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AP Physics 1 : Electricity and Waves

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Example Questions

Example Question #211 : Electricity And Waves

You have the following circuit. The values for the components are:

$R1=3k\Omega$

$R2=R3=4k\Omega$

$V_{SOURCE}=10V$

Circuit

What is the current passing though R3?

Possible Answers:

2.5A

1.25A

Correct answer:

Explanation:

The first step is to find the equivalent resistance of the entire circuit. Note that R2 and R3 are in parallel, and they are in series with R1. Therefore, in order to find the equivalent resistance:

$(R2\left | \right |R3)+R1=\frac{R2*R3}{R2+R3}+R1=\frac{4k\Omega*4k\Omega}{4k\Omega+4k\Omega}+3k\Omega$

$R_{equivalent}=2k\Omega+3k\Omega$

$R_{equivalent}=5k\Omega$

Then, in order to find the total current, use Ohm's Law:

$V=IR\rightarrow I=\frac{V}{R}\rightarrow I=\frac{10}{5}A=2A$

That means that will pass through R1, and will split when it reaches the junction between R2 and R3 based on the ratio of the two resistance:

$I_{R3}=\frac{R2}{R2+R3}I_{source}$

$I_{R3}=\frac{1}{2}(2A)=1A$

of current passes through the resistor R3.

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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:

$0.54\Omega$

$0.29\Omega$

$1.5\Omega$

$2.6\Omega$

$0.77\Omega$

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 cross-sectional area of the cord

Decreasing the cross-sectional area of the cord

Increasing the length of the cord

Increasing the resistivity

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 #1 : 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:

960m

10cm

9.6m

38m

2.6cm

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 #61 : Circuits

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 #111 : Circuits

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:

3mm

1mm

3000mm

150mm

15mm

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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AP Physics 1 : Electricity and Waves

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #211 : Electricity And Waves

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 #1 : Resistivity

Example Question #61 : Circuits

Example Question #111 : Circuits

All AP Physics 1 Resources

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