*Question E9F12:
What impedance does a 1/4-wavelength transmission line present to an RF generator when the line is open at the far end?*  

*Correct Answer is  D: Very low impedance*  

### AI Response: (may contain errors)  

To understand the impedance presented by a 1/4-wavelength transmission line that is open at the far end, we need to consider how transmission lines work.

### Correct Answer: C - Very high impedance

1. **Why the correct answer is correct:**
   - A 1/4-wavelength transmission line transforms the load impedance at its far end (which is open, meaning infinite impedance) to a very high impedance at the input (the generator side). This transformation occurs because of the properties of transmission lines, where the input impedance $Z_{in}$ can be calculated using the formula:
   $$
   Z_{in} = \frac{Z_0^2}{Z_L}
   $$
   where:
   - $Z_0$ = characteristic impedance of the line
   - $Z_L$ = load impedance (in this case, open circuit, which is infinite)

   Since $Z_L$ is infinite, the input impedance $Z_{in}$ approaches infinity, indicating a very high impedance.

2. **Why the other choices are incorrect:**
   - **A: The same as the characteristic impedance of the line**  
     This is incorrect because the input impedance is not equal to the characteristic impedance when the line is terminated in an open circuit. Instead, it transforms to a very high impedance.
   
   - **B: The same as the input impedance to the generator**  
     This is misleading. The input impedance is transformed by the line, and in this case, it becomes very high due to the open circuit at the far end.
   
   - **D: Very low impedance**  
     This is incorrect because an open circuit presents a very high impedance, not low. A low impedance would occur if the line were shorted at the far end.

### Summary:
- A 1/4-wavelength transmission line with an open circuit at the far end presents a very high impedance to the RF generator due to the transformation properties of transmission lines. The key formula to remember is:
$$
Z_{in} = \frac{Z_0^2}{Z_L}
$$
where $Z_L$ is infinite for an open circuit, leading to a very high input impedance.