*Question G5B06:
What is the PEP produced by 200 volts peak-to-peak across a 50-ohm dummy load?*  

*Correct Answer is  B: 100 watts*  

### AI Response: (may contain errors)  

To calculate the Peak Envelope Power (PEP) from a peak-to-peak voltage across a load, we can use the following steps:

1. **Convert peak-to-peak voltage to RMS voltage:**

   The RMS voltage ($V_{\text{RMS}}$) is related to the peak voltage ($V_{\text{peak}}$) by the formula:
   $$V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}}$$

   First, find the peak voltage from the peak-to-peak voltage:
   $$V_{\text{peak}} = \frac{V_{\text{peak-to-peak}}}{2} = \frac{200}{2} = 100 \text{ volts}$$

   Then, calculate the RMS voltage:
   $$V_{\text{RMS}} = \frac{100}{\sqrt{2}} \approx 70.7 \text{ volts}$$

2. **Calculate the power using the RMS voltage:**

   The power ($P$) across a resistive load can be calculated using Ohm's Law:
   $$P = \frac{V_{\text{RMS}}^2}{R}$$

   Substituting the known values:
   $$P = \frac{70.7^2}{50} \approx 100 \text{ watts}$$

Therefore, the correct answer is **B: 100 watts**.

### Why the other choices are incorrect:

- **A: 1.4 watts**  
  This value is far too low for the given voltage and load resistance.

- **C: 353.5 watts**  
  This value is too high. It might result from a miscalculation or misunderstanding of the formula.

- **D: 400 watts**  
  This value is also too high and does not match the calculated PEP for the given conditions.

By following these steps and using the correct formulas, you can accurately determine the PEP for a given peak-to-peak voltage across a resistive load.