*Question G5B14:
What is the output PEP of 500 volts peak-to-peak across a 50-ohm load?*  

*Correct Answer is  B: 625 watts*  

### AI Response: (may contain errors)  

To calculate the output Peak Envelope Power (PEP) across a 50-ohm load with a peak-to-peak voltage of 500 volts, we can use the following formula:

$$
P = \frac{V_{rms}^2}{R}
$$

Where:
- $P$ = power in watts
- $V_{rms}$ = root mean square voltage
- $R$ = load resistance in ohms

### Step 1: Convert Peak-to-Peak Voltage to RMS Voltage

The relationship between peak-to-peak voltage ($V_{pp}$) and RMS voltage ($V_{rms}$) for a sine wave is given by:

$$
V_{rms} = \frac{V_{pp}}{2\sqrt{2}}
$$

Substituting $V_{pp} = 500$ volts:

$$
V_{rms} = \frac{500}{2\sqrt{2}} = \frac{500}{2 \times 1.414} \approx 176.78 \text{ volts}
$$

### Step 2: Calculate Power

Now, we can substitute $V_{rms}$ into the power formula:

$$
P = \frac{(176.78)^2}{50}
$$

Calculating this gives:

$$
P \approx \frac{31200.0484}{50} \approx 624.00 \text{ watts}
$$

Rounding this, we find:

$$
P \approx 625 \text{ watts}
$$

### Explanation of the Correct Answer

- **Correct Answer: 625 watts**  
  This is the calculated output PEP based on the given peak-to-peak voltage and load resistance.

### Explanation of Incorrect Choices

- **A: 8.75 watts**  
  This value is too low and does not reflect the calculations based on the given voltage and load.

- **C: 2500 watts**  
  This value is significantly higher than what the calculations yield, indicating a misunderstanding of the voltage-to-power relationship.

- **D: 5000 watts**  
  This value is also too high and does not correspond to the calculated PEP based on the provided voltage and load.

### Summary

To summarize, the correct method to find the output PEP involves converting the peak-to-peak voltage to RMS voltage and then using the power formula with the load resistance. The correct answer is 625 watts, while the other choices do not align with the calculations.