An electric heating element rated at 1000\,\text{W} and 220\,\text{V} is connected to a 11… — JAMB UTME

An electric heating element rated at $1000\,\text{W}$ and $220\,\text{V}$ is connected to a $110\,\text{V}$ power line. Assuming the electrical resistance of the heating element remains constant, what is the power dissipated by the heater when operating at this reduced voltage?

$250\,\text{W}$ Answer
B
$500\,\text{W}$
C
$1000\,\text{W}$
D
$2000\,\text{W}$

Answer

250\,\text{W}

The electrical power rating of an appliance defines its fixed resistance via

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

. With

R = 48.4\,\Omega

, connecting the appliance to a

110\,\text{V}

supply yields

P = \frac{110^2}{48.4} = 250\,\text{W}

. Alternatively, because

P \propto V^2

for a constant resistance, halving the voltage reduces the power by a factor of

(1/2)^2 = 1/4

, giving

1000\,\text{W} \times \frac{1}{4} = 250\,\text{W}

Step-by-Step Solution

Calculate the resistance

R

of the heater from its rated values.

R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{220^2}{1000} = \frac{48400}{1000} = 48.4\,\Omega

The resistance of a heating element is determined by its physical design and rated specifications.

Calculate the power

P_{\text{new}}

dissipated when connected to the

110\,\text{V}

supply.

P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{110^2}{48.4} = \frac{12100}{48.4} = 250\,\text{W}

Electric power dissipated across a constant resistance varies with the square of the applied voltage.

Key Concept

Relationship between voltage, resistance, and electrical power dissipation

Estimated Time:1m 30s

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