---
title: "70 km, 40 MW line: regulation and zero-regulation pf"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:70km-40mw-regulation"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/70km-40mw-regulation
site: BCT Engineering
---

# 70 km, 40 MW line: regulation and zero-regulation pf

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2070 Bhadra, 6+2 marks
- 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/70km-40mw-regulation
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

At the receiving end of a 70 km long transmission line a 40 MW load at 0.8 lagging power factor is connected at 66 kV. The series resistance and inductive reactance of the conductor per phase are 0.03 ohm/km and 0.15 ohm/km respectively. Calculate (i) voltage regulation of the line and sending end power factor (ii) receiving end power factor to make the voltage regulation zero.

## Short answer

For the short line, voltage regulation is 9.47% and sending-end power factor is 0.753 lagging. To achieve zero regulation, the receiving-end power factor must be about 0.981 leading (exact 0.969 leading).

## Answer

Short line (70 km), shunt capacitance neglected.

#### Data
$$
\begin{aligned}
R &= 0.03\times70 = 2.1\ \Omega,\quad X = 0.15\times70 = 10.5\ \Omega \\
V_R &= 66\,000/\sqrt3 = 38\,105.1\ \text{V} \\
I &= \frac{40\times10^6}{\sqrt3\times66\,000\times0.8} = 437.39\angle-36.87^\circ\ \text{A}
\end{aligned}
$$

#### (i) Regulation and sending-end pf
$$
\begin{aligned}
IZ &= 437.39\angle-36.87^\circ\times10.708\angle78.69^\circ = 4683.5\angle41.82^\circ\ \text{V} \\
V_S &= 38\,105.1 + 4683.5\angle41.82^\circ = 41\,712.5\angle4.29^\circ\ \text{V}
\end{aligned}
$$
Line value $= \sqrt3\times41.713 = 72.25$ kV.
$$
\%VR = \frac{72.25 - 66}{66}\times100 = 9.47\%
$$
Angle between $V_S$ and $I$ $= 4.29^\circ + 36.87^\circ = 41.16^\circ$:
$$
\text{pf}_S = \cos41.16^\circ = 0.753\ \text{lagging}
$$

#### (ii) Receiving-end pf for zero regulation
Approximate condition (leading pf):
$$
\begin{aligned}
IR\cos\phi - IX\sin\phi &= 0 \\
\tan\phi &= \frac{R}{X} = \frac{2.1}{10.5} = 0.2 \\
\cos\phi &= 0.981\ \text{leading}
\end{aligned}
$$
(The exact condition $|V_S| = |V_R|$ with $P$ = 40 MW gives $Q = -10.13$ MVAr, i.e. pf 0.969 leading.)

**Answer: (i) Regulation ≈ 9.47 %, sending-end pf ≈ 0.753 lag; (ii) receiving-end pf ≈ 0.981 leading (exact 0.969 leading).**

## Check yourself

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

1. How is regulation found for this 70 km line?
   A. (66 - 72.25)/66
   B. V_S = V_R - I Z
   C. I R/V_R
   D. V_S = V_R + I Z, then (72.25 - 66)/66 = 9.47 %
   Answer: D. Short line: V_S line = 72.25 kV. (Key point: Regulation)

2. How is the sending pf found?
   A. cos(78.69)
   B. cos(4.29 + 36.87) = cos(41.16) = 0.753 lagging
   C. cos(36.87) = 0.8
   D. cos(4.29)
   Answer: B. Angle between V_S and I = delta + phi_R. (Key point: Sending pf)

3. What pf gives zero regulation (approx)?
   A. unity
   B. 0.5 leading
   C. tan(phi) = R/X = 0.2, cos(phi) = 0.981 leading
   D. 0.8 lagging
   Answer: C. R cos(phi) = X sin(phi). (Key point: Zero regulation)

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