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AP Physics 1 : Circuit Power

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Example Question #21 : Circuit Power

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 will stay the same

Power will double

Power wil be four times larger

Power will triple

Correct answer:

Power wil be four times larger

Explanation:

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

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

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

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

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

$P=\frac{V^{2}}{R} \right$

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

Expressed mathematically,

If V=2V

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

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Example Question #22 : Circuit Power

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

Possible Answers:

6050J

12500J

600J

35J

Correct answer:

6050J

Explanation:

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

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

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

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

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

$P=\frac{V^{2}}{R} \right$

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

Substituting theequation for power into the equation for energy gives

$E=P\cdot t=\frac{V^{2}}{R} t$

For our problem

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

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Example Question #21 : Circuit Power

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

Possible Answers:

1000Amps

2Amps

10Amps

1Amp

Correct answer:

2Amps

Explanation:

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

Solving the power equation for current gives

$I=\sqrt{\frac{P}{R}}$

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

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Example Question #181 : Electricity

How much energy does a 75W lightbulb give off in 5min ?

Possible Answers:

22500J

125J

Not enough information

12000J

Correct answer:

22500J

Explanation:

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

Time needs to be in seconds.

$5min\times \frac{60sec}{1min}=300 sec$

For our problem, energy is

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

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Example Question #1401 : Ap Physics 1

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

Possible Answers:

50W

7500W

278 W

2500W

Correct answer:

2500W

Explanation:

Recall the formula for circuit power:

$P=\frac{V^2}{R}$

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

$\frac{1}{R_{total}}=\frac{1}{3 \Omega}+\frac{1}{3 \Omega}+\frac{1}{3 \Omega}$

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

$R_{total}=1\Omega$

From our formula we know that the circuit power is:

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

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Example Question #183 : Electricity

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

Possible Answers:

32.4 W

0.9 W

16.2 W

108.0W

Correct answer:

16.2 W

Explanation:

We can write the power of a circuit as:

P=IV

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

$V=IR\Rightarrow I= \frac{V}{R}$

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

$P = \frac{V^2}{R}$

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

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

Therefore, the power of the circuit is:

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

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Example Question #184 : Electricity

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

Possible Answers:

4.1 W

0.34 W

50.35 W

4.20W

Correct answer:

50.35 W

Explanation:

The power of a circuit is given by:

P=IV .

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

$P=\frac{V^2}{R}$

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

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

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

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

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Example Question #185 : Electricity

You are given the following circuit powered by a 10V battery:

Circuit

$R1=5k\Omega$

$R2=8k\Omega$

$R3=12k\Omega$

Find the power supplied by the battery.

Possible Answers:

4mW

12mW

8mW

2.5mW

Correct answer:

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:

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

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

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

The battery supplies 4mW of power.

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Example Question #186 : Electricity

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

Possible Answers:

40 W

400 W

200 W

100 W

20 W

Correct answer:

40 W

Explanation:

The equation for power in a circuit is given as:

$P = I^{2}R$

is the power in Watts, is the current in amps, and is the resistance in ohms. By solving the equation and substituting our values, we obtain:

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

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Example Question #1411 : Ap Physics 1

A capacitor with a capacitance of 10 F stores 500 J of energy. What is the voltage across this capacitor?

Possible Answers:

50 V

500 V

10 V

100 V

20 V

Correct answer:

10 V

Explanation:

The correct answer is 10 V .

We can obtain this answer with the following equation regarding capacitors:

$U= \frac{1}{2}CV^{2}$

is the energy stored on the capacitor, is the capacitance, and is the voltage.

$(500J)=\frac{1}{2}(10F)(V^{2})$

Solving for we obtain:

V= 10 V

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AP Physics 1 : Circuit Power

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #21 : Circuit Power

Example Question #22 : Circuit Power

Example Question #21 : Circuit Power

Example Question #181 : Electricity

Example Question #1401 : Ap Physics 1

Example Question #183 : Electricity

Example Question #184 : Electricity

Example Question #185 : Electricity

Example Question #186 : Electricity

Example Question #1411 : Ap Physics 1

All AP Physics 1 Resources

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