A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current?

The rms current in an AC circuit in which the current varies sinusoidally is given by

\(\begin{array}{l}{{\bf{I}}_{{\bf{rms}}}}{\bf{ = }}\frac{{{{\bf{I}}_{{\bf{max}}}}}}{{\sqrt {\bf{2}} }}\\{{\bf{I}}_{{\bf{rms}}}}{\bf{ = 0}}{\bf{.707}}\;{{\bf{I}}_{{\bf{max}}}}\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\end{array}\)

Where\({{\rm{I}}_{{\rm{max}}}}\)is the maximum current in the circuit

The peak current in the circuit is found by solving the equation\(\left( 1 \right)\),

\({{\rm{I}}_{\max }} = \sqrt 2 {I_{rms}}\)

Entering the value for\({I_{rms}}\)we obtain:

\(\begin{array}{c}{I_{\max }} = \sqrt 2 (15.0\;{\rm{A}})\\ = 21.2\;{\rm{A}}\end{array}\)

Therefore, the value of peak current is\({\rm{21}}{\rm{.2\;A}}\).