Question E9F09:
What impedance does a 1/4-wavelength transmission line present to an RF generator when the line is shorted at the far end?  

Choices:  
A: Very high impedance  
B: Very low impedance  
C: The same as the characteristic impedance of the transmission line  
D: The same as the generator output impedance    

Correct Answer is  A: Very high impedance  

To a 1/4-wave line, a short looks like an open. Here’s why.

Why A (Very high impedance) is correct:
- Input impedance of a lossless transmission line:
  $$Z_{in} = Z_0 \frac{Z_L + j Z_0 \tan(\beta l)}{Z_0 + j Z_L \tan(\beta l)}$$
- For a shorted load, $Z_L = 0$, so:
  $$Z_{in} = j Z_0 \tan(\beta l)$$
- For $l = \lambda/4$, $\beta l = \pi/2$ and $\tan(\pi/2) \to \infty$, so $|Z_{in}| \to \infty$ (an open circuit, very high impedance).

Why the other choices are incorrect:
- B: Very low impedance  
  That would be true for an open-circuited 1/4-wave line (open transforms to short), not for a shorted 1/4-wave line.
- C: Same as the characteristic impedance  
  You only see $Z_{in} = Z_0$ when the load is matched ($Z_L = Z_0$). A shorted load is not matched.
- D: Same as the generator output impedance  
  The line’s input impedance depends on $Z_0$, $Z_L$, and length, not the generator’s impedance.

Useful formulas and rules of thumb:
- General (lossless) input impedance:
  $$Z_{in} = Z_0 \frac{Z_L + j Z_0 \tan(\beta l)}{Z_0 + j Z_L \tan(\beta l)}$$
- Quarter-wave transformer:
  $$Z_{in}\big|_{l=\lambda/4} = \frac{Z_0^2}{Z_L}$$
  So a short ($Z_L = 0$) becomes an open ($Z_{in} \to \infty$), and an open becomes a short.
- Half-wave repeat:
  $$Z_{in}\big|_{l = n\lambda/2} = Z_L$$
  The line repeats the load every half-wavelength.