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High School Physics : Resistors

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Example Question #1 : Calculating Total Resistance

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

Possible Answers:

$2.57\Omega$

$2.18\Omega$

$0.153\Omega$

$0.389\Omega$

$6.55\Omega$

Correct answer:

$6.55\Omega$

Explanation:

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

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

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

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

$R_{eq}=6.55\Omega$

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Example Question #11 : Resistors

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

Possible Answers:

$0.389\Omega$

$6.55\Omega$

$0.458\Omega$

$0.153\Omega$

$2.57\Omega$

Correct answer:

$0.389\Omega$

Explanation:

The formula for resistance in parallel is:

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

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

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

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

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

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Example Question #1 : Calculating Total Resistance

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

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

Possible Answers:

$10\Omega$

$100\Omega$

$67.1\Omega$

$13.7\Omega$

$53.4\Omega$

Correct answer:

$67.1\Omega$

Explanation:

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

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

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

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

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

Use the resistance equation to solve.

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

$R=67.1\Omega$

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Example Question #41 : Electric Circuits

What is the total resistance of a $14\Omega$ , $2.2\Omega$ , and $13\Omega$ series circuit?

Possible Answers:

$29.2\Omega$

$0.03\Omega$

$0.6\Omega$

$1.66\Omega$

$9.7\Omega$

Correct answer:

$29.2\Omega$

Explanation:

The formula for resistors in series is $R_{eq}=R_1+R_2+R_3...$ .

Plug in our given values.

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

$R_{eq}=14\Omega+2.2\Omega+13\Omega$

$R_{eq}=29.2\Omega$

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Example Question #2 : Electricity And Magnetism

There are three resistors in parallel in a circuit with resistances of $10\Omega$ , $20\Omega$ , and $30\Omega$ .

What is the equivalent resistance?

Possible Answers:

$5.45\Omega$

$1\Omega$

$6000\Omega$

$0.017\Omega$

$60\Omega$

Correct answer:

$5.45\Omega$

Explanation:

The equation for resistors in parallel is:

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

We are given the values of the resistors. Using this formula, we can solve for the equivalent resistance.

Plug in the given values and solve.

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

$\frac{1}{R_{eq}}=\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{30\Omega}...$

$\frac{1}{R_{eq}}=\frac{11}{60\Omega}$

$R_{eq}=\frac{60\Omega}{11}$

$R_{eq}=5.45\Omega$

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Example Question #949 : High School Physics

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

Possible Answers:

$30\Omega$

$3,000\Omega$

$3\Omega$

$30,000\Omega$

$300\Omega$

Correct answer:

$3,000\Omega$

Explanation:

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

$R_{eq}=R1+R2+R3...$

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

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

$R_{eq}=3000\Omega)$

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Example Question #950 : High School Physics

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

Possible Answers:

$2.57\Omega$

$6.55\Omega$

$0.153\Omega$

$0.389\Omega$

$0.458\Omega$

Correct answer:

$0.389\Omega$

Explanation:

The formula for resistance in parallel is:

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

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

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

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

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

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Example Question #1 : Resistivity

An electric circuit is set up in series with five resistors. If the resistors remain the same, but the circuit is now set up with the resistors in parallel, how would this affect the total resistance?

Possible Answers:

Resistance would remain unchanged

We need to know the numerical values to solve

Resistance would decrease substantially

Resistance would increase substantially

Correct answer:

Resistance would decrease substantially

Explanation:

Think of resistors as doors, preventing the flow of people (electrons). Imagine the following scenarios: a large group of people are in a room and all try to leave at once. If the five resistors are in series, that's like having all of these people trying to go through all five doors before they can leave. In a circuit, all the electrons in the current mast pass through every resistor.

If the resistors are in parallel, it's like having five separate doors from the room. All of a sudden, the group can leave MUCH faster, encountering less resistance to their flow out of the room. The path of the electrons can split, allowing each particle to pass through only one resistor.

From a formula perspective, the resistors in series are simply summed together to find the equivalent resistance.

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

In parallel, however, the reciprocals are summed to find the reciprocal equivalent resistance.

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

Adding whole numbers will always give you a much greater result than adding fractions. For the exact same set of resistors, arrangement in series will have a greater total resistance than arrangement in parallel.

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Example Question #951 : High School Physics

When current in a circuit crosses a resistor, energy is lost. What form does this lost energy most commonly take?

Possible Answers:

The energy is converted into heat

The energy is not converted; it simply disappears

The energy is converted into motion

The energy is converted into sound

The energy is converted into light

Correct answer:

The energy is converted into heat

Explanation:

In basic resistors, energy lost due to resistance is converted into heat. In some cases, other conversions also take place (such as generation of light in a lightbulb), but heat is still dissipated along with any alternative conversations. Lightbulbs, batteries, and other types of resistors will become hot as current passes through them.

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Example Question #71 : Electric Circuits

You have a long, 0.5cm diameter copper wire that has an electric current running through it. Which of the following would decrease the wire's overall resistivity?

Possible Answers:

None of these

Decreasing the diameter of the wire

Increasing the length of the wire

Increasing the diameter of the wire

Decreasing the length of the wire

Correct answer:

None of these

Explanation:

The formula for resistance is:

$R=\rho LA$

Above, is length, is the cross-sectional area of the wire, and $\rho$ is the resistivity of the material, and is a property of that material. The resistivity is constant for a given material, and thus cannot be changed by altering the dimensions of the wire.

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High School Physics : Resistors

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Calculating Total Resistance

Example Question #11 : Resistors

Example Question #1 : Calculating Total Resistance

Example Question #41 : Electric Circuits

Example Question #2 : Electricity And Magnetism

Example Question #949 : High School Physics

Example Question #950 : High School Physics

Example Question #1 : Resistivity

Example Question #951 : High School Physics

Example Question #71 : Electric Circuits

All High School Physics Resources

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