---
title: "Surge on bifurcated 150 Ω and 90 Ω lines"
subject: "High Voltage Engineering"
chapter: 3
key: "semester_8/high_voltage_engineering:bifurcated-line-surge"
source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/bifurcated-line-surge
site: BCT Engineering
---

# Surge on bifurcated 150 Ω and 90 Ω lines

High Voltage Engineering · Chapter 3: Overvoltages in Power System

- Asked in: 2071 Magh, 7 marks
- Video: [PS65 Wave Reflection at Bifurcated (Forked) Line](https://www.youtube.com/watch?v=0hjckQfioxs) — Lectures in Electrical Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/bifurcated-line-surge
- Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-3

## Question

An overhead line of 300Ω surge impedance bifurcates into two lines of 150Ω and 90Ω surge impedances respectively. If a step wave of 150 kV is launched on the line, estimate the magnitude of waves transmitted on the bifurcated lines.

## Short answer

The two outgoing lines act in parallel, equivalent surge impedance 56.25 ohm. Transmitted voltage is 2*56.25/(300+56.25)*150 = 47.37 kV on both branches. Branch currents are 315.8 A in 150 ohm and 526.3 A in 90 ohm; reflected wave is -102.63 kV.

## Answer

At the bifurcation point, the two outgoing lines are in parallel for the incoming wave. The same transmitted voltage travels on both.

**Data:** $V = 150$ kV, $Z_1 = 300\ \Omega$, $Z_2 = 150\ \Omega$, $Z_3 = 90\ \Omega$.

#### Parallel impedance

$$
Z_p = \frac{150 \times 90}{150 + 90} = 56.25\ \Omega
$$

#### Transmitted voltage

$$
\begin{aligned}
V_t &= \frac{2Z_p}{Z_1 + Z_p}V = \frac{2 \times 56.25}{300 + 56.25} \times 150 \\
&= 0.3158 \times 150 = 47.37\ \text{kV}
\end{aligned}
$$

#### Transmitted currents

$$
\begin{aligned}
I_2 &= \frac{47.37}{150} = 0.3158\ \text{kA} = 315.8\ \text{A} \\
I_3 &= \frac{47.37}{90} = 0.5263\ \text{kA} = 526.3\ \text{A}
\end{aligned}
$$

#### Reflected wave and check

- $V_r = V_t - V = 47.37 - 150 = -102.63$ kV
- Incident current $= 150/300 = 500$ A; reflected current $= 102.63/300 = 342.1$ A
- Line current at junction $= 500 + 342.1 = 842.1$ A $= 315.8 + 526.3$ ✓

```text
 Z1=300 ohm            +--- 150 ohm: 47.37 kV, 315.8 A
 150 kV ----->  J -----+
                       +---  90 ohm: 47.37 kV, 526.3 A
```

**Answer:** a **47.37 kV** wave travels on each branch, with currents **315.8 A** (150 Ω line) and **526.3 A** (90 Ω line). A −102.6 kV wave is reflected back.

## Check yourself

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

1. For 150 Ω and 90 Ω branches, the equivalent seen by the incoming wave is:
   A. 240 Ω
   B. 120 Ω
   C. 56.25 Ω
   D. 300 Ω
   Answer: C. 150x90/240 = 56.25 Ω, since both lines get the same voltage. (Key point: Parallel impedance)

2. The voltage transmitted into each branch is:
   A. 150 kV
   B. 47.37 kV
   C. 102.6 kV
   D. 75 kV
   Answer: B. 2x56.25/(300+56.25) x 150 = 47.37 kV. (Key point: Transmitted voltage)

3. The branch currents are:
   A. 315.8 A in 150 Ω and 526.3 A in 90 Ω
   B. 526.3 A in 150 Ω and 315.8 A in 90 Ω
   C. Both 421 A
   D. 500 A each
   Answer: A. I = 47.37 kV/Z: 315.8 A and 526.3 A; the lower-impedance line carries more. (Key point: Branch currents)

4. The reflected wave on the incoming line is:
   A. +102.6 kV
   B. -47.37 kV
   C. -150 kV
   D. -102.6 kV
   Answer: D. Vr = 47.37 - 150 = -102.63 kV. (Key point: Reflected wave)

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