---
title: "1.5λ line, power and minimum voltage"
subject: "Electromagnetism"
chapter: 5
key: "semester_3/electromagnetism:one-point-five-lambda-line-power"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/one-point-five-lambda-line-power
site: BCT Engineering
---

# 1.5λ line, power and minimum voltage

Electromagnetism · Chapter 5: Transmission lines

- Asked in: 2075 Chaitra, 4+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/one-point-five-lambda-line-power
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5

## Question

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.

## 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.

## Answer

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

```text
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

```text
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

```text
|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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What effect does the 1.5 lambda length have?
   A. It transforms the load by Z0^2/R_L
   B. It inverts Gamma
   C. It is a multiple of lambda/2, so Z_in = R_L
   D. It changes the load power
   Answer: C. A half-wave multiple repeats the impedance. Gamma = 1/3 and S = 2. (Key point: Effect of 1.5 lambda)

2. What is the average power delivered to the 100 ohm load at V_L = 40 V?
   A. |V|^2/R_L = 16 W
   B. |V|^2/(2 R_L) = 8 W
   C. |V|^2/(2 Z0) = 16 W
   D. |V|^2 R_L/2 = 80 kW
   Answer: B. The load is resistive, so P = 1600/200 = 8 W. Check: P_inc - P_ref = 9 - 1 = 8 W. (Key point: Load power)

3. What is |V|_min on the line?
   A. 20 V, since the load end is a maximum (R_L > Z0) and V_min = V_max/S
   B. 40 V
   C. 30 V
   D. 13.3 V
   Answer: A. Gamma is positive, so the load is at a maximum of 40 V. S = 2 gives 20 V. (Key point: Minimum voltage)

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