*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, you use the following formula:

$$
\text{Efficiency} = \frac{\text{RF output power}}{\text{DC input power}}
$$

This formula shows that efficiency is the ratio of the useful RF power output to the total DC power input. It tells you how effectively the amplifier converts DC power into RF power.

### Explanation of the Correct Answer:
- **Correct Answer (B):** Divide the RF output power by the DC input power.  
  This is the standard method for calculating the efficiency of an RF power amplifier. It directly measures how much of the input power is converted into the desired RF output.

### Explanation of Incorrect Choices:
- **Choice A:** Divide the DC input power by the DC output power.  
  This choice is incorrect because it does not involve the RF output power, which is essential for determining the efficiency of an RF amplifier.

- **Choice C:** Multiply the RF input power by the reciprocal of the RF output power.  
  This choice is incorrect because it does not relate to the efficiency calculation. Efficiency is not determined by manipulating RF input and output powers in this manner.

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

### Key Points:
- Efficiency is a measure of how well an amplifier converts input power into output power.
- The correct formula involves the RF output and DC input, not just DC or RF powers alone.