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High School Physics : Calculating Total Resistance

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Example Questions

Example Question #1 : Calculating Total Resistance

Ten resistors, each with $300\Omega$ $\displaystyle 300\Omega$ resistance, are set up in series. What is their equivalent resistance?

Possible Answers:

$30,000\Omega$ $\displaystyle 30,000\Omega$

$300\Omega$ $\displaystyle 300\Omega$

$30\Omega$ $\displaystyle 30\Omega$

$3\Omega$ $\displaystyle 3\Omega$

$3,000\Omega$ $\displaystyle 3,000\Omega$

Correct answer:

$3,000\Omega$ $\displaystyle 3,000\Omega$

Explanation:

For resistors aligned in series, the equivalent resistance is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$ $\displaystyle R_{eq}=R_1+R_2+R_3...$

Since all the resistors in this problem are equal, we can simplify with multiplication.

$R_{eq}=10 (300\Omega)$ $\displaystyle R_{eq}=10 (300\Omega)$

$R_{eq}=3000\Omega$ $\displaystyle R_{eq}=3000\Omega$

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Example Question #2 : Calculating Total Resistance

Ten resistors, each with $300\Omega$ $\displaystyle 300\Omega$ resistance, are set up in parallel. What is the equivalent resistance?

Possible Answers:

$30\Omega$ $\displaystyle 30\Omega$

$3,000\Omega$ $\displaystyle 3,000\Omega$

$30,000\Omega$ $\displaystyle 30,000\Omega$

$300\Omega$ $\displaystyle 300\Omega$

$3\Omega$ $\displaystyle 3\Omega$

Correct answer:

$30\Omega$ $\displaystyle 30\Omega$

Explanation:

For resistors in parallel, the equivalent resistance can be found by summing the reciprocals of the individual resistances, then taking the reciprocal of the resultant sum.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

Since all the resistors in this problem are equal, we can simplify by multiplying.

$\frac{1}{R_{eq}}=10( \frac{1}{300\Omega})$ $\displaystyle \frac{1}{R_{eq}}=10( \frac{1}{300\Omega})$

$\frac{1}{R_{eq}}=\frac{10}{300\Omega}$ $\displaystyle \frac{1}{R_{eq}}=\frac{10}{300\Omega}$

$\frac{1}{R_{eq}}=\frac{1}{30\Omega}$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{30\Omega}$

$R_{eq}=30\Omega$ $\displaystyle R_{eq}=30\Omega$

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Example Question #3 : Calculating Total Resistance

Twelve resistors of equal resistance are set up in series. If the total resistance is $600\Omega$ $\displaystyle 600\Omega$, what is the resistance of each resistor?

Possible Answers:

$5\Omega$ $\displaystyle 5\Omega$

$588\Omega$ $\displaystyle 588\Omega$

$50\Omega$ $\displaystyle 50\Omega$

$0.2\Omega$ $\displaystyle 0.2\Omega$

$500\Omega$ $\displaystyle 500\Omega$

Correct answer:

$50\Omega$ $\displaystyle 50\Omega$

Explanation:

The total resistance of resistors in series is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$ $\displaystyle R_{eq}=R_1+R_2+R_3...$

In this case, we know the total resistance and the number of resistors, and we are told that all resistors are of equal strength. That means we can simplify this problem using multiplication.

$R_{eq}=12R_$ $\displaystyle R_{eq}=12R_$

We can use the total resistance to solve for the individual value.

$600\Omega=12R$ $\displaystyle 600\Omega=12R$

$\frac{600\Omega}{12}=R$ $\displaystyle \frac{600\Omega}{12}=R$

$50\Omega=R$ $\displaystyle 50\Omega=R$

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Example Question #1 : Calculating Total Resistance

Three resistors are in a series. They have $10\Omega$ $\displaystyle 10\Omega$, $18.3\Omega$ $\displaystyle 18.3\Omega$, and $21.22\Omega$ $\displaystyle 21.22\Omega$ of resistance respectively. What is the total resistance?

Possible Answers:

$49.52\Omega$ $\displaystyle 49.52\Omega$

$16.51\Omega$ $\displaystyle 16.51\Omega$

$1.43\Omega$ $\displaystyle 1.43\Omega$

$0.61\Omega$ $\displaystyle 0.61\Omega$

$0.20\Omega$ $\displaystyle 0.20\Omega$

Correct answer:

$49.52\Omega$ $\displaystyle 49.52\Omega$

Explanation:

When working in a series, the total resistance is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$ $\displaystyle R_{eq}=R_1+R_2+R_3...$

Use the given values for each individual resistor to solve for the total resistance.

