Question

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

Short answer

Value of peak current is \({{\rm{I}}_{{\rm{max}}}}{\rm{ = 21}}{\rm{.2\;A}}\).

Step-by-step solution

Relation between rms current and its corresponding peak value.

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

Calculation for the peak current using the rms current.

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\({\rm{21}}{\rm{.2\;A}}\).