A direct-current (DC) power supply maintains a constant potential difference of 15\text{ V… — Armed Services Vocational Aptitude Battery (ASVAB)

A direct-current (DC) power supply maintains a constant potential difference of $15\text{ V}$ across a circuit resistor that dissipates $4.5\text{ W}$ of electrical power. What is the electrical resistance of the resistor in ohms ( $\Omega$ )?

Answer: 50 ohms

Answer

The resistance of the resistor is

50\ \Omega

According to Watt's Law, power dissipated by a resistor with a known voltage across it is given by

P = \frac{V^2}{R}

. Solving for resistance gives

R = \frac{V^2}{P} = \frac{15^2}{4.5} = \frac{225}{4.5} = 50\ \Omega

Step-by-Step Solution

State the known values and identify the target variable.

Voltage

V = 15\text{ V}

, Power

P = 4.5\text{ W}

, Resistance

R = ?

Establishing the known electrical parameters determines the appropriate form of Watt's Law to apply.

Select and rearrange the Watt's Law power equation that relates power, voltage, and resistance.

R = \frac{V^2}{P}

Since voltage and power are known, using

P = \frac{V^2}{R}

allows direct solving for resistance.

Calculate the squared voltage.

V^2 = 15^2 = 225\text{ V}^2

Evaluating the numerator is the first calculation step in solving for

R

Divide the squared voltage by the power dissipation.

R = \frac{225}{4.5} = 50\ \Omega

Carrying out the division yields the final resistance value in ohms.

Key Concept

Watt's Law relates electrical power, voltage, and resistance through the formula

P = \frac{V^2}{R}

, which rearranges algebraically to

R = \frac{V^2}{P}

Alternative Method

First find the circuit current using

I = \frac{P}{V} = \frac{4.5\text{ W}}{15\text{ V}} = 0.3\text{ A}

. Then apply Ohm's Law to determine the resistance:

R = \frac{V}{I} = \frac{15\text{ V}}{0.3\text{ A}} = 50\ \Omega

Estimated Time:1m 0s

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Step-by-Step Solution

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