Question

An AC generator with a frequency of 60.0Hz60.0 \mathrm{Hz} is connected to a 375Ω375-\Omega. If the average power dissipated in the resistor is 2.25W2.25 \mathrm{W}, what is the maximum voltage of the generator?

Solution

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Answered 2 years ago
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The average power dissipated on the resistor is given by the relation:

Pavg=Irms2RPavg=(Vmax2R)2RPavg=(Vmax)22R\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:

(Vmax)2=2PavgRVmax=2PavgR\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:

Vmax=22.25 W375 Ω\begin{align*} V_{max}&=\sqrt{2\cdot{2.25\text{ W}}\cdot{375\text{ $\Omega$}}} \end{align*}

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

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