---
title: "ABCD with shunt capacitor at line end"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:abcd-with-shunt-capacitor"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/abcd-with-shunt-capacitor
site: BCT Engineering
---

# ABCD with shunt capacitor at line end

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2071 Magh, 5 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/abcd-with-shunt-capacitor
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

The ABCD constants of a 3-phase overhead line are as follows: A = D = 0.9955∠0.086°, B = 31.552∠71.608° Ω, C = 3 × 10⁻⁴∠90° S. If a shunt capacitor with admittance 2.5 × 10⁻⁴∠90° S is connected at the end of the line, determine the equivalent ABCD constants of the system.

## Short answer

By cascading line and shunt capacitor matrices: A' = A + B Y_sh = 0.9880∠0.231°, B' = B = 31.552∠71.608° Ω, C' = C + D Y_sh = 5.489×10⁻⁴∠90.04° S, D' = D = 0.9955∠0.086°.

## Answer

**Given:** $A = D = 0.9955\angle 0.086^\circ$, $B = 31.552\angle 71.608^\circ\ \Omega$, $C = 3\times10^{-4}\angle 90^\circ$ S; shunt capacitor $Y_{sh} = 2.5\times10^{-4}\angle 90^\circ$ S at the receiving end of the line.

#### Method
The shunt capacitor is a two-port with constants $\begin{bmatrix} 1 & 0 \\ Y_{sh} & 1 \end{bmatrix}$. It is in cascade (series) after the line, so the overall matrix is the product:

```text
 VS    +------------+         VR
 o-----|  Line      |----+-----o
       |  A B C D   |    |
       +------------+  [Ysh]
 o-----------------------+-----o
```

$$
\begin{bmatrix} A' & B' \\ C' & D' \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix}\begin{bmatrix} 1 & 0 \\ Y_{sh} & 1 \end{bmatrix} = \begin{bmatrix} A + BY_{sh} & B \\ C + DY_{sh} & D \end{bmatrix}
$$

#### Calculation

$$
\begin{aligned}
BY_{sh} &= 31.552\angle 71.608^\circ\times2.5\times10^{-4}\angle 90^\circ = 7.888\times10^{-3}\angle 161.608^\circ \\
&= -0.007485 + j0.002489 \\
A' &= (0.995499 + j0.001494) + (-0.007485 + j0.002489) \\
&= 0.98801 + j0.00398 = 0.9880\angle 0.231^\circ \\
B' &= B = 31.552\angle 71.608^\circ\ \Omega \\
DY_{sh} &= 0.9955\angle 0.086^\circ\times2.5\times10^{-4}\angle 90^\circ = 2.4888\times10^{-4}\angle 90.086^\circ \\
C' &= j3\times10^{-4} + (-3.7\times10^{-7} + j2.4887\times10^{-4}) \\
&= 5.489\times10^{-4}\angle 90.04^\circ\ \text{S} \\
D' &= D = 0.9955\angle 0.086^\circ
\end{aligned}
$$

Check: $A'D' - B'C' = 1.000$ (to 5 decimal places).

**Answer:** $A' = 0.9880\angle 0.231^\circ$, $B' = 31.552\angle 71.608^\circ\ \Omega$, $C' = 5.489\times10^{-4}\angle 90.04^\circ$ S, $D' = 0.9955\angle 0.086^\circ$.

Note: the combined network is no longer symmetrical ($A' \ne D'$). If the capacitor were at the sending end, $A$ and $D$ would swap roles: $A' = 0.9955\angle 0.086^\circ$, $D' = 0.9880\angle 0.231^\circ$.

## Check yourself

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

1. What is the ABCD matrix of a shunt capacitor of admittance Ysh?
   A. [1 Ysh; 0 1]
   B. [1 0; Ysh 1]
   C. [Ysh 0; 0 Ysh]
   D. [1 Ysh; Ysh 1]
   Answer: B. A shunt admittance has A = D = 1, B = 0 and C = Ysh. (Key point: ABCD of shunt element)

2. How are the equivalent constants found when the capacitor is at the receiving end of the line?
   A. Add the matrices
   B. Multiply the line matrix by the capacitor matrix (line first)
   C. Multiply the line ABCD values element by element
   D. Average the two matrices
   Answer: B. Cascaded two-ports multiply in order: [line][capacitor] = [A + B Ysh, B; C + D Ysh, D]. (Key point: Cascading rule)

3. Which equivalent constants change when the capacitor is added at the receiving end?
   A. A and C only (B and D unchanged)
   B. B and D only
   C. All four
   D. Only C
   Answer: A. A' = A + B Ysh and C' = C + D Ysh; B' = B and D' = D. (Key point: Which constants change)

## Related in Power System Analysis I

- [Series R, X and shunt admittance from ABCD](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aline-constants-from-abcd) — Power System Analysis I, chapter 5
- [150 km line with mid-point shunt capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150km-mid-shunt-capacitor) — Power System Analysis I, chapter 6
- [220 kV line, 200 MW: sending end from ABCD](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-220kv-200mw) — Power System Analysis I, chapter 6
- [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
- [Two lines in parallel: ABCD constants](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aparallel-lines-abcd-numerical) — Power System Analysis I, chapter 5

## Related in other subjects

- [Fault at load end of overhead line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-load-end-33kv-line) — Power System Analysis II, chapter 3
- [Shunt reactors for 750 kV line ABCD](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Ashunt-reactor-abcd-750kv) — High Voltage Engineering, chapter 3