---
title: "Line with 50+j50 load, first maximum, power"
subject: "Electromagnetism"
chapter: 5
key: "semester_3/electromagnetism:line-50-j50-load-first-maximum"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/line-50-j50-load-first-maximum
site: BCT Engineering
---

# Line with 50+j50 load, first maximum, power

Electromagnetism · Chapter 5: Transmission lines

- Asked in: 2070 Chaitra, 4+4 marks
- Video: [Transmission Line Example on VSWR, Reflection Coefficient & Load Impedance in Microwave Engineering](https://www.youtube.com/watch?v=7Je7c5EHfxg) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/line-50-j50-load-first-maximum
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5

## Question

A lossless transmission line with Z0 = 50 Ω with length 1.5 m connects a voltage source V_g = 60∠0° V to a terminal load Z_L = (50 + j50) Ω. If the operating frequency f = 100 MHz, generator impedance Z_g = 50 Ω and speed of wave equal to the speed of light, find the distance of the first voltage maximum from the load. What is the power delivered to the load?

## Short answer

A half-wave line repeats the load, and the reflection-coefficient phase gives the first voltage maximum at 0.264 m (0.088λ) from the load. Power delivered is 7.2 W.

## Answer

Given: Z₀ = 50 Ω, l = 1.5 m, V_g = 60∠0° V, Z_g = 50 Ω, Z_L = 50 + j50 Ω, f = 100 MHz, v = c.

```text
λ = c/f = 3×10⁸/10⁸ = 3 m
β = 2π/3 = 2.094 rad/m
l = 1.5 m = 0.5λ   (half-wave line)
```

#### (a) Distance of first voltage maximum from the load

```text
Γ_L = (Z_L − Z₀)/(Z_L + Z₀) = j50/(100 + j50)
    = 50∠90° / 111.8∠26.57°
    = 0.447 ∠63.43°
S = (1 + 0.447)/(1 − 0.447) = 2.618
```

A voltage maximum occurs where the phase of Γ is zero: θ_Γ − 2βd = 0.

```text
d_max = θ_Γ / (2β) = (63.43° × π/180) / (2 × 2.094)
      = 1.1071 / 4.189 = 0.264 m  (= 0.088 λ)
```

**First voltage maximum at 0.264 m from the load.**

#### (b) Power delivered to the load

A half-wave line repeats the load impedance: tan βl = tan 180° = 0, so Z_in = Z_L = 50 + j50 Ω.

```text
I_in = V_g/(Z_g + Z_in) = 60/(100 + j50) = 0.5367 ∠−26.57° A
P_in = ½ |I_in|² Re(Z_in) = ½ (0.5367)² × 50 = 7.2 W
```

The line is lossless, so P_L = P_in.

Check: V_in = V_g Z_in/(Z_g + Z_in) = 60(50 + j50)/(100 + j50) = 37.9∠18.4° V, and P = ½ Re(V_in I_in*) = ½ × 37.9 × 0.5367 × cos(45°) = 7.2 W. ✓

**Answer: d_max = 0.264 m (0.088λ); P_L = 7.2 W.**

## Check yourself

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

1. The line is 1.5 m at 100 MHz (lambda = 3 m). What does this length mean for Z_in?
   A. l = lambda/4, so Z_in = Z0^2/Z_L
   B. l = lambda, so Z_in = 0
   C. l = 0.5 lambda, so Z_in = Z0
   D. l = 0.5 lambda, so Z_in = Z_L = 50 + j50 ohm
   Answer: D. A half-wave line repeats the load impedance. (Key point: Half-wave line)

2. How is the first voltage maximum located?
   A. d = lambda/4 = 0.75 m
   B. d = theta_Gamma/beta = 0.53 m
   C. Where the phase of Gamma is zero: d = theta_Gamma/(2 beta) = 0.264 m
   D. d = lambda/2 = 1.5 m
   Answer: C. Gamma_L = 0.447 angle 63.43 degrees, so 2 beta d = 63.43 degrees giving 0.264 m. (Key point: First voltage maximum)

3. What power is delivered to the load?
   A. P = V_g^2/(8 Z_g) = 9 W
   B. P = (1/2)|I_in|^2 Re(Z_in) = 7.2 W
   C. P = (1/2)|I|^2 |Z_in| = 14.4 W
   D. P = V_g^2/(2 Z_g) = 36 W
   Answer: B. I_in = 60/(100 + j50), magnitude 0.5367 A. P = 0.5 x 0.288 x 50 = 7.2 W; a lossless line delivers it all. (Key point: Power)

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