---
title: "Reducing Ferranti effect mathematically"
subject: "High Voltage Engineering"
chapter: 3
key: "semester_8/high_voltage_engineering:ferranti-effect-reduction"
source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/ferranti-effect-reduction
site: BCT Engineering
---

# Reducing Ferranti effect mathematically

High Voltage Engineering · Chapter 3: Overvoltages in Power System

- Asked in: 2074 Bhadra, 8 marks
- Video: [Voltage Rise | Ferranti Effect and Mitigation](https://www.youtube.com/watch?v=qaEu276nL0Q) — Engineering Simple
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/ferranti-effect-reduction
- Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-3

## Question

How the Ferranti effect can be reduced in over voltage issues? Explain mathematically.

## Short answer

Ferranti effect is reduced by shunt reactors, which absorb charging current. With a reactor XL at the receiving end, Vr/Vs=1/[cos βl+(Zc/XL) sin βl]; selecting XL=Zc cot(βl/2) makes Vr=Vs.

## Answer

The **Ferranti effect** is the rise of receiving-end voltage above the sending-end voltage on a long line at no load or light load. The line's charging current flows through its series inductance and raises the voltage along the line. It is a main cause of temporary overvoltage.

#### Mathematical explanation

For a lossless line of length $l$, phase constant $\beta = \omega\sqrt{LC}$ and surge impedance $Z_c = \sqrt{L/C}$:

$$
V_s = V_r\cos\beta l + jZ_c I_r\sin\beta l
$$

At no load $I_r = 0$:

$$
\frac{V_r}{V_s} = \frac{1}{\cos\beta l}
$$

At 50 Hz, $\beta \approx 6^\circ$ per 100 km. For example, a 400 km line with $L = 1$ mH/km and $C = 10$ nF/km gives $\beta l = 0.397$ rad and $V_r = 1.085\,V_s$. A 1500 km line ($\beta l = 90^\circ$) gives an infinite voltage.

#### Reduction by shunt reactor compensation

Connect a shunt reactor $jX_L$ at the receiving end. At no load, the reactor current is

$$
I_r = \frac{V_r}{jX_L}
$$

Substituting:

$$
\begin{aligned}
V_s &= V_r\cos\beta l + jZ_c\sin\beta l \cdot \frac{V_r}{jX_L} \\
\frac{V_r}{V_s} &= \frac{1}{\cos\beta l + \dfrac{Z_c}{X_L}\sin\beta l}
\end{aligned}
$$

The extra positive term $\dfrac{Z_c}{X_L}\sin\beta l$ in the denominator reduces $V_r$. For $V_r = V_s$:

$$
\cos\beta l + \frac{Z_c}{X_L}\sin\beta l = 1 \;\Rightarrow\; X_L = \frac{Z_c\sin\beta l}{1 - \cos\beta l} = Z_c\cot\frac{\beta l}{2}
$$

Physically, the reactor draws a lagging current that cancels part of the line's leading charging current. A smaller net current flows through the series inductance, so the voltage rise is smaller.

#### Other methods

1. **Shunt reactors** at the ends and at intermediate stations; switched or controlled reactors for varying load.
2. **Static VAR compensators (SVC) and STATCOMs**, which absorb reactive power under light load.
3. **Series capacitors are not used for this.** They reduce the effective electrical length for power transfer but do not cure the no-load rise.
4. **Sectionalising long lines** with intermediate substations, so that $\beta l$ of each section is small.
5. **Generator under-excitation** (absorbing VArs), within stability limits.
6. Proper **switching sequence**: energise the line from the stronger end and connect the reactors before or with the line.

## Check yourself

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

1. Why does the receiving-end voltage rise on a long unloaded line?
   A. Line resistance drops voltage
   B. Charging current flows through the series inductance and raises the voltage
   C. Generator speed increases
   D. Corona adds voltage
   Answer: B. The leading charging current through line inductance causes a voltage rise towards the receiving end. (Key point: Ferranti effect)

2. For a lossless line at no load, Vr/Vs equals:
   A. cos βl
   B. 1/sin βl
   C. 1/cos βl
   D. 1 + βl
   Answer: C. With Ir = 0, Vs = Vr cos βl, so Vr/Vs = 1/cos βl. (Key point: No-load rise formula)

3. With a shunt reactor jXL at the receiving end, the denominator becomes cos βl + (Zc/XL) sin βl. Why is Vr reduced?
   A. The extra positive term makes the denominator larger
   B. The reactor increases Zc
   C. The reactor reduces β
   D. The term is negative
   Answer: A. A larger denominator means a smaller Vr/Vs; physically the reactor absorbs the lagging current that cancels charging current. (Key point: Effect of shunt reactor)

4. The reactor value that makes Vr = Vs is:
   A. XL = Zc tan(βl/2)
   B. XL = Zc cot(βl/2)
   C. XL = Zc sin βl
   D. XL = Zc cos βl
   Answer: B. XL = Zc sin βl / (1 - cos βl) = Zc cot(βl/2). (Key point: Condition Vr = Vs)

## Related in High Voltage Engineering

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