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AP Physics 1 : Ohm's Law

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Example Questions

Example Question #1 : Ohm's Law

Consider the circuit:

Circuit_4

If each resistor has a value of $2\Omega$ $\displaystyle 2\Omega$, how much current is flowing through the circuit?

Possible Answers:

24A $\displaystyle 24A$

$\displaystyle 6A$

12A $\displaystyle 12A$

$\displaystyle 3A$

48A $\displaystyle 48A$

Correct answer:

24A $\displaystyle 24A$

Explanation:

First we need to calculate the equivalent resistance of the circuit using the following expression for condensing parallel resistors:

$\frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{2\Omega}+\frac{1}{2\Omega}+\frac{1}{2\Omega}+\frac{1}{2\Omega}$ $\displaystyle \frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{2\Omega}+\frac{1}{2\Omega}+\frac{1}{2\Omega}+\frac{1}{2\Omega}$

$R_{eq}=0.5\Omega$ $\displaystyle R_{eq}=0.5\Omega$

Now we can use Ohm's law to calculate the current flowing through the circuit:

V = IR $\displaystyle V = IR$

$I = \frac{V}{R}=\frac{12V}{0.5\Omega} = 24A$ $\displaystyle I = \frac{V}{R}=\frac{12V}{0.5\Omega} = 24A$

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Example Question #1 : Circuits

A light bulb requires 60 W to function properly. If it is connected to a powersupply of 120 A and functions properly, then what is the resitance of the light bulb?

Possible Answers:

$7m\Omega$ $\displaystyle 7m\Omega$

$0.5\Omega$ $\displaystyle 0.5\Omega$

$1\Omega$ $\displaystyle 1\Omega$

$1m\Omega$ $\displaystyle 1m\Omega$

$4m\Omega$ $\displaystyle 4m\Omega$

Correct answer:

$4m\Omega$ $\displaystyle 4m\Omega$

Explanation:

First, identify the given information:

P=60W $\displaystyle P=60W$

I=120A $\displaystyle I=120A$

Two equations are required for this problem:

1.) Ohm's law, V=IR $\displaystyle V=IR$

2.) Electrical power P=IV $\displaystyle P=IV$

Using the equation for electrical power, we can rearrange to solve for $\displaystyle V$:

P=IV $\displaystyle P=IV$

$V=\frac{P}{I}$ $\displaystyle V=\frac{P}{I}$

At this point, we can substitute in the known values and determine the voltage:

$V=\frac{60W}{120A}=0.5V$ $\displaystyle V=\frac{60W}{120A}=0.5V$

Ohm's law can then be rearranged to solve for the resistence of the light bulb:

V=IR $\displaystyle V=IR$

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

The known voltage value then can be substituted into Ohm's law to determine the resistance of the light bulb:

$R=\frac{0.5V}{120A}=0.00416\Omega\approx 4m\Omega$ $\displaystyle R=\frac{0.5V}{120A}=0.00416\Omega\approx 4m\Omega$

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Example Question #1261 : Ap Physics 1

What is the resistance of a resistor if the current going through it is $\displaystyle 2A$ and the voltage across it it is 10V $\displaystyle 10V$?

Possible Answers:

$5\Omega$ $\displaystyle 5\Omega$

$20\Omega$ $\displaystyle 20\Omega$

$12\Omega$ $\displaystyle 12\Omega$

$0.2\Omega$ $\displaystyle 0.2\Omega$

Correct answer:

$5\Omega$ $\displaystyle 5\Omega$

Explanation:

Use Ohm's law.

V=IR $\displaystyle V=IR$

Plug in known values and solve for resistance.

$R=\frac{10V}{2A}=5\Omega$ $\displaystyle R=\frac{10V}{2A}=5\Omega$

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Example Question #1 : Circuits

What is the voltage across a resistor with a resistance of $10\Omega$ $\displaystyle 10\Omega$ that has a current of 20A $\displaystyle 20A$ going through it?

Possible Answers:

30V $\displaystyle 30V$

20V $\displaystyle 20V$

$\displaystyle 2V$

200V $\displaystyle 200V$

Correct answer:

200V $\displaystyle 200V$

Explanation:

Use Ohm's Law.

V=IR $\displaystyle V=IR$

$V=20A\cdot10\Omega =200V$ $\displaystyle V=20A\cdot10\Omega =200V$

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Example Question #1 : Ohm's Law

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

Possible Answers:

46A $\displaystyle 46A$

$\displaystyle 2A$

24A $\displaystyle 24A$

1.5A $\displaystyle 1.5A$

Correct answer:

1.5A $\displaystyle 1.5A$

Explanation:

Use Ohm's law.

