VSWR of a perfectly short-circuited line A transmission line with characteristic impedance Z0 = 300 ∠ 0 Ω is terminated in a perfect short circuit. What is the standing wave ratio (VSWR) on the line?

Introduction:
VSWR quantifies the severity of standing waves caused by impedance mismatch. Extreme cases—open or short circuits—yield the largest possible mismatch magnitude. Recognizing these limits is essential for interpreting line measurements and avoiding damage due to high voltage/current antinodes.

Given Data / Assumptions:

• Lossless/low-loss line with Z0 = 300 Ω (purely real).
• Load is an ideal short circuit: ZL = 0 Ω.
• Single-frequency steady-state excitation.

Concept / Approach:

The reflection coefficient is Γ = (ZL − Z0)/(ZL + Z0). For ZL = 0, Γ = −1, so |Γ| = 1. VSWR relates to |Γ| by VSWR = (1 + |Γ|)/(1 − |Γ|). When |Γ| = 1, the denominator becomes zero, implying VSWR → ∞. Physically, voltage nodes and antinodes repeat every λ/2 with zero voltage at the load and very large maxima along the line (limited by losses and source constraints).

Step-by-Step Solution:

Verification / Alternative check:

Smith Chart: a short is at the leftmost extreme; the SWR circle radius corresponds to |Γ| = 1, confirming infinite VSWR.

Why Other Options Are Wrong:

VSWR of 1 denotes a perfect match; finite values like 2 or 10 imply partial mismatch; 0 is nonphysical for SWR.

Common Pitfalls:

Confusing SWR with return loss; thinking that “very large” is still finite for an ideal short—mathematically it is infinite.

Final Answer:

∞.