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AP Physics 1 : Circuit Power

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Example Questions

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Example Question #31 : Circuit Power

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

Possible Answers:

406W

92W

12W

192W

Correct answer:

192W

Explanation:

We calculate the power in a circuit using the following equation

P=I^2R

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

P=(4)^2(12)=192W

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Example Question #189 : Electricity

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

Possible Answers:

192W

4050W

1350W

560W

Correct answer:

1350W

Explanation:

We calculate the power in a circuit using the following equation

P=I^2R

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

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 and 15A current?

Possible Answers:

75W

90W

45W

192W

Correct answer:

90W

Explanation:

We calculate the power in a circuit using the following equation

P=IV

Given I=15A and V=6V

P=(15)(6)=90W

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Example Question #191 : Electricity

What is the power of a circuit with a voltage of and 12A current?

Possible Answers:

88W

108W

192W

92W

Correct answer:

108W

Explanation:

We calculate the power in a circuit using the following equation

P=IV

Given I=12A and V=9V

P=(12)(9)=108W

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Example Question #192 : Electricity

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

Possible Answers:

1.92W

4.67W

6.75W

8.54W

Correct answer:

6.75W

Explanation:

We calculate the power in a circuit using the following equation

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

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

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

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Example Question #193 : Electricity

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

Possible Answers:

22.5W

42.6W

32W

19.2W

Correct answer:

22.5W

Explanation:

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

We calculate the power in a circuit using the following equation

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

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

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

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Example Question #194 : Electricity

Series curcuit

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

Possible Answers:

25W

45W

10W

Correct answer:

25W

Explanation:

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

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

For this problem

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

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

$V=IR_{eq}\Rightarrow$

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

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

P=IV

P=IV=1*25=25W

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Example Question #191 : Electricity

Parallel circuit

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

Possible Answers:

91.32W

192.067W

19.384W

104.167W

Correct answer:

104.167W

Explanation:

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

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

For this problem

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

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

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

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

$V=IR_{eq}\Rightarrow$

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

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

P=IV

to find the power in the circuit.

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

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Example Question #31 : Circuit Power

Combo circuit

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

Possible Answers:

412W

300W

60W

192W

Correct answer:

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:

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

For this problem:

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

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

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

For this problem

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

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

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

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

$V=IR_{eq}\Rightarrow$

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

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

P=IV

to find the power in the circuit.

P=IV=6*25=300W

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AP Physics 1 : Circuit Power

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Circuit Power

Example Question #189 : Electricity

Example Question #261 : Electricity And Waves

Example Question #191 : Electricity

Example Question #192 : Electricity

Example Question #193 : Electricity

Example Question #194 : Electricity

Example Question #191 : Electricity

Example Question #31 : Circuit Power

All AP Physics 1 Resources

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