V=IR $\displaystyle V=IR$

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

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Example Question #1 : Circuits

If the current through a $10\Omega$ $\displaystyle 10\Omega$ resistor is $\displaystyle 6A$, what is the voltage across the resistor?

Possible Answers:

10V $\displaystyle 10V$

$\frac{5}{3}V$ $\displaystyle \frac{5}{3}V$

60V $\displaystyle 60V$

$\displaystyle 6V$

Correct answer:

60V $\displaystyle 60V$

Explanation:

UseOhm's law.

V=IR $\displaystyle V=IR$

Plug in known values and solve.

$V=6A\cdot 10\Omega$ $\displaystyle V=6A\cdot 10\Omega$

V=60V $\displaystyle V=60V$

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Example Question #7 : Circuits

Basic circuit

$V_{in}=12V$ $\displaystyle V_{in}=12V$

I=2 A $\displaystyle I=2 A$

The voltage measured from a point between Z_1 $\displaystyle Z_1$ and Z_2 $\displaystyle Z_2$ to the ground is $\displaystyle 8V$

What is the resistance of Z_1 $\displaystyle Z_1$ ?

Possible Answers:

$4\Omega$ $\displaystyle 4\Omega$

$12\Omega$ $\displaystyle 12\Omega$

$1\Omega$ $\displaystyle 1\Omega$

$3\Omega$ $\displaystyle 3\Omega$

$2\Omega$ $\displaystyle 2\Omega$

Correct answer:

$2\Omega$ $\displaystyle 2\Omega$

Explanation:

Begin by finding the total resistance in the circuit.

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

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

$\displaystyle IZ_2=V_D$

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

$Z_2=4\Omega$ $\displaystyle Z_2=4\Omega$

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

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

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Example Question #1 : Circuits

Basic circuit

$V_{in}=12V$ $\displaystyle V_{in}=12V$

I=2 A $\displaystyle I=2 A$, and the voltage measured from a point between Z_1 $\displaystyle Z_1$ and Z_2 $\displaystyle Z_2$ to the ground is $\displaystyle 8V$

In the circuit above, what is the resistance of Z_2 $\displaystyle Z_2$?

Possible Answers:

$2\Omega$ $\displaystyle 2\Omega$

$1\Omega$ $\displaystyle 1\Omega$

$4\Omega$ $\displaystyle 4\Omega$

$12\Omega$ $\displaystyle 12\Omega$

$6\Omega$ $\displaystyle 6\Omega$

Correct answer:

$4\Omega$ $\displaystyle 4\Omega$

Explanation:

Begin by finding the total resistance in the circuit.

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

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

$\displaystyle IZ_2=V_D$

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

$Z_2=4\Omega$ $\displaystyle Z_2=4\Omega$

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Example Question #9 : Circuits

Basic circuit

$V_{in}=12V$ $\displaystyle V_{in}=12V$

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

Z_2 $\displaystyle Z_2$ is a single $6\Omega$ $\displaystyle 6\Omega$ resistor.

In the circuit above, what is the current?

Possible Answers:

1.50 A $\displaystyle 1.50 A$

2.96A $\displaystyle 2.96A$

0.75 A $\displaystyle 0.75 A$

1.20A $\displaystyle 1.20A$

1.48A $\displaystyle 1.48A$

Correct answer:

1.48A $\displaystyle 1.48A$

Explanation:

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

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

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

$Z_1=2.1\Omega$ $\displaystyle Z_1=2.1\Omega$

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

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

Use Ohm's law to find the current.

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

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

I=1.48A $\displaystyle I=1.48A$

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Example Question #1 : Circuits

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

Possible Answers:

10V $\displaystyle 10V$

$\frac{5}{2}V$ $\displaystyle \frac{5}{2}V$

$\frac{2}{5}V$ $\displaystyle \frac{2}{5}V$

20 V $\displaystyle 20 V$

Correct answer:

20 V $\displaystyle 20 V$

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 $\displaystyle V=IR$

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

$\displaystyle V=5*4=20V$

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AP Physics 1 : Ohm's Law

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Ohm's Law

Example Question #1 : Circuits

Example Question #1261 : Ap Physics 1

Example Question #1 : Circuits

Example Question #1 : Ohm's Law

Example Question #1 : Circuits

Example Question #7 : Circuits

Example Question #1 : Circuits

Example Question #9 : Circuits

Example Question #1 : Circuits

All AP Physics 1 Resources

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