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AP Physics 1 : Electricity

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Example Questions

Example Question #11 : Equivalent Resistance

Three identical resistors connected in parallel have an equivalent resistance equal to $2\: \Omega$ . What is the resistance of each of the individual resistors in this circuit?

Possible Answers:

$3\:\Omega$

$6\:\Omega$

$9\:\Omega$

$12\:\Omega$

Correct answer:

$6\:\Omega$

Explanation:

In the question stem, we're told that a circuit containing three identical resistors connected in parallel has an equivalent resistance equal to $2\:\Omega$ . We are then asked to solve for the resistance of each individual resistor.

To start with, it's important to remember that resistors in parallel add inversely. Thus, the inverse of the equivalent resistance is equal to the sum of the inverse of each individual resistor. Put another way:

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+....\frac{1}{R_{n}}$

Since we know the three resistors we're dealing with are identical, we can assign each of them a value of .

$\frac{1}{2\: \Omega }=\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x}$

And rearranging, we obtain:

$x=(2\: \Omega)(3)=6\: \Omega$

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Example Question #11 : Equivalent Resistance

Two resistors, $R_{1}=15\Omega$ and $R_{2}=10\Omega$ , are connected in parallel. What is the equivalent resistance of this setup?

Possible Answers:

$25\Omega$

$5\Omega$

$10\Omega$

$6\Omega$

Correct answer:

$6\Omega$

Explanation:

The equivalent resistance of resistors connected in parallel is given by the following equation,

$\frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$

In our problem,

$\frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}=\frac{2}{30}+\frac{3}{30}$

$\frac{1}{R_{p}}=\frac{5}{30}$

$R_{p}=\frac{30}{5}=6\Omega$

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Example Question #1291 : Ap Physics 1

Two resistors, $R_{1}=15\Omega$ and $R_{2}=10\Omega$ , are connected in series. What is the equivalent resistance of this setup?

Possible Answers:

$5\Omega$

$10\Omega$

$15\Omega$

$25\Omega$

Correct answer:

$25\Omega$

Explanation:

The equivalent resistance of resistors connected in series is the sum of the resistance values of each resistor, or

$R_{s}=\sum R_{n}=R_{1}+R_{2}+...$

In our problem,

$R_{s}=R_{1}+R_{2}=15+10=25\Omega$

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Example Question #32 : Circuits

Three resistors, $R_{1}=15\Omega$ , $R_{2}=10\Omega$ , and $R_{3}=5\Omega$ , are connected in series. What is the equivalent resistance of this setup?

Possible Answers:

$60\Omega$

$15\Omega$

$5\Omega$

$30\Omega$

Correct answer:

$30\Omega$

Explanation:

The equivalent resistance of resistors connected in series is the sum of the resistance values of each resistor, or

$R_{s}=\sum R_{n}=R_{1}+R_{2}+...$

In our problem,

$R_{s}=R_{1}+R_{2}+R_{5}=15+10+5=30\Omega$

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Example Question #12 : Equivalent Resistance

Three resistors, $R_{1}=15\Omega$ , $R_{2}=10\Omega$ , and $R_{3}=5\Omega$ , are connected in parallel. What is the equivalent resistance of this setup?

Possible Answers:

$2.73\Omega$

$30\Omega$

$15\Omega$

$5\Omega$

Correct answer:

$2.73\Omega$

Explanation:

The equivalent resistance of resistors connected in parallel is given by the following equation,

$\frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$

In our problem,

$\frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}+\frac{1}{5}=\frac{2}{30}+\frac{3}{30}+\frac{6}{30}$

$\frac{1}{R_{p}}=\frac{11}{30}$

$R_{p}=\frac{30}{11}=2.73\Omega$

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Example Question #11 : Equivalent Resistance

Equiv

$R_{1}=10\Omega, R_{2}=15\Omega, R_{3}=25\Omega.$

Three resistors are connected to a battery. What is the equivalent resistance of the given circuit?

