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

Question

Quizplus · Accepted Answer

The maximum current can be calculated using the power formula $P = I^2R$, where $P = 0.80 \, \text{W}$ and $R = 400 \, \Omega$. Solving for $I$, we get $I = \sqrt{\frac{P}{R}} = \sqrt{\frac{0.80}{400}} = 0.045 \, \text{A}$ or $45 \, \text{mA}$.

If the Power Rating of A 400−Ω400 - \Omega400−Ω Resistor Is 0.80 W0.80 \mathrm {~W}0.80 W

If the Power Rating of A $400 - \Omega$ Resistor Is $0.80 \mathrm {~W}$