The electric power P dissipated in a resistor varies directly as the square of the current… — JAMB UTME

The electric power $P$ dissipated in a resistor varies directly as the square of the current $I$ flowing through it. If a current of $3\text{ A}$ produces a power of $45\text{ W}$ , what is the power dissipated, in watts, when the current is $5\text{ A}$ ?

Answer: 125 W

Answer

The power dissipated when the current is 5 A is 125 W.

Because electric power varies directly as the square of the current, the formula is

P = k I^2

. Substituting the initial conditions gives

45 = k(3^2) = 9k

, so

k = 5

. Evaluating at

I = 5\text{ A}

gives

P = 5(5^2) = 125\text{ W}

Step-by-Step Solution

Set up the variation equation

P = k I^2

Power varies directly as the square of current.

Find the constant of variation

k

k = 5

Substitute

P = 45

and

I = 3

into the variation equation:

45 = k(3^2) \implies 45 = 9k \implies k = 5

Calculate the required power for

I = 5\text{ A}

P = 125\text{ W}

Substitute

k = 5

and

I = 5

into

P = k I^2

P = 5(5^2) = 125

Key Concept

Direct variation involving a squared quantity

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