---
title: "Lossless line, L Z0 β Γ"
subject: "Electromagnetism"
chapter: 5
key: "semester_3/electromagnetism:lossless-line-velocity-capacitance-30pf"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/lossless-line-velocity-capacitance-30pf
site: BCT Engineering
---

# Lossless line, L Z0 β Γ

Electromagnetism · Chapter 5: Transmission lines

- Asked in: 2075 Asoj, 2+2+2+2 marks
- Video: [Transmission Line Example on Characteristics Impedance, Attenuation Constant & Phase Constant](https://www.youtube.com/watch?v=3LP1Lu7MwUs) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/lossless-line-velocity-capacitance-30pf
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5

## Question

The velocity of propagation in a lossless transmission line is 2.5×10⁸ m/s. If the capacitance of the line is 30 pF/m, find: (a) inductance of the line (b) characteristic impedance (c) phase constant at 100 MHz (d) reflection coefficient if the line is terminated with a resistive load of 50 Ω.

## Short answer

L = 1/(v²C) = 0.533 μH/m. Z₀ = √(L/C) = 133.3 Ω. β = ω/v = 2.51 rad/m at 100 MHz. Γ = (50 − 133.3)/(50 + 133.3) = −0.455.

## Answer

Given: v = 2.5×10⁸ m/s, C = 30 pF/m, lossless line.

For a lossless line v = 1/√(LC) and Z₀ = √(L/C).

#### (a) Inductance

```text
L = 1/(v² C) = 1 / [(2.5×10⁸)² × 30×10⁻¹²]
  = 1 / (6.25×10¹⁶ × 3×10⁻¹¹)
  = 1 / (1.875×10⁶)
  = 0.5333×10⁻⁶ H/m
```

**L = 0.533 μH/m**

#### (b) Characteristic impedance

```text
Z₀ = √(L/C) = √(0.5333×10⁻⁶ / 30×10⁻¹²) = √17778 = 133.3 Ω
```

(Also Z₀ = 1/(vC) = 1/(2.5×10⁸ × 30×10⁻¹²) = 133.3 Ω ✓.)

**Z₀ = 133.3 Ω**

#### (c) Phase constant at 100 MHz

```text
β = ω/v = 2π × 100×10⁶ / 2.5×10⁸ = 2.513 rad/m
```

**β = 2.51 rad/m** (λ = 2.5 m)

#### (d) Reflection coefficient for R_L = 50 Ω

```text
Γ = (R_L − Z₀)/(R_L + Z₀) = (50 − 133.3)/(50 + 133.3) = −83.3/183.3
  = −0.4545
```

**Γ = −0.455** (magnitude 0.455, angle 180°), and S = (1 + 0.4545)/(1 − 0.4545) = 2.67.

## Check yourself

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

1. How is L found from v and C?
   A. L = v^2 C = 1.9x10^6 H/m
   B. L = 1/(v C) = 133 H/m
   C. L = C/v^2
   D. L = 1/(v^2 C) = 0.533 uH/m
   Answer: D. v = 1/sqrt(LC) so L = 1/(v^2 C). (Key point: Inductance)

2. What is Z0?
   A. sqrt(LC) = 4 ohm
   B. L/C = 17,778 ohm
   C. sqrt(L/C) = 133.3 ohm
   D. 1/sqrt(LC) = 2.5x10^8 ohm
   Answer: C. Z0 = sqrt(L/C) = 1/(vC). (Key point: Characteristic impedance)

3. What is beta at 100 MHz?
   A. w*v rad/m
   B. w/v = 2.51 rad/m
   C. 1/(w v)
   D. 2.5x10^8/100x10^6 = 2.5 rad/s
   Answer: B. beta = 2 pi x 10^8/(2.5x10^8) = 2.513 rad/m. (Key point: Phase constant)

4. Why is Gamma negative for R_L = 50 ohm?
   A. Because R_L < Z0, giving Gamma = -0.455
   B. Because R_L > Z0
   C. Because the line is lossless
   D. Because beta is positive
   Answer: A. Gamma = (50 - 133.3)/(50 + 133.3) = -0.455; a load smaller than Z0 gives a negative Gamma. (Key point: Reflection coefficient)

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