Possible Answers:

$R_{eq}=50\Omega$

$R_{eq}=22.5\Omega$

$R_{eq}=8\Omega$

$R_{eq}=19.4\Omega$

Correct answer:

$R_{eq}=19.4\Omega$

Explanation:

Remember the rules of adding resistors. For resistors in series,

$R_{eq}=R_{1}+R_{2}+ ... +R_{n}$

and for resistors in parallel,

$R_{eq}=\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}}+...+\frac{1}{R_{n}}\right )^{-1}$

The idea is to start from the side furthest away from the battery and work back toward it. Notice that $R_{2}$ and $R_{3}$ are in parallel. We can add them in parallel so that they have an equivalent resistance $R_{23}$ ,

Equiv2

This can be calculated, but for 3 resistors we can leave it in equation form until the end. This looks like:

$R_{23}=\left ( \frac{1}{15\Omega}+\frac{1}{25\Omega}\right )^{-1}=9.375\Omega$

Notice that $R_{1}$ is now in series with $R_{23}$ ,

$R_{eq}=R_{1}+R_{23}=R_{1}+\left ( \frac{1}{15\Omega}+\frac{1}{25\Omega}\right )^{-1}=19.375\Omega$

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Example Question #41 : Circuits

A circuit has 10 identical resistors in parallel with a battery of 10V , and a total resistance of $\frac{1}{20} \Omega$ . Determine the voltage drop across one resistor.

Possible Answers:

2 V

20V

10V

$\frac{1}{20}V$

Correct answer:

10V

Explanation:

Since the resistors are in parallel to a 10V battery, the voltage drop across each resistor has to be equal to the voltage gain across the battery. Therefore, the voltage drop for any of the resistors will be 10V .

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Example Question #41 : Circuits

Calculate the equivalent resistance, of four resistors in parallel, which values $R_1 = 5 \Omega, R_2 = 10 \Omega, R_3 = 1 \Omega, R_4 = 6 \Omega$ .

Possible Answers:

$11.0\Omega$

$1.47 \Omega$

$22.0 \Omega$

$0.68 \Omega$

Correct answer:

$0.68 \Omega$

Explanation:

In order to find the equivalent resistance of resistors in parallel, we add the inverses of their values, as shown below

$\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{5 \Omega}+\frac{1}{10 \Omega}+\frac{1}{1\Omega}+\frac{1}{6 \Omega}=1.467 \Omega ^{-1}$

Finally

$\frac{1}{R_{tot}}=1.467 \Omega ^{-1} \Rightarrow R_{tot}=\frac{1}{1.467 \Omega ^{-1}}=0.68 \Omega$

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Example Question #13 : Equivalent Resistance

How does adding an additional resistor to a parallel circuit affect the equivalence resistance, or total resistance of the circuit?

Possible Answers:

Cannot be determined unless the resistance of the new resistor is known

Has no effect on the equivalent resistance

Decreases the equivalent resistance

Increases the equivalent resistance

Correct answer:

Decreases the equivalent resistance

Explanation:

We can answer this by looking at how the equivalent resistance of a parallel circuit is calculated.

$\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+...\Rightarrow R_{tot}= \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...}$ .

What we can see is that because the inverses are being added, any additional resistors will make contributions to the denominator of the fraction. Thus, lowering, or decreasing the equivalent resistance of the circuit.

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Example Question #14 : Equivalent Resistance

Consider the following resistors in a circuit: $R_1 = 5 \Omega, R_2 = 5 \Omega, R_3 = 5 \Omega$ . How should the resistors be connected in order to maximize the equivalent resistance of the circuit?

Possible Answers:

Parallel

Series

Some combination of parallel and series

Cannot be determined without knowing the voltage of the source

Correct answer:

Series

Explanation:

The equivalent resistance of resistors in series is calculated by $R_{tot}= R_1 + R_2 + R_3$ , while for parallel resistors, $R_{tot}= \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$ . This clearly shows that resistors in a series configuration will generate a much higher, or maximum, equivalent resistance value.

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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 #11 : Equivalent Resistance

Example Question #11 : Equivalent Resistance

Example Question #1291 : Ap Physics 1

Example Question #32 : Circuits

Example Question #12 : Equivalent Resistance

Example Question #11 : Equivalent Resistance

Example Question #41 : Circuits

Example Question #41 : Circuits

Example Question #13 : Equivalent Resistance

Example Question #14 : Equivalent Resistance

All AP Physics 1 Resources

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