Question

Suppose a 1.0-kQ resistor has a "power rating" of 500 mW. Meaning: the resistor can only dissipate 1/2 W before failure. What is the maximum current I in mA that it can handle? MA

Answer 1

Step-by-step explanation:

Power = I * I * R

. 5 A = I * I * 1000 Ω

.5 / 1000 = I * I

I = 22.3 mA