---
title: "220 kV line, 200 MW: sending end from ABCD"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:abcd-220kv-200mw"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/abcd-220kv-200mw
site: BCT Engineering
---

# 220 kV line, 200 MW: sending end from ABCD

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2073 Magh, 10 marks
- Video: [PS25 Numerical on Medium Transmission Lines](https://www.youtube.com/watch?v=Dt4Y5BlylMo) — Lectures in Electrical Engineering
- 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/abcd-220kv-200mw
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A three phase 220 kV, 50 Hz transmission line supplies a power of 200 MW at a power factor of 0.8 lagging and the line has ABCD parameters as follows: A = 0.9∠0.001°, B = B Ω (value not printed), C = 1.14 × 10⁻³∠90°, D = 0.9∠0.001°. Determine the following: (i) sending end voltage and current (ii) real and reactive power at sending end at full load (iii) real and reactive power loss in line at full load (iv) voltage regulation and line efficiency at full load.

## Short answer

The sending-end line voltage is 346.5 kV and current is 516.8 A. Sending-end power is 200.03 MW and 237.03 MVAr. Real power loss is 0.03 MW, reactive loss 87.03 MVAr; voltage regulation is 75% and efficiency is 99.99%.

## Answer

**Assumption:** $B$ is not printed, so it is found from the reciprocity condition $AD - BC = 1$:
$$
\begin{aligned}
B &= \frac{AD - 1}{C} = \frac{0.81\angle0.002^\circ - 1}{1.14\times10^{-3}\angle90^\circ} \\
&= \frac{-0.19}{1.14\times10^{-3}\angle90^\circ} = 166.67\angle89.99^\circ\ \Omega \approx j166.67\ \Omega
\end{aligned}
$$

#### Receiving-end quantities
$$
\begin{aligned}
V_R &= 220/\sqrt3 = 127.02\angle0^\circ\ \text{kV} \\
I_R &= \frac{200\times10^6}{\sqrt3\times220\times10^3\times0.8} = 656.08\angle-36.87^\circ\ \text{A}
\end{aligned}
$$

#### (i) Sending-end voltage and current
$$
\begin{aligned}
AV_R &= 114.32\angle0^\circ\ \text{kV},\quad BI_R = 109.35\angle53.12^\circ\ \text{kV} \\
V_S &= AV_R + BI_R = 200.07\angle25.93^\circ\ \text{kV (phase)} \\
CV_R &= 144.80\angle90^\circ\ \text{A},\quad DI_R = 590.47\angle-36.87^\circ\ \text{A} \\
I_S &= CV_R + DI_R = 516.75\angle-23.92^\circ\ \text{A}
\end{aligned}
$$
Sending-end line voltage $= \sqrt3\times200.07 = 346.53$ kV.

#### (ii) Sending-end power
$$
S_S = 3V_SI_S^* = 3(200.07\angle25.93^\circ)(0.51675\angle23.92^\circ) = 200.03 + j237.03\ \text{MVA}
$$
$P_S$ = 200.03 MW, $Q_S$ = 237.03 MVAr (pf 0.645 lag).

#### (iii) Losses
Receiving end: $S_R = 200 + j150$ MVA.
$$
\begin{aligned}
P_{loss} &= 200.03 - 200 = 0.03\ \text{MW} \\
Q_{loss} &= 237.03 - 150 = 87.03\ \text{MVAr}
\end{aligned}
$$
The active loss is almost zero because the given $A$, $D$ have nearly zero angle, i.e. the line is practically lossless.

#### (iv) Regulation and efficiency
$$
\begin{aligned}
\%VR &= \frac{|V_S|/|A| - |V_R|}{|V_R|}\times100 = \frac{200.07/0.9 - 127.02}{127.02}\times100 = 75.02\% \\
\eta &= \frac{200}{200.03}\times100 = 99.99\%
\end{aligned}
$$

**Answer: $V_S$ = 346.5 kV (line), $I_S$ = 516.8 A; $P_S$ = 200.03 MW, $Q_S$ = 237.03 MVAr; losses 0.03 MW and 87.03 MVAr; regulation ≈ 75 %, efficiency ≈ 99.99 %.** (The very high regulation shows the line is loaded far above its surge impedance loading of about 127 MW.)

## Check yourself

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

1. How is the missing B determined?
   A. B = A/C
   B. B = C/D
   C. B = 1/C
   D. B = (AD - 1)/C from AD - BC = 1
   Answer: D. Reciprocity gives B = (0.81 - 1)/(j1.14e-3) = about j166.67 ohm. (Key point: Finding B)

2. Which equations give V_S and I_S?
   A. V_S = V_R + I_R B only
   B. V_S = A V_R + B I_R and I_S = C V_R + D I_R
   C. V_S = D V_R + C I_R
   D. I_S = A V_R + B I_R
   Answer: B. ABCD parameters: V_S = A V_R + B I_R, I_S = C V_R + D I_R; here V_S = 200.07 kV phase, I_S = 516.75 A. (Key point: Sending quantities)

3. How is the reactive power loss found, and what is it?
   A. 237.03 + 150
   B. Q_R/Q_S
   C. Q_S - Q_R = 237.03 - 150 = 87.03 MVAr, with S_R = 200 + j150
   D. P_S - P_R
   Answer: C. Receiving Q = 200 x 0.75 = 150 MVAr; difference with sending Q is the line's loss. (Key point: Reactive loss)

4. Why is regulation so high (about 75 %) while efficiency is about 99.99 %?
   A. The line has large resistance
   B. The line is practically lossless (angles near zero) but loaded far above SIL, so the voltage drop is large
   C. The power factor is leading
   D. The load is below SIL
   Answer: B. Active loss is only 0.03 MW but heavy reactive flow (load 200 MW vs SIL about 127 MW) causes a big drop. (Key point: High regulation)

## Related in Power System Analysis I

- [150 mile, 60 Hz line: ABCD and sending end](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150-mile-60hz-line-abcd) — Power System Analysis I, chapter 5
- [110 km, 110 kV line: sending end, regulation, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A110km-line-performance) — Power System Analysis I, chapter 6, asked 2×
- [Receiving and sending end powers from ABCD, 132/134 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-power-132kv-134kv) — Power System Analysis I, chapter 6
- [Sending end active power from ABCD, 132/130.5 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asending-power-132kv-130-5kv) — Power System Analysis I, chapter 6
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