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AP Physics C Electricity : Using Ohm's Law

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Example Questions

Example Question #1 : Using Ohm's Law

A simple circuit consists of a $2\ \text{M}\Omega$ resistor that's connected to a battery. How much power is dissipated by the resistor?

Possible Answers:

$5.5*10^{-6}W$

$4*10^{-5}W$

$2*10^{-2}W$

$7.2*10^{-2}W$

$8*10^{-4}W$

Correct answer:

$4*10^{-5}W$

Explanation:

One of the formulas for power is $P=\frac{V^2}{R}$ . We are given the values of the voltage and resistance.

$V=9\ \text{V}$

$R=2\ \text{M}\Omega=2*10^6\ \Omega$

Using these values, we can solve for power.

$P=\frac{(9\ \text{V})^2}{2*10^6\ \Omega}$

$P=4*10^{-5}\ \text{W}$

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Example Question #1 : Using Ohm's Law

Ps0_resistorcircuit

$R_1=15\Omega,\ R_2=20\Omega,\ R_3=20\Omega$

V=50V

What is the current delivered to the curcuit by the battery?

Possible Answers:

4.6A

1.1A

0.91A

2.0A

Correct answer:

2.0A

Explanation:

Using Ohm's law to solve for the value of the current from the battery requires calculation of the equivalent resistance of the circuit.

$V=IR_{eq}$

The resistors R₂ and R₃ are in parallel with one another. Once combined, their requivalent resistor (R₂₃) is in series with R₁.

Ps0_resistorcircuit_2

The equivalent resistance is thus given by:

$R_{eq}=\left ( \frac{1}{R_{2}} + \frac{1}{R_{3}} \right )^{-1} + R_{1}$

$R_{eq}=\left ( \frac{1}{20\Omega} + \frac{1}{20\Omega} \right )^{-1} + 15\Omega$

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

Use this value and the given voltage to solve for the current in the circuit:

$I = \frac{V}{R_{eq}}$

$I=\frac{50V}{25\Omega}=2.0A$

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Using Ohm's Law

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #1 : Using Ohm's Law

Example Question #1 : Using Ohm's Law

All AP Physics C Electricity Resources

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