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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 delet
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Answers ( )
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.