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1.5λ line, power and minimum voltage

Short answer

Average load power is |V_L|²/(2R_L) = 40²/(2×100) = 8 W. Since the 100 Ω load is a voltage maximum, |V|_min = |V⁺|(1 − |Γ|) = 30 × (1 − 1/3) = 20 V.

A lossless 50-Ω line is 1.5λ long and is terminated with a pure resistance of 100 Ω. The load voltage is 40∠60° V. Find: (a) the average power delivered to the load; (b) the magnitude of the minimum voltage on the line.

Given: Z₀ = 50 Ω, l = 1.5λ, R_L = 100 Ω, V_L = 40∠60° V.

Γ_L = (R_L − Z₀)/(R_L + Z₀) = (100 − 50)/(100 + 50) = 1/3
S   = (1 + 1/3)/(1 − 1/3) = 2

The line length 1.5λ is a multiple of λ/2, so the input impedance simply repeats the load (Z_in = R_L = 100 Ω).

(a) Average power delivered to the load

The load is a pure resistance, so

P_L = |V_L|² / (2 R_L) = 40² / (2 × 100) = 1600/200 = 8 W

Answer (a): 8 W.

(b) Minimum voltage magnitude on the line

At the load: V_L = V⁺(1 + Γ_L) → V⁺ = V_L/(1 + Γ_L) = 40∠60° / (4/3) = 30∠60° V.

Since R_L > Z₀, Γ_L is positive, so the load end is a voltage maximum (|V| = 40 V = |V⁺|(1 + |Γ|)). The minimum is

|V|_min = |V⁺| (1 − |Γ|) = 30 × (1 − 1/3) = 20 V

(Check: |V|_max/|V|_min = 40/20 = 2 = S.)

Power check using the incident wave: P_inc = |V⁺|²/2Z₀ = 900/100 = 9 W; P_ref = Γ² P_inc = 1 W; net = 8 W ✓.

Answer (b): |V|_min = 20 V (located λ/4 from the load, i.e. at 0.25λ, 0.75λ, ... from the load).

✓ Answer checked by Claude Opus 5.5 (high effort)

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