A headlight has a cold resistance of 0.350 ohm and 12.0 V is applied at the instant it is switched on. What power does it dissipate then?

1. Voltage and resistance are known, so use P = V squared / R. 2. Square the voltage: (12.0 V) squared = 144 V squared. 3. Divide by the cold resistance: P = 144 V squared / 0.350 ohm. 4. That gives P = 411 W at the instant of switch-on. 5. Once hot, the same lamp draws 2.50 A at 12.0 V and dissipates only P = I V = 30.0 W. 6. 30.0 W is therefore the steady value, not the switch-on value, and picking it misses the whole point. 7. 34.3 A is the cold current in amperes, so quoting 34.3 W confuses current with power. _Source: OpenStax College Physics (CC BY 4.0), Ch 20 "Electric Current, Resistance, and Ohm's Law", section Power in Electric Circuits_