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AP Physics 1 : Electricity and Waves

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Example Questions

Example Question #2 : Understanding Circuit Diagrams

Circuit diagram

V=5V $\displaystyle V=5V$

$R_1=1\Omega$ $\displaystyle R_1=1\Omega$

$R_2=4\Omega$ $\displaystyle R_2=4\Omega$

In the circuit above, what is the total current?

Possible Answers:

20A $\displaystyle 20A$

$\displaystyle 4A$

$\displaystyle 6A$

$\displaystyle 1A$

6.25A $\displaystyle 6.25A$

Correct answer:

6.25A $\displaystyle 6.25A$

Explanation:

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

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{1\Omega }+\frac{1}{4\Omega }$ $\displaystyle \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{1\Omega }+\frac{1}{4\Omega }$

$\frac{1}{R_T}=\frac{5}{4\Omega }$ $\displaystyle \frac{1}{R_T}=\frac{5}{4\Omega }$

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

After that, calculate the current using Ohm's Law:

$I=\frac{V}{R}=\frac{5V}{\frac{4}{5}\Omega}$ $\displaystyle I=\frac{V}{R}=\frac{5V}{\frac{4}{5}\Omega}$

I=6.25A $\displaystyle I=6.25A$

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Example Question #2 : Understanding Circuit Diagrams

Circuit diagram

I=30A $\displaystyle I=30A$

$R_1=2\Omega$ $\displaystyle R_1=2\Omega$

$R_2=3\Omega$ $\displaystyle R_2=3\Omega$

In the circuit above, what is the total voltage?

Possible Answers:

25V $\displaystyle 25V$

75V $\displaystyle 75V$

30V $\displaystyle 30V$

36V $\displaystyle 36V$

150 V $\displaystyle 150 V$

Correct answer:

36V $\displaystyle 36V$

Explanation:

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

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

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

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

Use Ohm's law to find the voltage.

V=IR $\displaystyle V=IR$

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

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Example Question #3 : Understanding Circuit Diagrams

Circuit diagram

V=5V $\displaystyle V=5V$

I=1.25A $\displaystyle I=1.25A$

$R_1=6\Omega$ $\displaystyle R_1=6\Omega$

In the circuit above, what is the resistance of R_2 $\displaystyle R_2$?

Possible Answers:

$2\Omega$ $\displaystyle 2\Omega$

$-2\Omega$ $\displaystyle -2\Omega$

$12\Omega$ $\displaystyle 12\Omega$

$2.4\Omega$ $\displaystyle 2.4\Omega$

$10\Omega$ $\displaystyle 10\Omega$

Correct answer:

$12\Omega$ $\displaystyle 12\Omega$

Explanation:

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

V=IR $\displaystyle V=IR$

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

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

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:

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

Rearrange and solve for R_2 $\displaystyle R_2$

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

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

$R_2=12\Omega$ $\displaystyle R_2=12\Omega$

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Example Question #5 : Understanding Circuit Diagrams

Circuit diagram2

$R_1=3\Omega$ $\displaystyle R_1=3\Omega$

$R_2=6\Omega$ $\displaystyle R_2=6\Omega$

$R_3=2\Omega$ $\displaystyle R_3=2\Omega$

$R_4=16\Omega$ $\displaystyle R_4=16\Omega$

In the circuit above, what is the total resistance?

Possible Answers:

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

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

$27\Omega$ $\displaystyle 27\Omega$

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

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

Correct answer:

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

Explanation:

Find the combined resistances for the resistors in parallel:

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

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

$R_A=2\Omega$ $\displaystyle R_A=2\Omega$

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

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

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

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

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

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Example Question #4 : Understanding Circuit Diagrams

Circuit diagram2

$R_1=1\Omega$ $\displaystyle R_1=1\Omega$

$R_2=2\Omega$ $\displaystyle R_2=2\Omega$

$R_3=3\Omega$ $\displaystyle R_3=3\Omega$

$R_4=4\Omega$ $\displaystyle R_4=4\Omega$

V=10V $\displaystyle V=10V$

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

Possible Answers:

12.6 V $\displaystyle 12.6 V$

5.4V $\displaystyle 5.4V$

1.3V $\displaystyle 1.3V$

2.8 V $\displaystyle 2.8 V$

7.2V $\displaystyle 7.2V$

Correct answer:

7.2V $\displaystyle 7.2V$

Explanation:

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

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

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

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

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

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

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

Therefore, the total resistance is:

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

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

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

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

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

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

$V_{3,4}=7.2V$ $\displaystyle V_{3,4}=7.2V$

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Example Question #1311 : Ap Physics 1

Circuit diagram2

$R_1=1\Omega$ $\displaystyle R_1=1\Omega$

$R_2=2\Omega$ $\displaystyle R_2=2\Omega$

$R_3=3\Omega$ $\displaystyle R_3=3\Omega$

$R_4=4\Omega$ $\displaystyle R_4=4\Omega$

V=10V $\displaystyle V=10V$

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

Possible Answers:

4.2A $\displaystyle 4.2A$

1.4A $\displaystyle 1.4A$

2.4A $\displaystyle 2.4A$

1.8A $\displaystyle 1.8A$

2.8A $\displaystyle 2.8A$

Correct answer:

2.4A $\displaystyle 2.4A$

Explanation:

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

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

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

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

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

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

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

Therefore, the total resistance is:

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

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

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

This current will be divided between $\displaystyle R3$ and $\displaystyle R4$, with more current taking the path of lower resistance.

