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AP Physics 1 : Electricity

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Example Questions

Example Question #3 : Understanding Circuit Diagrams

Circuit diagram

I=30A

$R_1=2\Omega$

$R_2=3\Omega$

In the circuit above, what is the total voltage?

Possible Answers:

75V

25V

150 V

36V

30V

Correct answer:

36V

Explanation:

To find the voltage, first find the combined resistances of the resistors in parallel:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2\Omega }+\frac{1}{3\Omega }$

$\frac{1}{R_T}=\frac{5}{6\Omega }$

$R_T=\frac{6}{5}\Omega$

Use Ohm's law to find the voltage.

V=IR

$V=30A\left(\frac{6}{5}\Omega\right)=36V$

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Example Question #3 : Understanding Circuit Diagrams

Circuit diagram

V=5V

I=1.25A

$R_1=6\Omega$

In the circuit above, what is the resistance of R_2 ?

Possible Answers:

$2\Omega$

$-2\Omega$

$12\Omega$

$2.4\Omega$

$10\Omega$

Correct answer:

$12\Omega$

Explanation:

Find the total resistance of the circuit, which can be determined using Ohm's law.

V=IR

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

$R_T=\frac{5}{1.25}\Omega=4\Omega$

Now, the resistance of the second resistor can be found. Since the two resistors are in parallel, they're related to the total resistance as follows:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}$

Rearrange and solve for R_2

$\frac{1}{R_2}=\frac{1}{R_T}-\frac{1}{R_1}$

$\frac{1}{R_2}=\frac{1}{4\Omega }-\frac{1}{6\Omega }=\frac{6-4}{24\Omega }=\frac{1}{12\Omega }$

$R_2=12\Omega$

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Example Question #5 : Understanding Circuit Diagrams

Circuit diagram2

$R_1=3\Omega$

$R_2=6\Omega$

$R_3=2\Omega$

$R_4=16\Omega$

In the circuit above, what is the total resistance?

Possible Answers:

$\frac{9}{32}\Omega$

$\frac{34}{5}\Omega$

$27\Omega$

$\frac{27}{2}\Omega$

$\frac{34}{9}\Omega$

Correct answer:

$\frac{34}{9}\Omega$

Explanation:

Find the combined resistances for the resistors in parallel:

$\frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_A}=\frac{1}{3\Omega}+\frac{1}{6\Omega}=\frac{9}{18\Omega}$

$R_A=2\Omega$

$\frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\frac{1}{R_B}=\frac{1}{2\Omega}+\frac{1}{16\Omega}=\frac{18}{32\Omega}$

$R_B=\frac{16}{9}\Omega$

Combine these two combined series resistors to find the total resistance:

$R_T=R_A+R_B=2\Omega+\frac{16}{9}\Omega=\frac{34}{9}\Omega$

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Example Question #4 : Understanding Circuit Diagrams

Circuit diagram2

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

V=10V

In the circuit above, what is the voltage drop across R_3 ?

Possible Answers:

12.6 V

5.4V

1.3V

2.8 V

7.2V

Correct answer:

7.2V

Explanation:

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

$\frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_A}=\frac{1}{1\Omega}+\frac{1}{2\Omega}=\frac{3}{2\Omega}$

$R_A=\frac{2}{3}\Omega$

$\frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\frac{1}{R_B}=\frac{1}{3\Omega}+\frac{1}{4\Omega}=\frac{7}{12\Omega}$

$R_B=\frac{12}{7}\Omega$

Therefore, the total resistance is:

$R_T=R_A+R_B=\frac{2}{3}\Omega+\frac{12}{7}\Omega=\frac{50}{21}\Omega$

Now, note that since R_3 and R_4 are in parallel, the voltage drop across them is the same. Use Ohm's law to relate current in terms of voltage and resistance.

$I=\frac{V_T}{R_T}$

Substitute into Ohm's law for the resistance across R_3 :

$V_{3,4}=\frac{V}{R_T}R_B$

$V_{3,4}=10V\left(\frac{21}{50\Omega}\right)\left(\frac{12\Omega}{7}\right)$

$V_{3,4}=7.2V$

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Example Question #1311 : Ap Physics 1

Circuit diagram2

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

V=10V

In the circuit above, what is the current passing through R_3 ?

Possible Answers:

4.2A

1.4A

2.4A

1.8A

2.8A

Correct answer:

2.4A

Explanation:

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

$\frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_A}=\frac{1}{1\Omega}+\frac{1}{2\Omega}=\frac{3}{2\Omega}$

$R_A=\frac{2}{3}\Omega$

$\frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\frac{1}{R_B}=\frac{1}{3\Omega}+\frac{1}{4\Omega}=\frac{7}{12\Omega}$

$R_B=\frac{12}{7}\Omega$

Therefore, the total resistance is:

$R_T=R_A+R_B=\frac{2}{3}\Omega+\frac{12}{7}\Omega=\frac{50}{21}\Omega$

From Ohm's law, we know that $\frac{V}{R_T}$ is the current traveling through the circuit.

