AP Physics 1 : Circuits

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #11 : Circuit Power

What is the power of a circuit whose voltage is \(\displaystyle 3V\)  and equivalent resistance is \(\displaystyle 10\Omega\)?

Possible Answers:

\(\displaystyle 10W\)

\(\displaystyle 0.9W\)

\(\displaystyle 3W\)

\(\displaystyle 100W\)

Correct answer:

\(\displaystyle 0.9W\)

Explanation:

The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Since we are given resistance and voltage, we will also need Ohm's law, \(\displaystyle V=IR\), where \(\displaystyle V\) is the voltage, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving Ohm's law for current gives us \(\displaystyle I=\frac{V}{R}\).

Substituting this form of Ohm's law into the power equation gives us

\(\displaystyle P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right\)

The power equation is now in a form that we can solve with the information we are given.

\(\displaystyle P=\frac{V^{2}}{R} \right=\frac{3^{2}}{10} \right=0.9W\)

Example Question #141 : Circuits

If the voltage of a circuit is doubled, how is the power of a circuit changed?  Assume the resistance of the circuit stays the same.

Possible Answers:

Power wil be four times larger

Power will double

Power will stay the same

Power will triple

Correct answer:

Power wil be four times larger

Explanation:

The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

To relate power to voltage, we will also need Ohm's law, \(\displaystyle V=IR\), where \(\displaystyle V\) is the voltage, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving Ohm's law for current gives us \(\displaystyle I=\frac{V}{R}\).

Substituting this form of Ohm's law into the power equation gives us

\(\displaystyle P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right\)

\(\displaystyle P=\frac{V^{2}}{R} \right\)

Assuming the current stays the same, if the voltage is doubled, the power will be four times larger.

Expressed mathematically,

If \(\displaystyle V=2V\)

\(\displaystyle P=\frac{(2V)^{2}}{R} \right=\frac{4(V)^{2}}{R} \right=4P\)

Example Question #1401 : Ap Physics 1

How much energy does a heat coil produce with it is plugged into a \(\displaystyle 110V\) outlet for \(\displaystyle 5sec\) and has a resistance of \(\displaystyle 10\Omega\)?

Possible Answers:

\(\displaystyle 35J\)

\(\displaystyle 600J\)

\(\displaystyle 6050J\)

\(\displaystyle 12500J\)

Correct answer:

\(\displaystyle 6050J\)

Explanation:

The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

To relate power to voltage, we will also need Ohm's law, \(\displaystyle V=IR\), where \(\displaystyle V\) is the voltage, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving Ohm's law for current gives us \(\displaystyle I=\frac{V}{R}\).

Substituting this form of Ohm's law into the power equation gives us

\(\displaystyle P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right\)

\(\displaystyle P=\frac{V^{2}}{R} \right\)

Energy is related to power by the equation \(\displaystyle E=P\cdot t\), where \(\displaystyle E\) is energy, \(\displaystyle P\) is the power of the circuit, and \(\displaystyle t\) is time.

Substituting theequation for power into the equation for energy gives

\(\displaystyle E=P\cdot t=\frac{V^{2}}{R} t\)

For our problem

\(\displaystyle E=P\cdot t=\frac{V^{2}}{R} t=\frac{(110)^{2}}{10} (5)=6050 J\)

Example Question #1401 : Ap Physics 1

If a \(\displaystyle 60W\) lightbulb has a resistance of \(\displaystyle 15\Omega\), how much current is going across the lightbulb?

Possible Answers:

\(\displaystyle 1Amp\)

\(\displaystyle 1000Amps\)

\(\displaystyle 10Amps\)

\(\displaystyle 2Amps\)

Correct answer:

\(\displaystyle 2Amps\)

Explanation:

The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving the power equation for current gives 

\(\displaystyle I=\sqrt{\frac{P}{R}}\)

In this problem, \(\displaystyle I=\sqrt{\frac{P}{R}}=\sqrt{\frac{60}{15}}=\sqrt{4}=2Amps\)

Example Question #181 : Electricity

How much energy does a \(\displaystyle 75W\) lightbulb give off in \(\displaystyle 5min\)?

Possible Answers:

Not enough information

\(\displaystyle 22500J\)

\(\displaystyle 125J\)

\(\displaystyle 12000J\)

Correct answer:

\(\displaystyle 22500J\)

Explanation:

Energy is related to power by the equation \(\displaystyle E=P\cdot t\), where \(\displaystyle E\) is energy, \(\displaystyle P\) is the power of the circuit, and \(\displaystyle t\) is time.

Time needs to be in seconds.

