A current of 0.50 A passes through a lamp with a resistance of 600 $\Omega$.The power supplied to this lamp is
A)2.0 W
B)0.10 kW
C)0.12 kW
D)0.15 kW
E)0.30 kW

Question

Quizplus · Accepted Answer

The power supplied to the lamp can be calculated using the formula P = I^2 * R, where I is the current and R is the resistance. Substituting the given values, P = (0.50 A)^2 * 600 Ω = 0.25 * 600 = 150 W, which is equivalent to 0.15 kW.

A Current of 0 Ω\OmegaΩ The Power Supplied to This Lamp Is A)2

A Current of 0 $\Omega$ The Power Supplied to This Lamp Is
A)2