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 ${\mathrm{I}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=21.2\mathrm{A}$.

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

$\begin{array}{l}{\mathbf{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}{\sqrt{\mathbf{2}}}\\ {\mathbf{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.707}{\mathbf{I}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\end{array}$

Where${\mathrm{I}}_{\mathrm{m}\mathrm{a}\mathrm{x}}$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$\left(1\right)$,

${\mathrm{I}}_{max}=\sqrt{2}{I}_{rms}$

Entering the value for$I_{rms}$we obtain:

$\begin{array}{c}{I}_{max}=\sqrt{2}(15.0\mathrm{A})\\ =21.2\mathrm{A}\end{array}$

Therefore, the value of peak current is$21.2\mathrm{A}$.