An AC generator with a frequency of $60.0 \mathrm{Hz}$ is connected to a $375-\Omega$. If the average power dissipated in the resistor is $2.25 \mathrm{W}$, what is the maximum voltage of the generator?

Solution

VerifiedThe average power dissipated on the resistor is given by the relation:

$\begin{align*} P_{avg}&=I^2_{rms}\cdot{R}\\ P_{avg}&=\left(\frac{V_{max}}{\sqrt{2}\cdot{R}}\right)^2\cdot{R}\\ P_{avg}&=\frac{\left(V_{max}\right)^2}{2\cdot{R}} \end{align*}$

Let's express the maximum voltage from this equation:

$\begin{align*} \left(V_{max}\right)^2&=2\cdot{P_{avg}}\cdot{R}\\ V_{max}&=\sqrt{2\cdot{P_{avg}}\cdot{R}} \end{align*}$

Let's substitute and compute the value:

$\begin{align*} V_{max}&=\sqrt{2\cdot{2.25\text{ W}}\cdot{375\text{ $\Omega$}}} \end{align*}$

$\boxed{V_{max}=41.08\text{ V}}$