$R_{eq}=10\Omega+18.3\Omega+21.22\Omega$ $\displaystyle R_{eq}=10\Omega+18.3\Omega+21.22\Omega$

$R_{eq}=49.52\Omega$ $\displaystyle R_{eq}=49.52\Omega$

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Example Question #5 : Calculating Total Resistance

Six resistors are arranged in a series. Their resistances are $32.4\Omega$ $\displaystyle 32.4\Omega$, $8.7\Omega$ $\displaystyle 8.7\Omega$, $0.77\Omega$ $\displaystyle 0.77\Omega$, $65.2\Omega$ $\displaystyle 65.2\Omega$, $10\Omega$ $\displaystyle 10\Omega$, and $26.33\Omega$ $\displaystyle 26.33\Omega$. What is their total resistance?

Possible Answers:

$0.04\Omega$ $\displaystyle 0.04\Omega$

$0.07\Omega$ $\displaystyle 0.07\Omega$

$43.4\Omega$ $\displaystyle 43.4\Omega$

$23.9\Omega$ $\displaystyle 23.9\Omega$

$143.4\Omega$ $\displaystyle 143.4\Omega$

Correct answer:

$143.4\Omega$ $\displaystyle 143.4\Omega$

Explanation:

When working in a series, the total resistance is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$ $\displaystyle R_{eq}=R_1+R_2+R_3...$

Use the given values for each individual resistor to solve for the total resistance.

$R_{eq}=32.4\Omega + 8.7\Omega +0.77\Omega + 65.2\Omega + 10\Omega + 26.33\Omega$ $\displaystyle R_{eq}=32.4\Omega + 8.7\Omega +0.77\Omega + 65.2\Omega + 10\Omega + 26.33\Omega$

$R_{eq}=143.4\Omega$ $\displaystyle R_{eq}=143.4\Omega$

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Example Question #2 : Calculating Total Resistance

Three resistors are in a parallel circuit. They have resistances of $10\Omega$ $\displaystyle 10\Omega$, $18.3\Omega$ $\displaystyle 18.3\Omega$, and $21.22\Omega$ $\displaystyle 21.22\Omega$, respectively. What is their total resistance?

Possible Answers:

$49.52\Omega$ $\displaystyle 49.52\Omega$

$4.96\Omega$ $\displaystyle 4.96\Omega$

$0.202\Omega$ $\displaystyle 0.202\Omega$

$1.86\Omega$ $\displaystyle 1.86\Omega$

$0.07\Omega$ $\displaystyle 0.07\Omega$

Correct answer:

$4.96\Omega$ $\displaystyle 4.96\Omega$

Explanation:

The formula for resistors in parallel is:

$\frac{1}{R_{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$ $\displaystyle \frac{1}{R_{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

Using the given individual resistances, we can find the total resistance of the circuit.

$\frac{1}{R_{eq}}}=\frac{1}{10\Omega}+\frac{1}{18.3\Omega}+\frac{1}{21.22\Omega}$ $\displaystyle \frac{1}{R_{eq}}}=\frac{1}{10\Omega}+\frac{1}{18.3\Omega}+\frac{1}{21.22\Omega}$

$\frac{1}{R_{eq}}}=0.2017\Omega$ $\displaystyle \frac{1}{R_{eq}}}=0.2017\Omega$

$R_{eq}=4.96\Omega$ $\displaystyle R_{eq}=4.96\Omega$

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Example Question #7 : Calculating Total Resistance

Two resistors in series have $1.8\Omega$ $\displaystyle 1.8\Omega$ and $32.4\Omega$ $\displaystyle 32.4\Omega$ resistance, respectively. What is the total resistance?

Possible Answers:

$1.71\Omega$ $\displaystyle 1.71\Omega$

$0.586\Omega$ $\displaystyle 0.586\Omega$

$0.029\Omega$ $\displaystyle 0.029\Omega$

$34.2\Omega$ $\displaystyle 34.2\Omega$

$30.6\Omega$ $\displaystyle 30.6\Omega$

Correct answer:

$34.2\Omega$ $\displaystyle 34.2\Omega$

Explanation:

For resistors in a series, total resistance is equal to the sum of each individual resistance.

$R_{eq}=R_1+R_2+R_3...$ $\displaystyle R_{eq}=R_1+R_2+R_3...$

We can use the individual resistances from the question to solve for the total resistance.

