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AP Physics 1 : Electricity

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Example Questions

Example Question #1 : Circuits

What is the current through a resistor if the resistor has a resistance of $40\Omega$ and the voltage across the resistor is 60V ?

Possible Answers:

46A

24A

1.5A

Correct answer:

1.5A

Explanation:

Use Ohm's law.

V=IR

$I=\frac{60V}{40\Omega}=1.5A$

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Example Question #1 : Circuits

If the current through a $10\Omega$ resistor is , what is the voltage across the resistor?

Possible Answers:

10V

$\frac{5}{3}V$

60V

Correct answer:

60V

Explanation:

UseOhm's law.

V=IR

Plug in known values and solve.

$V=6A\cdot 10\Omega$

V=60V

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Example Question #41 : Electricity

Basic circuit

$V_{in}=12V$

I=2 A

The voltage measured from a point between Z_1 and Z_2 to the ground is

What is the resistance of Z_1 ?

Possible Answers:

$2\Omega$

$4\Omega$

$3\Omega$

$12\Omega$

$1\Omega$

Correct answer:

$2\Omega$

Explanation:

Begin by finding the total resistance in the circuit.

$R_{tot}=\frac{V}{I}=\frac{12V}{2A}=6\Omega$

Now note that the voltage identified in the problem is the same as the voltage drop across the second resistor:

IZ_2=V_D

$Z_2=\frac{V_D}{I}=\frac{8V}{2A}$

$Z_2=4\Omega$

Now, since Z_1 and Z_2 combine to form the total resistance:

$Z_1=Z-Z_2=6\Omega -4\Omega=2\Omega$

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Example Question #1 : Ohm's Law

Basic circuit

$V_{in}=12V$

I=2 A , and the voltage measured from a point between Z_1 and Z_2 to the ground is

In the circuit above, what is the resistance of Z_2 ?

Possible Answers:

$6\Omega$

$4\Omega$

$1\Omega$

$12\Omega$

$2\Omega$

Correct answer:

$4\Omega$

Explanation:

Begin by finding the total resistance in the circuit.

$R_{tot}=\frac{V}{I}=\frac{12V}{2A}=6\Omega$

Now note that the voltage identified in the problem is the same as the voltage drop across the second resistor:

IZ_2=V_D

$Z_2=\frac{V_D}{I}=\frac{8V}{2A}$

$Z_2=4\Omega$

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Example Question #42 : Electricity

Basic circuit

$V_{in}=12V$

Z_1 is composted of two resistors in parallel, $3\Omega$ and $7\Omega$

Z_2 is a single $6\Omega$ resistor.

In the circuit above, what is the current?

Possible Answers:

2.96A

0.75 A

1.20A

1.50 A

1.48A

Correct answer:

1.48A

Explanation:

To find the current, first find the total resistance of the circuit. Begin by simplifying Z_1 , the two resistors in parallel as follows:

$\frac{1}{Z_1}=\frac{1}{3\Omega}+\frac{1}{7\Omega}$

$\frac{1}{Z_1}=\frac{10}{21\Omega}$

$Z_1=2.1\Omega$

Since Z_1 and Z_2 are in series, their combined resistance is:

$R_{tot}=Z_1+Z_2=2.1\Omega+6\Omega=8.1\Omega$

Use Ohm's law to find the current.

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

$I=\frac{12V}{8.1\Omega}$

I=1.48A

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Example Question #1 : Ohm's Law

A resistor with a resistance of $4 \Omega$ has a current flowing through it of 5A. What is the potential drop across the resistor?

Possible Answers:

10V

$\frac{2}{5}V$

$\frac{5}{2}V$

20 V

Correct answer:

20 V

Explanation:

Ohm's law states that the potential drop across a resistor is equal to the product of the current flowing through the resistor and the resistance of the resistor:

V=IR

We were given the current, I, and the resistance, R, so we simply multiply the two together to get our final answer.

V=5*4=20V

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Example Question #11 : Ohm's Law

If a closed circuit connected to a $12\:V$ battery has a resistance of $2\: \Omega$ , what is the current flowing through this circuit?

Possible Answers:

$24\: Amperes$

$144\: Amperes$

$0.16\:Amperes$

$6\:Amperes$

Correct answer:

$6\:Amperes$

Explanation:

This question can be solved by making use of Ohm's law, which states that the voltage difference across a circuit is proportional to the current flowing through the circuit, as well as to the resistance of the circuit. Written in equation form, we have:

V=iR

Solving for current, we can rearrange to obtain:
$i=\frac{V}{R}=\frac{12\: V}{2\: \Omega }=6\:Amperes$

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Example Question #43 : Electricity

Determine the voltage drop across wire that is connected to two resistors in series with resistances $5\Omega$ and $10\Omega$ , with a current flowing through the circuit of 2.5A ?

Possible Answers:

37.5V

$\frac{25}{3}V$

22.5V

Correct answer:

37.5V

Explanation:

By Ohm's law:

$\Delta V=IR_{total}$ , where $\Delta V$ is the voltage drop across the wire. is the current flowing through the wire, and $R_{total}$ is the total resistance within the circuit.

Since resistors are in series:

$R_{total}=R_1+R_2$ , where R_1 and R_2 are the resistances of the two resistors.

In our case:

$R_{total}=5\Omega+10\Omega=15 \Omega$

I=2.5A

Therefore:

$\Delta V=(2.5A)(15\Omega)=37.5V$

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Example Question #13 : Ohm's Law

A current of 10A passes through a circuit. A single resistor in this circuit has a resistance of $2.5 \Omega$ . What is the voltage drop across this resistor?

Possible Answers:

0.25V

25V

Correct answer:

25V

Explanation:

We need to use Ohm's law here which is given by:

V=IR

Where is the voltage in Volts, is the current in Amperes, and is the resistance in Ohms. We know the current in the circuit as well as the resistance of the resistor. We substitute our known values and solve for Voltage which will give us the voltage drop across the resistor.

$V = (10 A)(2.5\Omega )= 25V$

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Example Question #14 : Circuits

A circuit has a 1.5V power source and a $25\Omega$ resistance. What is the current in the circuit?

Possible Answers:

0.06

4.5

Correct answer:

0.06

Explanation:

We use Ohm's law, V=IR , to find the current in the circuit. In Ohm's law is the voltage in the circuit, is the current in the circuit and is the circuit's resistance.

Solving the equation for , we have

I=V/R=1.5/25=0.06Amps

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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 #1 : Circuits

Example Question #1 : Circuits

Example Question #41 : Electricity

Example Question #1 : Ohm's Law

Example Question #42 : Electricity

Example Question #1 : Ohm's Law

Example Question #11 : Ohm's Law

Example Question #43 : Electricity

Example Question #13 : Ohm's Law

Example Question #14 : Circuits

All AP Physics 1 Resources

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