*Question G7B08:
How is the efficiency of an RF power amplifier determined?*  

*Correct Answer is  B: Divide the RF output power by the DC input power*  

### AI Response: (may contain errors)  

To determine the efficiency of an RF power amplifier, we use the following formula:

$$
\text{Efficiency} (\eta) = \frac{P_{RF \, output}}{P_{DC \, input}} \times 100\%
$$

where:
- $P_{RF \, output}$ = RF output power in watts
- $P_{DC \, input}$ = DC input power in watts

### Explanation of the Correct Answer:
- **Correct Answer (B)**: Dividing the RF output power by the DC input power gives us the efficiency of the amplifier. This ratio indicates how effectively the amplifier converts DC power into RF power. A higher ratio means better efficiency.

### Explanation of Incorrect Choices:
- **Choice A**: Divide the DC input power by the DC output power.
  - This is incorrect because it would yield a value less than 1 (or 100% when multiplied by 100), which does not represent efficiency correctly. Efficiency is about how much output we get from the input, not the other way around.

- **Choice C**: Multiply the RF input power by the reciprocal of the RF output power.
  - This is incorrect because it does not relate to the definition of efficiency. Efficiency is not calculated this way; it should be a ratio of output to input.

- **Choice D**: Add the RF input power to the DC output power.
  - This is incorrect because adding powers does not provide a meaningful measure of efficiency. Efficiency is a ratio, not a sum.

### Summary:
- The efficiency of an RF power amplifier is determined by the ratio of RF output power to DC input power. This is crucial for understanding how well the amplifier performs in converting power. Always remember the formula for efficiency to help with similar questions on the exam.