A circuit breaker trips at an rms current of 15.0 A. What peak current does that correspond to?

1. The rms and peak currents are linked by I_rms = I_0 divided by the square root of two. 2. Rearranging for the peak gives I_0 = 1.414 times I_rms. 3. Substitute the trip value: I_0 = 1.414 (15.0 A). 4. That gives I_0 = 21.2 A. 5. 10.6 A is the result of dividing by 1.414 instead, which would place the peak below the rms value. 6. 30.0 A doubles the rms figure, treating the root two factor as a factor of two. 7. The breaker still responds to the sustained rms value, so a peak above the rating is not by itself a trip condition. _Source: OpenStax College Physics (CC BY 4.0), Ch 20 "Electric Current, Resistance, and Ohm's Law", section Alternating Current_