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AP Physics 1 : Electricity

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Example Questions

Example Question #212 : Electricity And Waves

Which of the following wires with a current running through them would have the least resistance?

Possible Answers:

They would all have equal resistance

Wire B: 1 V and 100 A

Wire C: 100 V and 1 A

Wire D: 1000 V and .1 A

Wire A: 10 V and 10 A

Correct answer:

Wire B: 1 V and 100 A

Explanation:

The correct answer is Wire B with 1 V and 100 A of current.

The formula for voltage, current and resistance is as follows:

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

Thus, the wire with the lowest ratio of voltage to current will have the least resistance.

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Example Question #11 : Current And Voltage

Series curcuit

What is the current in the given circuit if $R_1=10\Omega$ , $R_2=15\Omega$ and V=25V ?

Possible Answers:

25A

10A

Correct answer:

Explanation:

To find the current we must first find the equivalent resistance. For resistors in series, the equivalent resistance is

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

For this problem

$R_{eq}=R_1+R_2=10+15=25\Omega$

Now we use Ohm's law, $V=IR_{eq}$ , to find the current,

$V=IR_{eq}\Rightarrow I=\frac{V}{R_{eq}}=\frac{25}{25}=1A$

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Example Question #1363 : Ap Physics 1

Parallel circuit

What is the current in the given circuit if $R_1=10\Omega$ , $R_2=15\Omega$ and V=25V ?

Possible Answers:

2.153A

4.167A

8.166A

Correct answer:

4.167A

Explanation:

To find the current we must first find the equivalent resistance. For resistors in parallel, the equivalent resistance is

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

For this problem

$\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$

$\frac{1}{R_{eq}}=\frac{5}{30}$

$R_{eq}=\frac{30}{5}=6\Omega$

Now we use Ohm's law, $V=IR_{eq}$ , to find the current,

$V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{6}\approx 4.167A$

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Example Question #21 : Current And Voltage

What is the current in the given circuit if $R_1=10\Omega$ , $R_2=15\Omega$ , $R_3=5\Omega$ and V=25V ?

Possible Answers:

15A

Correct answer:

Explanation:

To find the current we must first find the equivalent resistance. This problem is a combination of resistors in series and parallel. First we find the equivalent resistance of resistors in series.

For resistors in series, the equivalent resistance is:

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

For this problem

$R_{eq1}=R_1+R_2=10+15=25$

The resistors in series can be treated as one $25\Omega$ resistor. Now we address the resistors in parallel. For resistors in parallel, the equivalent resistance is:

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

For this problem

$\frac{1}{R_{eq}}=\frac{1}{R_{eq1}}+\frac{1}{R_3}=\frac{1}{25}+\frac{1}{5}=\frac{1+5}{25}$

$\frac{1}{R_{eq}}=\frac{6}{25}$

$R_{eq}=\frac{25}{6} \approx 4.167\Omega$

Now we use Ohm's law, $V=IR_{eq}$ , to find the current,

$V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{4.167}=6A$

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Example Question #1 : Resistivity

Consider the circuit:

Circuit_4

$R1=2\Omega,\ R2=4\Omega,\ R3=R4$

The current flowing through the entire circuit is 40A . What is the value of R3?

Possible Answers:

$1.5\Omega$

$0.29\Omega$

$0.54\Omega$

$0.77\Omega$

$2.6\Omega$

Correct answer:

$0.77\Omega$

Explanation:

We know the voltage and total current of the circuit, so we can calculate an equivalent resistance using Ohm's law:

$V = IR_{eq}$

$R_{eq}=\frac{V}{I} = \frac{12V}{40A}= 0.3 \Omega$

Now we can use the expression for condensing parallel resistors to calculate R3:

$\frac{1}{R_{eq}}= \sum\frac{1}{R}= \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}$

$\frac{1}{0.3\Omega} = \frac{1}{2\Omega}+\frac{1}{4\Omega} +\frac{1}{R3}+\frac{1}{R4}=\frac{1}{2\Omega}+\frac{1}{4\Omega} +\frac{2}{R3}$

$R3 = 0.77\Omega$

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Example Question #1 : Resistivity

Which of the following factors would decrease the resistance through an electrical cord?