\(\displaystyle 5min\times \frac{60sec}{1min}=300 sec\)

For our problem, energy is

\(\displaystyle E=75\cdot(300)=22500J\)

Example Question #252 : Electricity And Waves

What is the circuit power of a circuit with a battery with potential difference \(\displaystyle 50 V\), and 3 resistors in parallel with equivalent resistances of \(\displaystyle 3 \Omega\)

Possible Answers:

\(\displaystyle 278 W\)

\(\displaystyle 50W\)

\(\displaystyle 7500W\)

\(\displaystyle 2500W\)

Correct answer:

\(\displaystyle 2500W\)

Explanation:

Recall the formula for circuit power:

\(\displaystyle P=\frac{V^2}{R}\) 

We know that \(\displaystyle V=50V\). To determine resistance of 3 resistors in parallel we set up the equation for total resistance for resistors in parallel:

\(\displaystyle \frac{1}{R_{total}}=\frac{1}{3 \Omega}+\frac{1}{3 \Omega}+\frac{1}{3 \Omega}\)

\(\displaystyle \frac{1}{R_{total}}=3*(\frac{1}{3 \Omega})\)

\(\displaystyle R_{total}=1\Omega\)

From our formula we know that the circuit power is:

\(\displaystyle P=\frac{(50V)^2}{1\Omega}=2500 W\)

Example Question #142 : Circuits

Consider a series circuit with a \(\displaystyle 18 V\) battery. The circuit is composed of three resistors, \(\displaystyle R_1=10 \Omega, R_2 = 3 \Omega, R_3 = 7 \Omega\). Calculate the power \(\displaystyle P\) (units of Watts \(\displaystyle W\)) of the circuit.

Possible Answers:

\(\displaystyle 16.2 W\)

\(\displaystyle 32.4 W\)

\(\displaystyle 0.9 W\)

\(\displaystyle 108.0W\)

Correct answer:

\(\displaystyle 16.2 W\)

Explanation:

We can write the power of a circuit as: 

\(\displaystyle P=IV\)

This however, does not help much because we do not know the current in the circuit. But by Ohm's law: 

\(\displaystyle V=IR\Rightarrow I= \frac{V}{R}\)

Substitution of this expression into the expression for power gives us:

\(\displaystyle P = \frac{V^2}{R}\)

Where \(\displaystyle R= R_{tot}\), the total resistance of the circuit. For a circuit in series, the total resistance is simply the sum of the resistors:

\(\displaystyle R_{tot}=R_1+R_2+R_3 = (10+3+7) \Omega=20 \Omega\) 

Therefore, the power of the circuit is:

\(\displaystyle P=\frac{(18V)^2}{20 \Omega}=16.2W\)

Example Question #184 : Electricity

Calculate the power \(\displaystyle P\) of a circuit composed of a \(\displaystyle 12V\) battery and three resistors in parallel. The resistors have resistance values of \(\displaystyle R_1 = 10 \Omega, R_2 = 20 \Omega, R_3 = 5 \Omega\)

Possible Answers:

\(\displaystyle 50.35 W\)

\(\displaystyle 0.34 W\)

\(\displaystyle 4.20W\)

\(\displaystyle 4.1 W\)

Correct answer:

\(\displaystyle 50.35 W\)

Explanation:

The power \(\displaystyle P\) of a circuit is given by:

\(\displaystyle P=IV\).

By substitution of Ohm's law, we can show that:

\(\displaystyle P=\frac{V^2}{R}\)

Where \(\displaystyle R=R_{tot}\), or the total resistance of the circuit. For resistors in parallel, we add the inverses of their values. This is shown more plainly as: 

\(\displaystyle \frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\) 

From the given values, we can calculate the total resistance as \(\displaystyle R_{tot}=2.86 \Omega\). This leads us to the final answer:

\(\displaystyle P=\frac{(12V)^2}{2.86 \Omega}=50.35 W\)

Example Question #141 : Circuits

You are given the following circuit powered by a \(\displaystyle 10V\) battery:

Circuit

\(\displaystyle R1=5k\Omega\)

\(\displaystyle R2=8k\Omega\)

\(\displaystyle R3=12k\Omega\)

Find the power supplied by the battery.

Possible Answers:

\(\displaystyle 8mW\)

\(\displaystyle 4mW\)

\(\displaystyle 12mW\)

\(\displaystyle 2.5mW\)

\(\displaystyle 12mW\)

Correct answer:

\(\displaystyle 4mW\)

Explanation:

The power provided by the battery is equal to the product of the current that passes through it and the voltage it provides. In order to find the current, you must find the equivalent resistance of the circuit.

Since R1 is in series with R2 and R3, who are in parallel, the equivalent resistance is given by: 

\(\displaystyle R_{equivalent}=R1+\frac{R2*R3}{R2+R3}=5k\Omega+8k\Omega+12k\Omega=25k\Omega\)

Then, to find power, you can use the following formula:

\(\displaystyle P=IV=\frac{V^2}{R}=\frac{10^2}{25k\Omega}=4mW\)

The battery supplies \(\displaystyle 4mW\) of power.

Example Question #1401 : Ap Physics 1

How much power does a circuit produce if there is an equivalent resistance of \(\displaystyle 10 \Omega\) and a current of \(\displaystyle 2 A\)?

Possible Answers:

\(\displaystyle 40 W\)

\(\displaystyle 20 W\)

\(\displaystyle 400 W\)

\(\displaystyle 100 W\)

\(\displaystyle 200 W\)

Correct answer:

\(\displaystyle 40 W\)

Explanation:

The equation for power in a circuit is given as:

\(\displaystyle P = I^{2}R\)

\(\displaystyle P\) is the power in Watts, \(\displaystyle I\) is the current in amps, and \(\displaystyle R\) is the resistance in ohms. By solving the equation and substituting our values, we obtain:

\(\displaystyle P = (2 A)^{2}(10\Omega ) = 40 W\)

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