Question

the voltage supplied by a wall socket varies with time, reversing its polarity with a constant frequency, as shown in (figure 1). part c what is the average power $p_{avg}$ dissipated in the resistor? express your answer in watts. view available hint(s)

Explanation:

Step1: Find the RMS voltage

The peak - voltage $V_{max}=200$ V. The root - mean - square (RMS) voltage for an AC sinusoidal voltage is given by $V_{rms}=\frac{V_{max}}{\sqrt{2}}$. So, $V_{rms}=\frac{200}{\sqrt{2}}\text{ V}$.

Step2: Assume the resistance value

Let's assume the resistance of the resistor is $R = 1\Omega$ (since the resistance value is not given in the problem, and for the formula $P=\frac{V^{2}}{R}$, we can assume a value to show the general calculation process. The power is proportional to the square of the voltage and inversely proportional to the resistance).

Step3: Calculate the average power

The formula for average power dissipated in a resistor in an AC - DC circuit is $P_{avg}=\frac{V_{rms}^{2}}{R}$. Substitute $V_{rms}=\frac{200}{\sqrt{2}}\text{ V}$ and $R = 1\Omega$ into the formula. $V_{rms}^{2}=\frac{200^{2}}{2}= 20000$. Then $P_{avg}=\frac{20000}{1}=20000$ W. In a general case, if the resistance is $R$, $P_{avg}=\frac{( \frac{V_{max}}{\sqrt{2}})^{2}}{R}=\frac{V_{max}^{2}}{2R}$.

Answer:

If we assume $R = 1\Omega$, $P_{avg}=20000$ W. In general formula $P_{avg}=\frac{V_{max}^{2}}{2R}$ (where $V_{max} = 200$ V is the peak - voltage of the AC source and $R$ is the resistance of the resistor).