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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 : Circuits

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 voltage drop across R_1 .

Possible Answers:

$\frac{9}{8}V$

$\frac{9}{4}V$

None of these

4.5V

Correct answer:

$\frac{9}{4}V$

Explanation:

First, find the total resistance of the circuit.

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

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

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

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

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 through the system.

V=IR

(V_1+V_2)=IR

$9V=I\cdot 8\Omega$

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

Since R_1 and R_2 are in parallel, they will have the same voltage drop accross them.

$V_{1+2}=\frac{9}{8}A \cdot R_{1+2}$

$V_{1+2}=\frac{9}{8}A\cdot 2\Omega$

$V_{1+2}=\frac{9}{4}V$

Report an Error

Example Question #2 : Circuit Properties

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 current through R_2 .

Possible Answers:

$\frac{3}{8}A$

$\frac{3}{4}A$

None of these

$\frac{9}{8}A$

Correct answer:

$\frac{3}{8}A$

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.

2I_2=I_1

$3I_2=\frac{9}{8}A$

$I_2=\frac{3}{8}A$

Report an Error

Example Question #2 : Circuits

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 current through R_3 .

Possible Answers:

$\frac{9}{4}A$

$\frac{9}{8}A$

$\frac{3}{8}A$

None of these

$\frac{3}{4}A$

Correct answer:

$\frac{9}{8}A$

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

$9V=I\cdot 8\Omega$

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

In series, all resistors will have the same current.

Thus, the current through R_3 is the same as through the rest of the circuit.

Report an Error

Example Question #2 : Ohm's Law

Combined circuit

V_1 = V_2 = 4.5V

$R_1=R_3=R4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the voltage drop across R_3 .

Possible Answers:

None of these

4.5V

$\frac{27}{8}V$

Correct answer:

$\frac{27}{8}V$

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

$9V=I\cdot 8\Omega$

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

R_1 and R_2 will have the same voltage drop across them, as they are in parallel, and are equivalent to the combined resistor $R_{1+2}$

$V=\frac{9}{8}A\cdot R_{3}$

$V=\frac{9}{8}A\cdot 3\Omega$

$V=\frac{27}{8}V$

Report an Error

Example Question #171 : Electricity And Magnetism

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 voltage drop across R_2 .

Possible Answers:

$\frac{9}{8}V$

4.5V

$\frac{9}{4}V$

None of these

Correct answer:

$\frac{9}{4}V$

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

$9V=I\cdot 8\Omega$

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

R_1 and R_2 will have the same voltage drop across them, as they are in parallel, and are equivalent to the combined resistor $R_{1+2}$

$V=\frac{9}{8}A\cdot R_{1+2}$

$V=\frac{9}{8}A\cdot 2\Omega$

$V=\frac{9}{4}V$

Report an Error

Example Question #1 : Ohm's Law

Three parallel resistors

R_1=R_4=R_5=2 ohms

R_2=5 ohms

R_3=6 ohms

What is the total resistance of the circuit?

Possible Answers:

$\frac{67}{13}\Omega$

$4\Omega$

$\frac{13}{15}\Omega$

None of these

$17\Omega$

Correct answer:

$\frac{67}{13}\Omega$

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:

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

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

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

Report an Error

Example Question #3 : Circuits

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_3 ?

Possible Answers:

.650 amps

.398 amps

None of these

.375 amps

.560 amps

Correct answer:

.560 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:

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

Solving for I_3

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

$\frac{I_{Total}}{1+\frac{6}{2}+\frac{6}{5}}=I_3$

.560 amps=I_3

Report an Error

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

Report an Error

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

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

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

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

Report an Error

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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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 : Circuits

Example Question #2 : Circuit Properties

Example Question #2 : Circuits

Example Question #2 : Ohm's Law

Example Question #171 : Electricity And Magnetism

Example Question #1 : Ohm's Law

Example Question #3 : Circuits

Example Question #11 : Ohm's Law

Example Question #11 : Circuits

Example Question #13 : Ohm's Law

All AP Physics 2 Resources

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