$I=10V\left(\frac{21}{50\Omega}\right)=4.2A$

This current will be divided between and , with more current taking the path of lower resistance.

Total voltage drop across $R_{3,4}$ :

$V_{3,4}=I_TR_{3,4}$

$V_{3,4}=4.2A\cdot\frac{12}{7}\Omega$

$V_{3,4}=7.2V$

V_3=V_4=7.2V

The current through R_3 is given by:

$I_3=\frac{V_3}{R_3}$

$I_3=\frac{7.2V}{3\Omega}=2.4A$

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Example Question #11 : Understanding Circuit Diagrams

Circuitdiagram3

$R_1=3\Omega$

$R_2=6\Omega$

$R_3=2\Omega$

$R_4=10\Omega$

In the circuit above, what is the total resistance?

Possible Answers:

$21\Omega$

$\frac{21}{108}\Omega$

$\frac{80}{21}\Omega$

$\frac{21}{80}\Omega$

$\frac{108}{21}\Omega$

Correct answer:

$\frac{80}{21}\Omega$

Explanation:

Begin by combining the resistors that are immediately in series:

$R_A=R_1+R_3=3\Omega+2\Omega=5\Omega$

$R_B=R_2+R_4=6\Omega+10\Omega=16\Omega$

Circuit diagramab

Now to find the total resistance, combine these two new resistance values, which are in parallel:

$\frac{1}{R_T}=\frac{1}{5\Omega}+\frac{1}{16\Omega}=\frac{21}{80\Omega}$

$R_T=\frac{80}{21}\Omega$

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Example Question #11 : Understanding Circuit Diagrams

Circuitdiagram3

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

V=10V

In the circuit above, what is the voltage drop across R_3 ?

Possible Answers:

3.3V

2.5 V

5.0V

6.7V

7.5V

Correct answer:

7.5V

Explanation:

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top (R_1,R_3) and the bottom (R_2,R_4) is the same and equal to the voltage across the circuit, 10V .

The voltage drop across R_3 can be found as:

$V_3=\frac{R_3}{R_1+R_3}V=\frac{3\Omega}{1\Omega+3\Omega}10V=\frac{3}{4}\cdot10V$

V_3=7.5V

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Example Question #12 : Understanding Circuit Diagrams

Circuitdiagram3

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

V=10V

In the circuit above, what is the current passing through R_3 ?

Possible Answers:

0.625A

1.667A

2.5A

1.875A

4.167A

Correct answer:

2.5A

Explanation:

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top (R_1,R_3) and the bottom (R_2, R_4) is the same and equal to the voltage across the circuit, 10V .

Furthermore, the current that passes through R_1 must be the same as the current that passes through R_3 .

Therefore, the current that passes through them can be found by rearranging Ohm's law, solving for current.

$I_{1,3}=\frac{V}{R_1+R_3}=\frac{10V}{1\Omega+3\Omega}$

$I_{1,3}=2.5A$

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Example Question #13 : Understanding Circuit Diagrams

Basic circuit2

In the circuit above:

$V_{in}=12V$

$Z_1=11\Omega$

$Z_2=21\Omega$

$Z_3=16\Omega$

What is the current across Z_3 ?

Possible Answers:

0.38A

2.25A

1.13A

1.50A

0.75A

Correct answer:

0.75A

Explanation:

The quickest way to approach this problem is to realize that the voltage drop across Z_3 is the same as the voltage drop across the combined resistances of Z_1 and Z_2 . Since this parallel combination is the only presence of resistance in the circuit, this voltage drop must be the total voltage of the circuit, 12V .

Therefore, the current across Z_3 is:

$I_3=\frac{V_3}{Z_3}$

$I_3=\frac{12V}{16\Omega}$

I_3=0.75A

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Example Question #14 : Understanding Circuit Diagrams

Basic circuit2

In the circuit above:

$V_{in}=12V$

$Z_1=11\Omega$

$Z_2=21\Omega$

$Z_3=16\Omega$

What is the current across Z_2 ?

Possible Answers:

0.375A

1.50A

0.750A

2.25A

1.125A

Correct answer:

0.375A

Explanation:

Realize that the voltage drop across the combined resistances of Z_1 and Z_2 must be equal to the voltage of the circuit, since the parallel combination is the only presence of resistance in the circuit. This voltage drop must be the total voltage of the circuit, 12V .

The current across Z_1 and Z_2 is the same, and is given as:

$I_1=I_2=\frac{12V}{Z_1+Z_2}$

$I_2=\frac{12V}{32\Omega}$

I_2=0.375A

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AP Physics 1 : Electricity

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #3 : Understanding Circuit Diagrams

Example Question #3 : Understanding Circuit Diagrams

Example Question #5 : Understanding Circuit Diagrams

Example Question #4 : Understanding Circuit Diagrams

Example Question #1311 : Ap Physics 1

Example Question #11 : Understanding Circuit Diagrams

Example Question #11 : Understanding Circuit Diagrams

Example Question #12 : Understanding Circuit Diagrams

Example Question #13 : Understanding Circuit Diagrams

Example Question #14 : Understanding Circuit Diagrams

All AP Physics 1 Resources

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