If the power rating of a 400-Ω resistor is 0.80 W,what is the maximum current it can safely draw?
A)45 mA
B)18 mA
C)4.4 mA
D)2.0 mA
E)320 mA

Question

Quizplus · Accepted Answer

The power rating of a resistor is given by the equation P = I^2 * R, where P is the power in watts, I is the current in amps, and R is the resistance in ohms. In this case, we have a 400-Ω resistor with a power rating of 0.80 W. To find the maximum current it can safely draw, we can rearrange the equation to solve for I:I = sqrt(P / R) = sqrt(0.80 W / 400 Ω) = 0.045 A = 45 mATherefore, the maximum current the resistor can safely draw is 45 mA.

If the Power Rating of a 400-Ω Resistor Is 0