How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers will be deleted, and reported. (Please show some work)

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Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω

Power = 55 W

Power is the product of current and voltage. That is

P = IV …… (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

55 = 60 × I^2

Make I^2 the subject of formula

I^2 = 55/60

I^2 = 0.92

I = sqr(0.92)

I = 0.957 A

Therefore, 0.96 A current must be applied.