To find the root mean square (RMS) voltage $V$ in terms of the peak voltage $V_0$, we follow these steps:

Part (B)

1. Definition of RMS Voltage: The RMS voltage is given by: $V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T V(t)^2 \, dt}$ where $T$ is the period of the signal, and $V(t)$ is the instantaneous voltage. In this case, we take the average over 1 second.

2. Instantaneous Voltage: If we assume the voltage $V(t)$ is sinusoidal (which is common in AC circuits): $V(t) = V_0 \sin(2 \pi f t)$ where $V_0$ is the peak voltage and $f$ is the frequency of the AC signal.

3. RMS Voltage Formula: The formula for the RMS value of a sinusoidal waveform is: $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$

Part (C)

Given:

• The RMS voltage in an American house is $V_{\text{rms}} = 110$ volts.

Using the formula: $V_0 = V_{\text{rms}} \times \sqrt{2}$

Substituting $V_{\text{rms}} = 110$: $V_0 = 110 \times \sqrt{2} \approx 110 \times 1.414 \approx 155.4 \text{ volts}$

Summary

• (B) The RMS voltage $V$ in terms of the peak voltage $V_0$ is $V = \frac{V_0}{\sqrt{2}}$.
• (C) If the RMS voltage is 110 volts, then the peak voltage $V_0$ is approximately 155.4 volts.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. What is the physical significance of the RMS voltage in AC circuits?
2. How does the RMS value differ for different waveforms (e.g., square, triangular)?
3. What is the peak-to-peak voltage for the given RMS voltage?
4. How would the RMS voltage formula change if the waveform were not sinusoidal?
5. Why is RMS voltage used instead of average voltage in AC power calculations?

Tip: RMS values are crucial in AC circuits because they allow direct comparison with DC circuits in terms of power delivery.