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The power in a lightbulb is given by the equation P =
I^(2)R, where I is the current flowing through the lightbulb and R is the resistance of the lightbulb. What is the current in a circuit that has a resistance of 30.0 (LOOK AT IMAGE ATTACHED FOR SYMBOL) and a power of 55.0 W??

A. 1.83 A
B. 0.740 A
C. 1.35 A
D. 0.550 A

The power in a lightbulb is given by the equation P = I^(2)R, where I is the current-example-1

1 Answer

4 votes

Answer

1.35 A


Step-by-step explanation

Power is the rate of doing work. It is given by:

Power = I²R

Power = 55.0 W and resistance = 30.0Ω

55.0 = I² × 30.0

I² = 55/30

= 1.833333

I = √1.83333

= 1.35 A

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