# Problem: What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A?

###### FREE Expert Solution

The rms voltage and the peak voltage are related by:

$\overline{){{\mathbf{V}}}_{\mathbf{r}\mathbf{m}\mathbf{s}}{\mathbf{=}}\frac{{\mathbf{V}}_{\mathbf{0}}}{\sqrt{\mathbf{2}}}}$, where V0 is the peak voltage.

Rewriting the equation, the peak voltage:

$\overline{){{\mathbf{V}}}_{{\mathbf{0}}}{\mathbf{=}}{{\mathbf{V}}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\sqrt{\mathbf{2}}}$

Similarly, the peak current is:"

$\overline{){{\mathbf{i}}}_{{\mathbf{0}}}{\mathbf{=}}{{\mathbf{i}}}_{{\mathbf{rms}}}\sqrt{\mathbf{2}}}$

The average AC power:

$\overline{){{\mathbf{P}}}_{\mathbf{a}\mathbf{v}\mathbf{e}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{i}}}_{{\mathbf{0}}}{{\mathbf{V}}}_{{\mathbf{0}}}}$

Substituting:

$\begin{array}{rcl}{\mathbf{P}}_{\mathbf{a}\mathbf{v}\mathbf{e}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}{{\mathbf{i}}}_{{\mathbf{rms}}}\sqrt{\mathbf{2}}{\mathbf{\right)}}{\mathbf{\left(}}{{\mathbf{V}}}_{{\mathbf{rms}}}\sqrt{\mathbf{2}}{\mathbf{\right)}}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\left(}{{\mathbf{i}}}_{{\mathbf{rms}}}{{\mathbf{V}}}_{{\mathbf{rms}}}{\mathbf{\right)}}\end{array}$ ###### Problem Details

What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A?