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美联社Physics 2 : Circuit Properties

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Example Question #1 :Circuits

Parallel circuit 1 jpeg

$R_1=10\Omega$

$R_2=20\Omega$

V=25V

What is the current through the battery in the above circuit?

Possible Answers:

0.833A

0.266A

3.75A

30A

10A

Correct answer:

3.75A

Explanation:

First, find the total resistance of the circuit. Since the resistors are in parallel, use the following formula:

$\frac{1}{R1}+\frac{1}{R2}=\frac{1}{R_{total}}$

Plug in known values.

$\frac{1}{10\Omega}+\frac{1}{20\Omega}=\frac{1}{R_{total}}$

$R_{total}=6.66\Omega$

Next, use Ohm's law to find current.

V=IR

Plug in known values.

$I=\frac{25V}{6.66\Omega}=3.75A$

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Example Question #2 :Circuits

There are 3 resistors in series. Their resistances are, in order, $2\Omega$ , $4\Omega$ , and $6\Omega$ . The total potential drop is 12V . What is the potential drop across the second resistor?

Possible Answers:

12V

Correct answer:

Explanation:

Use Ohm's law to find the current passing through each resistor. Because they are in series, they have the same amount of current. Once we get the current, we can plug in the resistance for each resistor to find its potential drop.

$I&=\frac{V_{tot}}{R_{tot}}$

$I=\frac{12V}{12\Omega}$

I= 1A

Now, find the potential drop across the $4\Omega$ resistor.

V&=IR

$V=(1A)(4\Omega)$

V= 4V

Therefore, the potential drop across the $4\Omega$ resistor is

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Example Question #3 :Circuits

A 4.5V battery produces a current of 0.90A in a piece of copper wire. What is the resistance of the copper wire?

Possible Answers:

$3\Omega$

There's not enough information to find the resistance

$4.5\Omega$

$5\Omega$

$6.4\Omega$

Correct answer:

$5\Omega$

Explanation:

Even though there is no resistor, Ohm's law still applies. Use it to find the resistance of the wire.

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

$R= \frac{4.5V}{0.90A}$

$R= 5\Omega$

The resistance of the copper wire is $5\Omega$

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Example Question #4 :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:

$\frac{9}{8}V$

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 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.

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

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

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

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Example Question #5 :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_2 .

Possible Answers:

$\frac{3}{8}A$

$\frac{9}{8}A$

$\frac{3}{4}A$

None of these

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

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

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

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Example Question #1 :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_3 .

Possible Answers:

$\frac{3}{4}A$

$\frac{9}{4}A$

$\frac{9}{8}A$

None of these

$\frac{3}{8}A$

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.

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Example Question #7 :Circuits

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 $R_{1+2}$

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

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

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

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Example Question #8 :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_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 $R_{1+2}$

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

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

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

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Example Question #9 :Circuits

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:

$17\Omega$

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

$4\Omega$

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

None of these

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:

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

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

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

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Example Question #10 :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:

.398 amps

.650 amps

.560 amps

None of these

.375 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

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美联社Physics 2 : Circuit Properties

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 :Circuits

Example Question #2 :Circuits

Example Question #3 :Circuits

Example Question #4 :Circuits

Example Question #5 :Circuits

Example Question #1 :Circuit Properties

Example Question #7 :Circuits

Example Question #8 :Circuits

Example Question #9 :Circuits

Example Question #10 :Circuits

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