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AP Physics 2 : Circuit Power

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Example Questions

Example Question #11 : Circuit Power

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

Calculate the power being dissipated by R_4

Possible Answers:

None of these

6.62 watts

3.56 watts

7.11 watts

3.38 watts

Correct answer:

6.62 watts

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.

Combine 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

It is true that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Using ohms law:

I_3R_3=I_4R_4

It is also true that:

$I_3+I_4=I_{Total}$

Using Subsitution:

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

Solving for I_4 :

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

Pluggin in values:

$\frac{3.43}{(1+\frac{1}{3})}=I_4$

2.57 amps=I_4

Using the definition of electric power, where is current and is the resistance of the component in question.

P=I^2R

P=2.57^2*1

P=6.62 watts

Report an Error

Example Question #251 : 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=3 V

Calculate the power being dissipated by R_5

Possible Answers:

None of these

8.47 watts

1.23 watts

9.85 watts

3.21 watts

Correct answer:

8.47 watts

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.

Combine 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

It is true that the voltage drop across parallel resistors must be the same, so:

$V_{R5}=V_{R6}$

Using ohms law

I_5R_5=I_6R_6

It is also true that:

$I_5+I_6=I_{Total}$

Using Subsitution:

$I_5(1+\frac{R_5}{R_6})=I_{Total}$

Solving for I_5 :

$\frac{I_{Total}}{(1+\frac{R_5}{R_6})}=I_5$

Plugging in values:

$\frac{3.43}{(1+\frac{2}{3})}=I_5$

I_5=2.06 amps

Using the definition of electrical power, where is current and is the resistance of the component in question.

P=I^2R

P=2.06^2*2

P=8.47 watts

Report an Error

Example Question #11 : Circuit Power

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

Calculate the power being dissipated by R_6

Possible Answers:

2.27 watts

1.38 watts

3.39 watts

5.65 watts

None of these

Correct answer:

5.65 watts

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.

Combine 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

It is true that the voltage drop across parallel resistors must be the same, so:

$V_{R5}=V_{R6}$

Using ohms law

I_5R_5=I_6R_6

It is also true that:

$I_5+I_6=I_{Total}$

Using Subsitution:

$I_6(1+\frac{R_6}{R_5})=I_{Total}$

Solving for I_6 :

$\frac{I_{Total}}{(1+\frac{R_6}{R_5})}=I_6$

Pluggin in values:

$\frac{3.43}{(1+\frac{3}{5})}=I_6$

1.37 amps=I_6

Using the definition of electrical power, where is current and is the resistance of the component in question:

P=I^2R

P=1.37^2*3

P=5.65 watts

Report an Error

Example Question #14 : Circuit Power

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the power being dissapaited by R_5 ?

Possible Answers:

5.9 watts

0 watts

16.9 watts

3.1 watts

None of these

Correct answer:

16.9 watts

Explanation:

R_1 , R_2 , and R_3 are in parallel, so they are added by using:

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

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

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

$R_{1,2,3}$ , R_4 , and R_5 are in series. So they are added conventionally:

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

Plugging in values:

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

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

First, it is necessary to find the total current of the circuit. Using Ohm's law:

V=IR

Solving for $\I$ :

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

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

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

The total current of the circuit is also the current through I_5

Using the definition of electric power, where is current and is the resistance of the component in question:

P=I^2R

$P=(\frac{195}{67})^2*2\Omega$

16.9 Watts=P

Report an Error

Example Question #11 : Circuit Power

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the power being dissapaited by R_3 ?

Possible Answers:

1.44 watts

None of these

2.11 watts

1.88 watts

2.51 watts

Correct answer:

1.88 watts

Explanation:

R_1 , R_2 , and R_3 are in parallel, so they are added by using:

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

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

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

$R_{1,2,3}$ , R_4 , and R_5 are in series. So they are added conventionally:

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

Plugging in values:

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

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

First, it is necessary to find the total current of the circuit. Using Ohm's law:

V=IR

Solving for $\I$ :

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

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

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

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 since they are in parallel.

V_1=V_2=V_3

Using Ohm's law again and substituting:

V=IR

I_1R_1=I_2R_2=I_3R_3

Using algebraic subsitution:

$I_3+I_3 \frac{R_3}{R_1}+I_3 \frac{R_3}{R_2}=I_{total}$

Solving for I_3 :

$\frac{I_{Total}}{1+\frac{R_3}{R_1}+\frac{R_3}{R_2}}=I_3$

Plugging in values:

$\frac{195/67}{1+\frac{6}{2}+\frac{6}{5}}=I_3$

.560 amps=I_3

Using the definition of electric power, where is current and is reistance.

