McSnowy
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The power in a lightbulb is given by the equation P- FR where /is the current flowing through the lightbulb and Ris the resistance of the lightbulb. What is the current in a circuit that has a resistance of 30.0 o and a power of 2.00 W?

A.) 15.0 A
B.) 3.87 A
C.) 0.258 A
D.) 0.067 A

(PLEASE HELP NEED ANSWER ASAP)

Respuesta :

POWER OF CIRCUIT

P=VI

While,

VOLTAGE OF CIRCUIT

V=IR

THEN,

P=I^R

P/R=I^

2.00/30.0=I^

I=0.258A.

Answer:0.258

Explanation:

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