What voltage is involved in a 1.44 kW short circuit through a $0.100\Omega$ resitance?

Video Transcript

This is College Physics Answers with Shaun Dychko. A short circuit with a resistance of 0.100 ohms dissipates a power of 1.44 kilowatts, which is 1.44 times 10 to the 3 watts and the question is what voltage is involved in this circuit to make this power it dissipated with this resistance? So the formula for power is voltage squared divided by resistance and we can solve this for V squared anyhow by multiplying both sides by R and you get V squared is power times R and then take the square root of both sides to solve for V— that's the square root of power and resistance. So that's square root of 1.44 times 10 to the 3 watts times 0.100 ohms which is 12.0 volts.