P=I^2R

$P=(.560amps)^2*6\Omega$

1.88 watts=P

Report an Error

Example Question #92 : Circuit Properties

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the power being dissapaited by R_2 ?

Possible Answers:

None of these

2.26 watts

6.99 watts

3.45 watts

1.19 watts

Correct answer:

2.26 watts

Explanation:

R_1 , R_2 , and R_3 are in parallel, so they are added by using:

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

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

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

$R_{1,2,3}$ , R_4 , and R_5 are in series. So they are added conventionally:

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

Plugging in values:

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

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

First, it is necessary to find the total current of the circuit. Using Ohm's law:

V=IR

Solving for $\I$ :

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

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

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

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 since they are in parallel.

V_1=V_2=V_3

Using Ohm's law again and substituting:

V=IR

I_1R_1=I_2R_2=I_3R_3

Using algebraic subsitution:

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

Solving for I_2

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

Plugging in values:

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

.672 amps=I_2

Using the definition of electrical power, where is the current and is the resistance of the component in question:

P=I^2R

$P=(.672amps)^2*5\Omega$

2.26 watts=P

Report an Error

Example Question #91 : Circuit Properties

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5 \Omega$

$R_3=6 \Omega$

What is the power being dissapaited by R_1 ?

Possible Answers:

None of these

3.14 watts

4.24 watts

6.11 watts

5.64 watts

Correct answer:

5.64 watts

Explanation:

R_1 , R_2 , and R_3 are in parallel, so they are added by using:

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

Plugging in given values:

$\frac{1}{2} + \frac{1}{5} + \frac{1}{6} = \frac{1}{R_{1,2,3}}}$

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

$R_{1,2,3}$ , R_4 , and R_5 are in series. So they are added conventionally:

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

Plugging in values:

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

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

First, it is necessary to find the total current of the circuit. Using Ohm's law:

V=IR

Solving for $\I$ :

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

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

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

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 since they are in parallel.

V_1=V_2=V_3

Using Ohm's law again and substituting:

V=IR

I_1R_1=I_2R_2=I_3R_3

Using algebraic subsitution:

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

Solving for I_1

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

Plugging in values

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

Using the definition of electrical power, where is current and is the resistance of the component in question:

P=I^2R

Plugging in values

P=1.68^2*2

5.64 watts=P

Report an Error

Example Question #91 : 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=3 V

Calculate the power being dissipated by R_2

Possible Answers:

8.44 watts

None of these

2.61 watts

4.19 watts

3.45 watts

Correct answer:

2.61 watts

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.

Combine 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

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solving for I_2 :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

Plugging in values

$\frac{1.14}{(1+\frac{2}{1})}=I_2$

I_2= 1.14 amps

Using the definition of electrical power, where is current and is the resistance of the component in question:

P=I^2R

P=1.14^2*2

P= 2.61 watts

Report an Error

Example Question #92 : 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=3 V

Calculate the power being dissipated by R_1

Possible Answers:

1.5 watts

0.1 watts

5.24 watts

3.75 watts

None of these

Correct answer:

5.24 watts

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

Using the definition of electrical power, where is current and is the resistance of the component in question:

P=I^2R

P=2.29^2*1

5.24 watts=P

Report an Error

Example Question #11 : Circuit Power

A single $9\textup{ V}$ battery is in series with several Ohmic resistors. How will the power output of the circuit change if a second $9\textup{ V}$ battery is added in series?

Possible Answers:

Tripled

Quartered

Quadrupled

Halved

Doubled

Correct answer:

Quadrupled

Explanation:

The power dissipated by the resistor will be

P=V*I

Using Ohm's Law:

V=IR

And changing the equation to be exclusively in terms of voltage and resistance:

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

From this, it can be seen that doubling the voltage will quadruple the power.

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

$P=\frac{4V^2}{R}$

Report an Error

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AP Physics 2 : Circuit Power

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

Example Question #11 : Circuit Power

Example Question #251 : Electricity And Magnetism

Example Question #11 : Circuit Power

Example Question #14 : Circuit Power

Example Question #11 : Circuit Power

Example Question #92 : Circuit Properties

Example Question #91 : Circuit Properties

Example Question #91 : Circuit Properties

Example Question #92 : Circuit Properties

Example Question #11 : Circuit Power

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