$R_{eq}=1.8\Omega+32.4\Omega$ $\displaystyle R_{eq}=1.8\Omega+32.4\Omega$

$R_{eq}=34.2\Omega$ $\displaystyle R_{eq}=34.2\Omega$

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Example Question #8 : Calculating Total Resistance

What is the total resistance of a series circuit with resistors of $2\Omega$ $\displaystyle 2\Omega$, $0.55\Omega$ $\displaystyle 0.55\Omega$, and $4\Omega$ $\displaystyle 4\Omega$?

Possible Answers:

$0.389\Omega$ $\displaystyle 0.389\Omega$

$2.57\Omega$ $\displaystyle 2.57\Omega$

$6.55\Omega$ $\displaystyle 6.55\Omega$

$0.153\Omega$ $\displaystyle 0.153\Omega$

$2.18\Omega$ $\displaystyle 2.18\Omega$

Correct answer:

$6.55\Omega$ $\displaystyle 6.55\Omega$

Explanation:

For a series circuit, the formula for total resistance is:

$R_{eq}=R_1+R_1+R_3...$ $\displaystyle R_{eq}=R_1+R_1+R_3...$

We are given the values of each resistance, allowing us to sum them to find the total resistance in the circuit.

$R_{eq}=2\Omega+0.55\Omega+4\Omega$ $\displaystyle R_{eq}=2\Omega+0.55\Omega+4\Omega$

$R_{eq}=6.55\Omega$ $\displaystyle R_{eq}=6.55\Omega$

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Example Question #1 : Calculating Total Resistance

What is the total resistance of a parallel circuit with resistors of $2\Omega$ $\displaystyle 2\Omega$, $0.55\Omega$ $\displaystyle 0.55\Omega$, and $4\Omega$ $\displaystyle 4\Omega$?

Possible Answers:

$0.458\Omega$ $\displaystyle 0.458\Omega$

$2.57\Omega$ $\displaystyle 2.57\Omega$

$0.389\Omega$ $\displaystyle 0.389\Omega$

$0.153\Omega$ $\displaystyle 0.153\Omega$

$6.55\Omega$ $\displaystyle 6.55\Omega$

Correct answer:

$0.389\Omega$ $\displaystyle 0.389\Omega$

Explanation:

The formula for resistance in parallel is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

We are given the values for each individual resistor, allowing us to solve for the total resistance.

$\frac{1}{R_{eq}}=\frac{1}{2\Omega}+\frac{1}{0.55\Omega}+\frac{1}{4\Omega}$ $\displaystyle \frac{1}{R_{eq}}=\frac{1}{2\Omega}+\frac{1}{0.55\Omega}+\frac{1}{4\Omega}$

$\frac{1}{R_{eq}}=\frac{113}{44}\Omega$ $\displaystyle \frac{1}{R_{eq}}=\frac{113}{44}\Omega$

$R_{eq}=\frac{44}{113}\Omega=0.389\Omega$ $\displaystyle R_{eq}=\frac{44}{113}\Omega=0.389\Omega$

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Example Question #2 : Calculating Total Resistance

Calculate the resistance of a copper wire with cross-sectional area of $\small 0.501cm^2$ $\displaystyle \small 0.501cm^2$ and length of $\small 2m$ $\displaystyle \small 2m$.

$\small rho_{Cu}=1.68*10^{-3}\Omega m$ $\displaystyle \small rho_{Cu}=1.68*10^{-3}\Omega m$

Possible Answers:

$67.1\Omega$ $\displaystyle 67.1\Omega$

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

$100\Omega$ $\displaystyle 100\Omega$

$53.4\Omega$ $\displaystyle 53.4\Omega$

$13.7\Omega$ $\displaystyle 13.7\Omega$

Correct answer:

$67.1\Omega$ $\displaystyle 67.1\Omega$

Explanation:

The resistance of a wire is given by the following equation:

$R=\rho\frac{L}{A}$ $\displaystyle R=\rho\frac{L}{A}$

We are given the resistivity ( $\small \rho$ $\displaystyle \small \rho$), cross-sectional area, and length. Using these values, we can solve for the resistance.

First, convert the cross-sectional area to square-meters.

$0.501cm^2*\frac{1m^2}{10000cm^2}=5.01*10^{-5}m^2$ $\displaystyle 0.501cm^2*\frac{1m^2}{10000cm^2}=5.01*10^{-5}m^2$

Use the resistance equation to solve.

$R=(1.68*10^{-3}\Omega m)\frac{2m}{5.01*10^{-5}m^2}$ $\displaystyle R=(1.68*10^{-3}\Omega m)\frac{2m}{5.01*10^{-5}m^2}$

$R=67.1\Omega$ $\displaystyle R=67.1\Omega$

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High School Physics : Calculating Total Resistance

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