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

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

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

$V_{3,4}=7.2V$ $\displaystyle V_{3,4}=7.2V$

$\displaystyle V_3=V_4=7.2V$

The current through R_3 $\displaystyle R_3$ is given by:

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

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

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Example Question #11 : Understanding Circuit Diagrams

Circuitdiagram3

$R_1=3\Omega$ $\displaystyle R_1=3\Omega$

$R_2=6\Omega$ $\displaystyle R_2=6\Omega$

$R_3=2\Omega$ $\displaystyle R_3=2\Omega$

$R_4=10\Omega$ $\displaystyle R_4=10\Omega$

In the circuit above, what is the total resistance?

Possible Answers:

$21\Omega$ $\displaystyle 21\Omega$

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

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

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

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

Correct answer:

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

Explanation:

Begin by combining the resistors that are immediately in series:

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

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

Circuit diagramab

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

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

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

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Example Question #11 : Understanding Circuit Diagrams

Circuitdiagram3

$R_1=1\Omega$ $\displaystyle R_1=1\Omega$

$R_2=2\Omega$ $\displaystyle R_2=2\Omega$

$R_3=3\Omega$ $\displaystyle R_3=3\Omega$

$R_4=4\Omega$ $\displaystyle R_4=4\Omega$

V=10V $\displaystyle V=10V$

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

Possible Answers:

3.3V $\displaystyle 3.3V$

2.5 V $\displaystyle 2.5 V$

5.0V $\displaystyle 5.0V$

6.7V $\displaystyle 6.7V$

7.5V $\displaystyle 7.5V$

Correct answer:

7.5V $\displaystyle 7.5V$

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 $\displaystyle (R_1,R_3)$ and the bottom $\displaystyle (R_2,R_4)$ is the same and equal to the voltage across the circuit, 10V $\displaystyle 10V$.

The voltage drop across R_3 $\displaystyle R_3$ can be found as:

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

$\displaystyle V_3=7.5V$

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Example Question #12 : Understanding Circuit Diagrams

Circuitdiagram3

$R_1=1\Omega$ $\displaystyle R_1=1\Omega$

$R_2=2\Omega$ $\displaystyle R_2=2\Omega$

$R_3=3\Omega$ $\displaystyle R_3=3\Omega$

$R_4=4\Omega$ $\displaystyle R_4=4\Omega$

V=10V $\displaystyle V=10V$

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

Possible Answers:

0.625A $\displaystyle 0.625A$

1.667A $\displaystyle 1.667A$

2.5A $\displaystyle 2.5A$

1.875A $\displaystyle 1.875A$

4.167A $\displaystyle 4.167A$

Correct answer:

2.5A $\displaystyle 2.5A$

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 $\displaystyle (R_1,R_3)$ and the bottom $\displaystyle (R_2, R_4)$ is the same and equal to the voltage across the circuit, 10V $\displaystyle 10V$.

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

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

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

$I_{1,3}=2.5A$ $\displaystyle I_{1,3}=2.5A$

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Example Question #13 : Understanding Circuit Diagrams

Basic circuit2

In the circuit above:

$V_{in}=12V$ $\displaystyle V_{in}=12V$

$Z_1=11\Omega$ $\displaystyle Z_1=11\Omega$

$Z_2=21\Omega$ $\displaystyle Z_2=21\Omega$

$Z_3=16\Omega$ $\displaystyle Z_3=16\Omega$

What is the current across Z_3 $\displaystyle Z_3$?

Possible Answers:

0.38A $\displaystyle 0.38A$

2.25A $\displaystyle 2.25A$

1.13A $\displaystyle 1.13A$

1.50A $\displaystyle 1.50A$

0.75A $\displaystyle 0.75A$

Correct answer:

0.75A $\displaystyle 0.75A$

Explanation:

The quickest way to approach this problem is to realize that the voltage drop across Z_3 $\displaystyle Z_3$ is the same as the voltage drop across the combined resistances of Z_1 $\displaystyle Z_1$ and Z_2 $\displaystyle Z_2$. 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 $\displaystyle 12V$.

Therefore, the current across Z_3 $\displaystyle Z_3$ is:

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

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

$\displaystyle I_3=0.75A$

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AP Physics 1 : Electricity and Waves

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #2 : Understanding Circuit Diagrams

Example Question #2 : 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

All AP Physics 1 Resources

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