Question

alternating voltages and currents
the voltage supplied by a wall socket varies with time, reversing its polarity with a constant frequency, as shown in (figure 1)
figure
1 of 2
part c
what is the average power $p_{avg}$ dissipated in the resistor?
express your answer in watts.
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$p_{avg}=$ w
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Explanation:

Step1: Identify peak - voltage

From the graph, the peak - voltage $V_{max}=200\ V$.

Step2: Calculate rms - voltage

The formula for rms - voltage is $V_{rms}=\frac{V_{max}}{\sqrt{2}}$. So, $V_{rms}=\frac{200}{\sqrt{2}}\ V$.

Step3: Assume resistance value (not given, assume $R = 1\ \Omega$ for power - calculation concept)

The formula for average power in an AC - DC circuit is $P_{avg}=\frac{V_{rms}^{2}}{R}$. Substituting $V_{rms}=\frac{200}{\sqrt{2}}\ V$ and $R = 1\ \Omega$, we get $P_{avg}=\frac{(\frac{200}{\sqrt{2}})^{2}}{1}$.

Step4: Calculate the power

$P_{avg}=\frac{\frac{40000}{2}}{1}=20000\ W$. In a real - world problem, you would need the actual value of the resistance $R$. If we assume a general form with $R$ as the resistance, $P_{avg}=\frac{V_{rms}^{2}}{R}=\frac{(\frac{V_{max}}{\sqrt{2}})^{2}}{R}=\frac{V_{max}^{2}}{2R}$.

Answer:

If we assume $R = 1\ \Omega$, $P_{avg}=20000\ W$. In general form with resistance $R$, $P_{avg}=\frac{V_{max}^{2}}{2R}$ (where $V_{max} = 200\ V$ from the graph).