---
title: "Making short-line regulation zero"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:zero-regulation-short-line"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/zero-regulation-short-line
site: BCT Engineering
---

# Making short-line regulation zero

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2072 Magh, 6 marks
- Video: [2. Phasor diagram for short transmission line](https://www.youtube.com/watch?v=MBdbCU-5viU) — Bookforum
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/zero-regulation-short-line
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

How can the voltage regulation of a short transmission line be made zero? Explain with necessary mathematical derivation.

## Short answer

Voltage regulation of a short line is made zero by operating the receiving end at a leading power factor equal to X/|Z| (tan φ = R/X). This is achieved by connecting a shunt capacitor or synchronous condenser to supply the required leading reactive power.

## Answer

The voltage regulation of a short line is zero when the receiving-end voltage at full load equals the sending-end voltage, i.e. $|V_S| = |V_R|$. This happens only for a suitable **leading** power factor at the receiving end.

#### Derivation
Short line: $V_S = V_R + I(R + jX)$. Take $V_R$ as reference and the current at angle $\phi$ (lagging taken positive):
$$
I = I\cos\phi - jI\sin\phi
$$
$$
\begin{aligned}
V_S &= V_R + (I\cos\phi - jI\sin\phi)(R + jX) \\
&= (V_R + IR\cos\phi + IX\sin\phi) + j(IX\cos\phi - IR\sin\phi)
\end{aligned}
$$
The quadrature part changes $|V_S|$ very little, so
$$
|V_S| \approx V_R + IR\cos\phi + IX\sin\phi
$$
$$
\%VR \approx \frac{I(R\cos\phi + X\sin\phi)}{V_R}\times100
$$
For a **leading** load the angle is negative, so $\sin\phi$ changes sign. Writing $\phi$ from here on for the **magnitude** of the leading angle:
$$
\%VR \approx \frac{I(R\cos\phi - X\sin\phi)}{V_R}\times100
$$

#### Condition for zero regulation
$$
\begin{aligned}
R\cos\phi - X\sin\phi &= 0 \\
\tan\phi &= \frac{R}{X} \\
\cos\phi &= \frac{X}{\sqrt{R^2 + X^2}} = \frac{X}{|Z|}\ \text{(leading)}
\end{aligned}
$$
Equivalently, the load angle $\phi = 90^\circ - \theta$, where $\theta = \tan^{-1}(X/R)$ is the line impedance angle.

```text
         I (leads Vr by phi)
        /      Vs (same length as Vr)
       /     .
      /   .   IZ (perpendicular to Vr+Vs bisector)
     / .
    O--------------> Vr
```

#### How it is achieved in practice
1. Connect a **shunt capacitor bank** or **synchronous condenser** at the receiving end so that the net receiving-end pf becomes $X/|Z|$ leading.
2. Required leading reactive power (approx.): $Q_{net} = -P\,R/X$, so capacitor rating $Q_C = P\tan\phi_{load} + P\,R/X$.
3. If the load is already more leading than this (negative regulation), a **shunt reactor** is added instead.

**Example:** $R = 4\ \Omega$, $X = 8\ \Omega$: zero regulation needs $\cos\phi = 8/\sqrt{80} = 0.894$ leading.

(For an exact value, solve $|V_R + Z(P - jQ)/V_R| = |V_R|$ for $Q$; it gives a slightly more leading pf than the approximate formula.)

## Check yourself

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

1. Under which condition is regulation zero?
   A. |V_S| = 2|V_R|
   B. |V_S| = |V_R| at full load, requiring a suitably leading pf
   C. Any lagging pf
   D. Unity pf
   Answer: B. Zero regulation means receiving voltage at load equals sending voltage; this needs leading pf. (Key point: Definition)

2. For leading load, %VR is approximately:
   A. I(R cos phi + X sin phi)/V_R x 100
   B. I(R cos phi - X sin phi)/V_R x 100
   C. I(X cos phi - R sin phi)/V_R
   D. I Z/V_R
   Answer: B. For leading phi the X sin(phi) term changes sign. (Key point: Approximate regulation)

3. What phi gives zero regulation?
   A. tan(phi) = X/R
   B. tan(phi) = R/X, cos(phi) = X/|Z| leading
   C. phi = 0
   D. phi = 90 degrees
   Answer: B. Setting R cos(phi) = X sin(phi) gives tan(phi) = R/X; equivalently phi = 90 degrees - theta. (Key point: Condition)

4. How is it achieved in practice?
   A. Series resistor
   B. Shunt capacitor or synchronous condenser at receiving end (shunt reactor if already too leading)
   C. Increase the load
   D. Reduce sending voltage only
   Answer: B. Compensation makes the net receiving pf X/|Z| leading; if load is more leading than needed a reactor is used. (Key point: Implementation)

## Related in Power System Analysis I

- [70 km, 40 MW line: regulation and zero-regulation pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-40mw-regulation) — Power System Analysis I, chapter 6
- [Capacitor for zero regulation, 70 km 220 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-220kv-zero-regulation-capacitor) — Power System Analysis I, chapter 6
- [15 km line: regulation and zero-regulation capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A15km-line-zero-regulation) — Power System Analysis I, chapter 6
- [12 km 33 kV line: losses, zero-regulation capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-33kv-losses-zero-regulation) — Power System Analysis I, chapter 6
- [33 kV, 12 MW leading load: efficiency, zero regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A33kv-12mw-leading-efficiency) — Power System Analysis I, chapter 6, asked 2×
- [Efficiency at 0.995 leading pf and zero regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aleading-pf-efficiency-zero-regulation) — Power System Analysis I, chapter 6

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