*Question E9F04:
What impedance does a 1/2-wavelength transmission line present to an RF generator when the line is shorted at the far end?*  

*Correct Answer is  B: Very low impedance*  

### AI Response: (may contain errors)  

When a 1/2-wavelength transmission line is shorted at the far end, it presents a very low impedance to the RF generator. Here's why:

1. **Correct Answer Explanation**:
   - A 1/2-wavelength transmission line has a unique property: it transforms the load impedance at the far end (in this case, a short circuit, which has an impedance of $0 \, \Omega$) to the input end. The relationship can be described by the formula:
   $$ Z_{in} = Z_0 \frac{Z_L}{Z_0} $$
   where:
   - $Z_{in}$ = input impedance at the generator end
   - $Z_0$ = characteristic impedance of the line
   - $Z_L$ = load impedance (shorted, so $Z_L = 0 \, \Omega$)

   Substituting $Z_L = 0$ gives:
   $$ Z_{in} = Z_0 \frac{0}{Z_0} = 0 \, \Omega $$
   Thus, the input impedance is very low, effectively presenting a short circuit to the RF generator.

2. **Incorrect Answer Explanations**:
   - **A: Very high impedance**: This is incorrect because a shorted line does not reflect a high impedance; instead, it reflects a low impedance back to the generator.
   - **C: The same as the characteristic impedance of the line**: This is incorrect because the characteristic impedance is only relevant when the line is terminated with a load that matches it. A short circuit does not match the characteristic impedance.
   - **D: The same as the output impedance of the RF generator**: This is incorrect because the output impedance of the generator does not determine the impedance seen at the input of the transmission line when it is shorted. The line's properties dictate the impedance seen at the generator.

3. **Useful Rules of Thumb**:
   - A 1/2-wavelength line transforms impedances in a way that can be calculated using the above formula.
   - A short circuit at the end of a transmission line results in a low impedance at the input, while an open circuit would result in a very high impedance.

This understanding is crucial for analyzing RF circuits and ensuring proper matching between components.