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AP Physics 2 : Electricity and Magnetism

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

Example Questions

Example Question #1 : Circuit Power

Combined circuit

V_1 = V_2 = 4.5V

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the power being dissipated by R_1 .

Possible Answers:

$\frac{27}{16}W$

None of these

$\frac{3}{4}W$

$\frac{9}{8}W$

Correct answer:

$\frac{27}{16}W$

Explanation:

First, find the total resistance of the circuit.

R_1 and R_2 are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

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

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

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

V=IR

(V_1+V_2)=IR

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

The current through R_1 and R_2 needs to add up to the total current, since they are in parallel.

$I_1 + I_2 =I_{total}$

$I_1 + I_2 =\frac{9}{8}A$

Also, the voltage drop across them need to be equal, since they are in parallel.

V_1=V_2

I_1R_1=I_2R_2

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

Set up a system of equations.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

$I_1 + I_2 =\frac{9}{8}A$

Solve.

I_2=.5I_1

$1.5I_1=\frac{9}{8}A$

$I_1=\frac{3}{4}A$

The equation for power is as follows:

P=IV=I^2R

$P=\left(\frac{3}{4}A \right )^2\cdot3\Omega$

$P=\frac{27}{16}W$

Report an Error

Example Question #2 : Circuit Power

Three parallel resistors

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What can be said about the power being dissipated by R_4 and R_5 ?

Possible Answers:

R_4 will dissipate more power than R_5

Their values will be equal

None of these

R_5 will dissipate more power than R_4

It is impossible to tell

Correct answer:

Their values will be equal

Explanation:

R_4 and R_5 are in series with each other, therefore, they will have the same current values. They also have the same resistance values. Thus, they will have the same power dissipated, as P=I^2R .

Report an Error

Example Question #4 : 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_3

Possible Answers:

9.15 watts

None of these

2.21 watts

37.5 watts

1.55 watts

Correct answer:

2.21 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

We know 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:

$I_3(1+\frac{R_3}{R_4})=I_{Total}$

Solving for I_3 :

$\frac{I_{Total}}{(1+\frac{R_3}{R_4})}=I_3$

Plugging in values:

$\frac{3.43}{(1+\frac{3}{1})}=I_3$

.858 amps=I_3

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

P=I^2R

P=.858^2*3

P= 2.21 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_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

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AP Physics 2 : Electricity and Magnetism

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Circuit Power

Example Question #2 : Circuit Power

Example Question #4 : Circuit Power

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

All AP Physics 2 Resources

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