Question

# 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

Verified
Step 1
1 of 2

The 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*}

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

## Recommended textbook solutions #### Physics

1st EditionISBN: 9780133256925Walker
3,338 solutions #### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (5 more)Randall D. Knight
3,508 solutions #### Mathematical Methods in the Physical Sciences

3rd EditionISBN: 9780471198260Mary L. Boas
3,355 solutions #### Fundamentals of Physics

10th EditionISBN: 9781118230718 (2 more)David Halliday, Jearl Walker, Robert Resnick
8,950 solutions