*Question G5B08:
What is the peak-to-peak voltage of a sine wave with an RMS voltage of 120 volts?*  

*Correct Answer is  D: 339.4 volts*  

### AI Response: (may contain errors)  

To find the peak-to-peak voltage of a sine wave given its RMS voltage, we use the following relationships:

1. **RMS to Peak Voltage**:  
   The RMS (Root Mean Square) voltage of a sine wave is related to its peak voltage by the formula:  
   $$V_{\text{peak}} = V_{\text{RMS}} \times \sqrt{2}$$

2. **Peak to Peak Voltage**:  
   The peak-to-peak voltage is twice the peak voltage:  
   $$V_{\text{p-p}} = 2 \times V_{\text{peak}}$$

Given:
- $V_{\text{RMS}} = 120$ volts

First, calculate the peak voltage:
$$V_{\text{peak}} = 120 \times \sqrt{2} \approx 120 \times 1.414 \approx 169.7 \text{ volts}$$

Next, calculate the peak-to-peak voltage:
$$V_{\text{p-p}} = 2 \times 169.7 \approx 339.4 \text{ volts}$$

Therefore, the correct answer is **D: 339.4 volts**.

### Explanation of Other Choices:
- **A: 84.8 volts**: This value is incorrect because it is less than the RMS voltage, which is not possible for a sine wave.
- **B: 169.7 volts**: This is the peak voltage, not the peak-to-peak voltage.
- **C: 240.0 volts**: This value does not correspond to any standard conversion from RMS to peak or peak-to-peak voltage.

By understanding these relationships, you can accurately convert between RMS, peak, and peak-to-peak voltages for sine waves.