

Given I=15A $\displaystyle I=15A$ and $R=6\Omega$ $\displaystyle R=6\Omega$

$\displaystyle P=(15)^2(6)=1350W$

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Example Question #261 : Electricity And Waves

What is the power of a circuit with a voltage of $\displaystyle 6V$ and 15A $\displaystyle 15A$ current?

Possible Answers:

90W $\displaystyle 90W$

75W $\displaystyle 75W$

45W $\displaystyle 45W$

192W $\displaystyle 192W$

Correct answer:

90W $\displaystyle 90W$

Explanation:

We calculate the power in a circuit using the following equation

P=IV $\displaystyle P=IV$

Given I=15A $\displaystyle I=15A$ and V=6V $\displaystyle V=6V$

$\displaystyle P=(15)(6)=90W$

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Example Question #262 : Electricity And Waves

What is the power of a circuit with a voltage of $\displaystyle 9V$ and 12A $\displaystyle 12A$ current?

Possible Answers:

92W $\displaystyle 92W$

192W $\displaystyle 192W$

108W $\displaystyle 108W$

88W $\displaystyle 88W$

Correct answer:

108W $\displaystyle 108W$

Explanation:

We calculate the power in a circuit using the following equation

P=IV $\displaystyle P=IV$

Given I=12A $\displaystyle I=12A$ and V=9V $\displaystyle V=9V$

$\displaystyle P=(12)(9)=108W$

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Example Question #263 : Electricity And Waves

What is the power of a circuit with a voltage of $\displaystyle 9V$ and resistance of $12\Omega$ $\displaystyle 12\Omega$?

Possible Answers:

1.92W $\displaystyle 1.92W$

4.67W $\displaystyle 4.67W$

8.54W $\displaystyle 8.54W$

6.75W $\displaystyle 6.75W$

Correct answer:

6.75W $\displaystyle 6.75W$

Explanation:

We calculate the power in a circuit using the following equation

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

Given V=9V $\displaystyle V=9V$ and $R=12\Omega$ $\displaystyle R=12\Omega$

$P=\frac{9^2}{12}=6.75W$ $\displaystyle P=\frac{9^2}{12}=6.75W$

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Example Question #264 : Electricity And Waves

What is the power of a circuit with a voltage of 150V $\displaystyle 150V$ and resistance of $1000\Omega$ $\displaystyle 1000\Omega$?

Possible Answers:

22.5W $\displaystyle 22.5W$

42.6W $\displaystyle 42.6W$

32W $\displaystyle 32W$

19.2W $\displaystyle 19.2W$

Correct answer:

22.5W $\displaystyle 22.5W$

Explanation:

What is the power of a circuit with a voltage of 150V $\displaystyle 150V$ and resistance of $1000\Omega$ $\displaystyle 1000\Omega$?

We calculate the power in a circuit using the following equation

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

Given V=150V $\displaystyle V=150V$ and $R=1000\Omega$ $\displaystyle R=1000\Omega$

$P=\frac{(150)^2}{(1000)}=22.5W$ $\displaystyle P=\frac{(150)^2}{(1000)}=22.5W$

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Example Question #191 : Electricity

Series curcuit

What is the power in the given circuit if $R_1=10\Omega$ $\displaystyle R_1=10\Omega$, $R_2=15\Omega$ $\displaystyle R_2=15\Omega$ and V=25V $\displaystyle V=25V$?

Possible Answers:

10W $\displaystyle 10W$

$\displaystyle 5W$

45W $\displaystyle 45W$

25W $\displaystyle 25W$

Correct answer:

25W $\displaystyle 25W$

Explanation:

To find the current we must first find the equivalent resistance. For resistors in series, the equivalent resistance is

$R_{eq}=R_1+R_2+...$ $\displaystyle R_{eq}=R_1+R_2+...$

For this problem

$R_{eq}=R_1+R_2=10+15=25\Omega$ $\displaystyle R_{eq}=R_1+R_2=10+15=25\Omega$

Now we use Ohm's law, $V=IR_{eq}$ $\displaystyle V=IR_{eq}$, to find the current, $\displaystyle I$

$V=IR_{eq}\Rightarrow$ $\displaystyle V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{25}=1A$ $\displaystyle I=\frac{V}{R_{eq}}=\frac{25}{25}=1A$

Now that we have found the current in the resistors we use the following equation to find the power in the circuit:

P=IV $\displaystyle P=IV$

$\displaystyle P=IV=1*25=25W$

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Example Question #266 : Electricity And Waves

Parallel circuit

What is the power in the below circuit if $R_1=10\Omega$ $\displaystyle R_1=10\Omega$, $R_2=15\Omega$ $\displaystyle R_2=15\Omega$ and V=25V $\displaystyle V=25V$?

