That same cold headlight dissipates 411 W across 0.350 ohm. What current flows at that moment?

1. Power, current and resistance are linked by P = I squared R. 2. Rearranging gives I = the square root of (P / R). 3. Substitute: I = the square root of (411 W / 0.350 ohm) = the square root of 1174. 4. That gives I = 34.3 A. 5. A cross-check with I = V / R gives 12.0 V / 0.350 ohm = 34.3 A, the same answer. 6. 2.50 A is the steady operating current once the filament is hot, more than thirteen times smaller. 7. Fuses and breakers are built to tolerate this brief surge, which is why slow blow types exist for motors. _Source: OpenStax College Physics (CC BY 4.0), Ch 20 "Electric Current, Resistance, and Ohm's Law", section Power in Electric Circuits_