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High School Physics : Electric Circuits

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Example Questions

Example Question #7 : 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:

$2.57\Omega$

$0.153\Omega$

$2.18\Omega$

$0.389\Omega$

$6.55\Omega$

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 #1 : Calculating Total Resistance

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

Possible Answers:

$0.389\Omega$

$0.458\Omega$

$0.153\Omega$

$2.57\Omega$

$6.55\Omega$

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 #8 : 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:

$10\Omega$

$53.4\Omega$

$100\Omega$

$13.7\Omega$

$67.1\Omega$

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 #11 : Calculating Total Resistance

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

Possible Answers:

$0.03\Omega$

$1.66\Omega$

$0.6\Omega$

$9.7\Omega$

$29.2\Omega$

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 #41 : Electric Circuits

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:

$5.45\Omega$

$60\Omega$

$0.017\Omega$

$1\Omega$

$6000\Omega$

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 #1 : Understanding Capacitance

Which of these is the correct SI unit for capacitance?

Possible Answers:

Newtons

Farad-Joules

Ohms

Farads

Volts

Correct answer:

Farads

Explanation:

The correct unit for capacitance is the Farad, named after the English physicist Michael Faraday.

Newtons are used to measure force. Ohms measure resistance and volts measure electrical energy or potential. Joules are a unit of energy, but Farad-Joules are not a recognizable unit.

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Example Question #42 : Electric Circuits

By using a dielectric, it is possible to store more energy in a capacitor. Why is this?

Possible Answers:

The dielectric provides a direct energy source to the capacitor

The dielectric replaces air, which breaks down when the electric fields get strong enough

The dielectric changes the distance between the two plates of the capacitor

The dielectric can store excess energy

Correct answer:

The dielectric replaces air, which breaks down when the electric fields get strong enough

Explanation:

Dielectrics are insulating materials that can be polarized, and can be used to increase the separability of the two plates of a capacitor. When a dielectric is absent, air is present between the capacitor plates. At high voltages, when the electric field between the plates is incredibly strong, air begins to break down and becomes unable to prevent the flow of electricity between the capacitor plates. When this occurs, the capacitor cannot store energy, nor can it properly function. The presence of a dielectric eliminates this problem, allowing the capacitor to store much more energy.

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Example Question #1 : Calculating Total Capacitance

Three capacitors are in series. They have capacitances of , , and , respectively. What is their total capacitance?

Possible Answers:

0.077F

13F

0.958F

1.11F

1.04F

Correct answer:

1.04F

Explanation:

For capacitors in series the formula for total capacitance is:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...$

Note that this formula is similar to the formula for total resistance in parallel. Using the values for each individual capacitor, we can solve for the total capacitance.

$\frac{1}{C_{eq}}=\frac{1}{3F}+\frac{1}{2F}+\frac{1}{8F}$

$\frac{1}{C_{eq}}=0.958F$

$C_{eq}=1.04F$

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Example Question #1 : Capacitors

Three capacitors are in parallel. They have capacitance values of , , and . What is their total capacitance?

Possible Answers:

0.958F

1.09F

13F

0.077F

1.04F

Correct answer:

13F

Explanation:

For capacitors in parallel the formula for total capacitance is:

$C_{eq}=C_1+C_2+C_3...$

Note that this formula is similar to the formula for total resistance in series. Using the values for each individual capacitor, we can solve for the total capacitance.

$C_{eq}=3F+2F+8F$

$C_{eq}=13F$

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Example Question #2 : Capacitors

Three capacitors, each with a capacity of are arranged in parallel. What is the total capacitance of this circuit?

Possible Answers:

64F

0.08F

12F

0.75F

1.33F

Correct answer:

12F

Explanation:

The formula for capacitors in parallel is:

$C_{eq}=C_1+C_2+C_3...$

Our three capacitors all have equal capacitance values. We can simply add them together to find the total capacitance.

$C_{eq}=4F+4F+4F$

$C_{eq}=12F$

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High School Physics : Electric Circuits

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #7 : Calculating Total Resistance

Example Question #1 : Calculating Total Resistance

Example Question #8 : Calculating Total Resistance

Example Question #11 : Calculating Total Resistance

Example Question #41 : Electric Circuits

Example Question #1 : Understanding Capacitance

Example Question #42 : Electric Circuits

Example Question #1 : Calculating Total Capacitance

Example Question #1 : Capacitors

Example Question #2 : Capacitors

All High School Physics Resources

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