If the power rating of a $400 - \Omega$ resistor is $0.800 \mathrm {~W}$, what is the maximum voltage that can safely be connected across the resistor?
A) $170 \mathrm {~V}$
B) $17.9 \mathrm {~V}$
C) $110 \mathrm {~V}$
D) $1.80 \mathrm {~V}$

Question

Quizplus · Accepted Answer

The power rating of the resistor can be used with the formula $P = \frac{V^2}{R}$ to find the maximum voltage. Rearranging for $V$, we get $V = \sqrt{P \times R} = \sqrt{0.800 \times 400} = \sqrt{320} \approx 17.9 \mathrm{~V}$.

If the Power Rating of A 400−Ω400 - \Omega400−Ω Resistor Is 0.800 W0.800 \mathrm {~W}0.800 W

If the Power Rating of A $400 - \Omega$ Resistor Is $0.800 \mathrm {~W}$