Short-circuited transmission line – Where is the voltage minimum? A lossless transmission line is shorted at the far end. Identify the location of the first voltage minimum on the line relative to the short-circuited termination.

Introduction:
Standing waves arise on transmission lines when the termination is not matched to the characteristic impedance. A perfect short circuit forces the voltage at the termination to zero, creating a fixed voltage minimum at that point and establishing a predictable sequence of minima and maxima along the line.

Given Data / Assumptions:

• Lossless or low-loss line so standing-wave patterns are clear.
• Short-circuited load at the far end.
• Single-frequency steady-state sinusoidal excitation.

Concept / Approach:

For a shorted line, boundary condition is V = 0 at z = 0 (the load). The standing-wave solution is V(z) = V+ (e^{-jβz} − e^{jβz}) which equals −2j V+ sin(βz). This is zero at z = 0 and at every half-wavelength increment. Current is maximum at the short, because I(z) ∝ cos(βz). Thus the first and strongest voltage minimum is exactly at the shorted end.

Step-by-Step Solution:

Verification / Alternative check:

Smith Chart view: a short corresponds to a reflection coefficient of −1. The normalized voltage standing-wave pattern shows a node at the load.

Why Other Options Are Wrong:

Option B: A node is not guaranteed at the source unless the length happens to place one there. Option C: Midway is not universal; it depends on electrical length. Option D: The position is definite due to boundary conditions. Option E: Voltage nodes recur every λ/2, not only at λ/4.

Common Pitfalls:

Mixing up positions of voltage and current extrema; assuming λ/4 behavior for all shorted lines; ignoring electrical length dependence of additional minima.

Final Answer:

At the far end (at the short itself).