An electric heater of resistance 20\,\Omega carries a current of 5\,\text{A} when operated… — JAMB UTME

An electric heater of resistance $20\,\Omega$ carries a current of $5\,\text{A}$ when operated. What is the electrical power dissipated by the heater?

$500\,\text{W}$ Answer
B
$100\,\text{W}$
C
$4\,\text{W}$
D
$2000\,\text{W}$

Answer

500\,\text{W}

According to Joule's law of electric heating, electrical power

P

dissipated in a resistor of resistance

R

carrying current

I

is given by

P = I^2 R

. Substituting the given values

I = 5\,\text{A}

and

R = 20\,\Omega

gives

P = (5\,\text{A})^2 \times 20\,\Omega = 25 \times 20 = 500\,\text{W}

Step-by-Step Solution

Identify the given quantities from the problem statement

Current

I = 5\,\text{A}

and resistance

R = 20\,\Omega

These parameters are required to calculate the power dissipated.

Select the appropriate formula for power in terms of current and resistance

P = I^2 R

Joule's heating law specifies that power dissipation is given by

P = I^2 R

Substitute the values and evaluate the numerical result

P = (5)^2 \times 20 = 25 \times 20 = 500\,\text{W}

Squaring

5

yields

25

, and multiplying by

20

gives the final power output of

500\,\text{W}

Key Concept

Electrical Power Dissipation in a Resistor

Estimated Time:45s

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