An electric heating element is rated to dissipate 1.8\text{ kW} of power when connected ac… — Armed Services Vocational Aptitude Battery (ASVAB)

An electric heating element is rated to dissipate $1.8\text{ kW}$ of power when connected across a standard $120\text{ V}$ DC supply line. If a control circuit reduces the direct current flowing through the heating element to $5\text{ A}$ while its internal resistance remains constant, what is the new power dissipated by the element?

A
$600\text{ W}$
$200\text{ W}$ Answer
C
$40\text{ W}$
D
$0.2\text{ W}$

Answer

200\text{ W}

The correct answer of

200\text{ W}

is found by first converting

1.8\text{ kW}

1800\text{ W}

. Using the initial ratings, the element's resistance is

R = \frac{V^2}{P} = \frac{120^2}{1800} = 8\text{ }\Omega

. With the current controlled to

5\text{ A}

, the dissipated power is

P = I^2 R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

Step-by-Step Solution

Convert the rated power from kilowatts to watts

P_1 = 1.8\text{ kW} = 1800\text{ W}

Metric prefixes must be converted to base SI units before performing Watt's Law calculations.

Calculate the constant internal resistance (

R

) of the heating element using the initial rating

R = \frac{V_1^2}{P_1} = \frac{(120\text{ V})^2}{1800\text{ W}} = \frac{14400}{1800} = 8\text{ }\Omega

The heating element's resistance remains unchanged and serves as the bridge to calculate power under the new operating current.

Calculate the new power dissipation with the reduced current (

I_2 = 5\text{ A}

)

P_2 = I_2^2 \times R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

Watt's Law specifies that power dissipation across a known resistance is directly proportional to the square of the current.

Key Concept

Watt's Law Power and Resistance Relationships (

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

and

P = I^2 R

)

Alternative Method

Determine the initial current

I_1 = \frac{P_1}{V_1} = \frac{1800\text{ W}}{120\text{ V}} = 15\text{ A}

. Since

P \propto I^2

for a constant resistance, scaling the power by the ratio of the currents squared gives

P_2 = P_1 \times \left(\frac{I_2}{I_1}\right)^2 = 1800\text{ W} \times \left(\frac{5\text{ A}}{15\text{ A}}\right)^2 = 1800 \times \left(\frac{1}{3}\right)^2 = \frac{1800}{9} = 200\text{ W}

Estimated Time:1m 30s

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