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AP Physics 1 : Circuits

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Example Question #7 : Equivalent Resistance

Consider the circuit:

Circuit_4

If the equivalent resistance of the circuit is $3\Omega$ , which of the following configuration of resistance values is possible?

Possible Answers:

$R1 = 1\Omega,\ R2 = 0.25\Omega,\ R3 = 1.25\Omega,\ R4 = 0.5\Omega$

$R1 = 8\Omega,\ R2 = 20\Omega,\ R3 = 8\Omega,\ R4 = 30\Omega$

$R1 = 3\Omega,\ R2 = 7\Omega,\ R3 = 15\Omega,\ R4 = 5\Omega$

$R1 = 5\Omega,\ R2 = 5\Omega,\ R3 = 6\Omega,\ R4 = 8\Omega$

None of these

Correct answer:

$R1 = 8\Omega,\ R2 = 20\Omega,\ R3 = 8\Omega,\ R4 = 30\Omega$

Explanation:

We will need to test the values of each answer to find the one that generates an equivalent resistance of $3\Omega$ .

We know that when condensing parallel resistors, the equivalent resistance will never be larger than the largest single resistance, and will always be smaller than the smallest resistance. Therefore, two of the answer options cen be eliminated immediately.

After we have narrowed our choices down to the other options answers, we just have to test them with the following formula:

$\frac{1}{R_{eq}}=\sum\frac{1}{R}$

We will test the incorrect answer first:

$\frac{1}{R_{eq}} = \frac{1}{5\Omega}+\frac{1}{5\Omega}+\frac{1}{6\Omega}+\frac{1}{8\Omega}$

$R_{eq} = 1.4\Omega$

Now for the correct answer:

$\frac{1}{R_{eq}} = \frac{1}{8\Omega}+\frac{1}{20\Omega}+\frac{1}{8\Omega}+\frac{1}{30\Omega}$

$\frac{1}{R_{eq}}=\frac{40}{120}\Omega=\frac{1}{3}\Omega$

$R_{eq}= 3\Omega$

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Example Question #8 : Equivalent Resistance

Equivalent resistance circuit

What is the equivalent resistance from Point A to Point B?

Possible Answers:

$50\Omega$

$100\Omega$

$200\Omega$

$150\Omega$

$25\Omega$

Correct answer:

$50\Omega$

Explanation:

Because this circuit is neither purely series or purely parallel, we must simplify it before we solve it. Replace the right branch, which is purely series, with its equivalent resistance:

$50\Omega+50\Omega=100\Omega$

Now we have a purely parallel circuit, each branch having a resistance of $100\Omega$ . Apply the parallel formula and solve:

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

$R_{tot}=\frac{1}{\frac{1}{100\Omega }+\frac{1}{100\Omega }}=50\Omega$

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Example Question #9 : Equivalent Resistance

What is the equivalent resistance of the following resistors, all in series: $3 \Omega, 2 \Omega, 6\Omega$ ?

Possible Answers:

$11 \Omega$

$1 \Omega$

$10 \Omega$

$5 \Omega$

Correct answer:

$11 \Omega$

Explanation:

For resistors all in series, the equivalent resistance is equal to the sum of the resistances.

$R_{eq}=3+2+6=11\Omega$

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Example Question #1 : Equivalent Resistance

What is the equivalent resistance of a circuit consisting of a group of resistors (all in parallel), with the following resistances: $5\Omega, 5\Omega, 5\Omega, 5\Omega, 5\Omega$ ?

Possible Answers:

$25 \Omega$

$1 \Omega$

$\frac{1}{25} \Omega$

$20 \Omega$

Correct answer:

$1 \Omega$

Explanation:

The reciprocal of the equivalent resistance for resistors in parallel is equal to the sum of the reciprocals of the resistances:

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

$\frac{1}{R_{eq}}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1\Omega$

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Example Question #31 : Circuits

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:

$12\:\Omega$

$9\:\Omega$

$6\:\Omega$

$3\:\Omega$

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:

$10\Omega$

$25\Omega$

$6\Omega$

$5\Omega$

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 #11 : Equivalent Resistance

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:

$10\Omega$

$5\Omega$

$15\Omega$

$25\Omega$

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 #11 : Equivalent Resistance

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:

$5\Omega$

$30\Omega$

$15\Omega$

$60\Omega$

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 #35 : Circuits

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:

$30\Omega$

$2.73\Omega$

$15\Omega$

$5\Omega$

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 #36 : Circuits

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:

$R_{eq}=50\Omega$

$R_{eq}=22.5\Omega$

$R_{eq}=19.4\Omega$

$R_{eq}=8\Omega$

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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AP Physics 1 : Circuits

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #7 : Equivalent Resistance

Example Question #8 : Equivalent Resistance

Example Question #9 : Equivalent Resistance

Example Question #1 : Equivalent Resistance

Example Question #31 : Circuits

Example Question #11 : Equivalent Resistance

Example Question #11 : Equivalent Resistance

Example Question #11 : Equivalent Resistance

Example Question #35 : Circuits

Example Question #36 : Circuits

All AP Physics 1 Resources

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