Possible Answers:

Increasing the resistivity

Increasing the length of the cord

Decreasing the cross-sectional area of the cord

Increasing the cross-sectional area of the cord

Correct answer:

Increasing the cross-sectional area of the cord

Explanation:

The equation for resistance is given by $R = \frac{\rho L}{A}$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

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Example Question #3 : Resistivity

An electrician wishes to cut a copper wire $(\rho = 1.724 * 10^{-8}\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Possible Answers:

2.6cm

960m

38m

9.6m

10cm

Correct answer:

960m

Explanation:

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = \frac{\rho L}{A}$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = \frac{RA}{\rho }$ . We must first solve for A using the radius, 0.725mm.

$A = \pi r^{2} = \pi (0.000725m)^{2}=1.65 * 10^{-6}m^{2}$

Plugging in our numbers gives the answer, 960m.

$L = \frac{10\Omega * 1.65 * 10^{-6}m^{2}}{1.724*10^{-8}\Omega m} = 957m \approx960m$

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Example Question #141 : Electricity

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Possible Answers:

1.5A

0.25A

Correct answer:

Explanation:

Resistors in serries add according to the formula:

$R_{f}=R_{1}+R_{2}$

Resistors in parallel add according to the formula:

$\frac{1}{R_{f}}=\frac{1}R_{1}{}+\frac{1}{R_{2}}$

We can find the total equivalent resistance:

$\frac{1}{R_{f}}=\frac{1}{2+3}+\frac{1}{5}=\frac{2}{5}$

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

V=IR

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

Solve:

$I = \frac{5 V}{2.5 \Omega }=2 A$

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Example Question #2 : Resistivity

Basic circuit

$V_{in}=12V$

$Z_1=3\Omega$

In the circuit above, your aim is to limit the current to , so you must design a resistor to serve in the place of Z_2 .

To make this resistor, you have a spool of mystery metal, which has a cross sectional area of 1mm^2 and a resistivity of, $\rho = 3\cdot 10^{-3}\Omega m$ . What length of wire should you cut?

Possible Answers:

150mm

3mm

15mm

1mm

3000mm

Correct answer:

3mm

Explanation:

First, find out how much total resistance should be in the circuit in order to get the desired current:

$R_{tot}=\frac{V}{I}=\frac{12V}{1A}=12\Omega$

Determine what the second resistance should be.

$Z_2=12\Omega-Z_1$

$Z_2=9\Omega$

Find the necessary length of wire.

$Z_2=\frac{\rho L }{A}$

$L=\frac{Z_2A}{\rho}$

$L=\frac{(9\Omega)(1mm^2)}{3\cdot 10^{-3}\Omega m}=\frac{(9\Omega)(10^{-6}m^2)}{3\cdot 10^{-3}\Omega m}$

$L=3\cdot 10^{-3}m=3mm$

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Example Question #1 : Resistivity

What is the resistance of a 100m length of round copper wire with a radius of 0.3mm ?

$\rho_{copper}=1.68\cdot 10^{-8}\Omega m$

Possible Answers:

$118.8\Omega$

$37.7\Omega$

$3.77\Omega$

$5.94\Omega$

$59.4\Omega$

Correct answer:

$5.94\Omega$

Explanation:

Resistance and resistivity are related as follows:

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

$A={\pi}(0.0003m)^2$

$A\approx 2.83\cdot 10^{-7}$

$R\approx \frac{1.68\cdot10^{-8}\Omega m(100m)}{2.83\cdot10^{-7}m^2}$

$R\approx 5.94\Omega$

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AP Physics 1 : Electricity

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #212 : Electricity And Waves

Example Question #11 : Current And Voltage

Example Question #1363 : Ap Physics 1

Example Question #21 : Current And Voltage

Example Question #1 : Resistivity

Example Question #1 : Resistivity

Example Question #3 : Resistivity

Example Question #141 : Electricity

Example Question #2 : Resistivity

Example Question #1 : Resistivity

All AP Physics 1 Resources

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