Possible Answers:

$\displaystyle 192.067W$

19.384W $\displaystyle 19.384W$

$\displaystyle 104.167W$

91.32W $\displaystyle 91.32W$

Correct answer:

$\displaystyle 104.167W$

Explanation:

To find the current we must first find the equivalent resistance. For resistors in parallel, the equivalent resistance is

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

For this problem

$\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$

$\frac{1}{R_{eq}}=\frac{5}{30}$ $\displaystyle \frac{1}{R_{eq}}=\frac{5}{30}$

$R_{eq}=\frac{30}{5}=6\Omega$ $\displaystyle R_{eq}=\frac{30}{5}=6\Omega$

Now we use Ohm's law, $V=IR_{eq}$ $\displaystyle V=IR_{eq}$, to find the current, $\displaystyle I$

$V=IR_{eq}\Rightarrow$ $\displaystyle V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{6}\approx 4.167A$ $\displaystyle I=\frac{V}{R_{eq}}=\frac{25}{6}\approx 4.167A$

Now that we have found the current in the resistors we use the equation

P=IV $\displaystyle P=IV$

to find the power in the circuit.

$P=IV=4.167*25\approx 104.167W$ $\displaystyle P=IV=4.167*25\approx 104.167W$

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Example Question #267 : Electricity And Waves

Combo circuit

What is the power in the given circuit if $R_1=10\Omega$ $\displaystyle R_1=10\Omega$, $R_2=15\Omega$ $\displaystyle R_2=15\Omega$, $R_3=5\Omega$ $\displaystyle R_3=5\Omega$ and V=25V $\displaystyle V=25V$?

Possible Answers:

192W $\displaystyle 192W$

60W $\displaystyle 60W$

300W $\displaystyle 300W$

412W $\displaystyle 412W$

Correct answer:

300W $\displaystyle 300W$

Explanation:

To find the current we must first find the equivalent resistance. This problem is a combination of resistors in series and parallel. First we find the equivalent resistance of resistors in series.

For resistors in series, the equivalent resistance is:

$R_{eq}=R_1+R_2+...$ $\displaystyle R_{eq}=R_1+R_2+...$

For this problem:

$R_{eq1}=R_1+R_2=10+15=25$ $\displaystyle R_{eq1}=R_1+R_2=10+15=25$

The resistors in series can be treated as one $25\Omega$ $\displaystyle 25\Omega$ resistor. Now we address the resistors in parallel. For resistors in parallel, the equivalent resistance is

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

For this problem

$\frac{1}{R_{eq}}=\frac{1}{R_{eq1}}+\frac{1}{R_3}=\frac{1}{25}+\frac{1}{5}=\frac{1+5}{25}$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_{eq1}}+\frac{1}{R_3}=\frac{1}{25}+\frac{1}{5}=\frac{1+5}{25}$

$\frac{1}{R_{eq}}=\frac{6}{25}$ $\displaystyle \frac{1}{R_{eq}}=\frac{6}{25}$

$R_{eq}=\frac{25}{6} \approx 4.167\Omega$ $\displaystyle R_{eq}=\frac{25}{6} \approx 4.167\Omega$

Now we use Ohm's law, $V=IR_{eq}$ $\displaystyle V=IR_{eq}$, to find the current, $\displaystyle I$

$V=IR_{eq}\Rightarrow$ $\displaystyle V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{4.167}=6A$ $\displaystyle I=\frac{V}{R_{eq}}=\frac{25}{4.167}=6A$

Now that we have found the current in the resistors we use the equation

P=IV $\displaystyle P=IV$

to find the power in the circuit.

$\displaystyle P=IV=6*25=300W$

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AP Physics 1 : Circuits

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #23 : Circuit Power

Example Question #31 : Circuit Power

Example Question #32 : Circuit Power

Example Question #261 : Electricity And Waves

Example Question #262 : Electricity And Waves

Example Question #263 : Electricity And Waves

Example Question #264 : Electricity And Waves

Example Question #191 : Electricity

Example Question #266 : Electricity And Waves

Example Question #267 : Electricity And Waves

All AP Physics 1 Resources

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