---
title: "Series R, X and shunt admittance from ABCD"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:line-constants-from-abcd"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/line-constants-from-abcd
site: BCT Engineering
---

# Series R, X and shunt admittance from ABCD

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2070 Bhadra, 4 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/line-constants-from-abcd
- 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. Determine the series resistance, series inductive reactance and shunt admittance of the line.

## Short answer

Assume nominal-pi: series impedance Z = B = 31.552 angle 71.608° Ω, giving R = 9.96 Ω and X_L = 29.94 Ω per phase. From A = 1 + ZY/2, shunt admittance Y = 2(A-1)/Z ≈ j3.0×10⁻⁴ S.

## 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.

Since $|A|$ is close to 1 and $C$ is purely capacitive, this is a medium line. The **nominal-Π** model is assumed, for which $B = Z$, $A = 1 + ZY/2$ and $C = Y(1 + ZY/4)$.

#### Series impedance

$$
\begin{aligned}
Z &= B = 31.552\angle 71.608^\circ \\
R &= 31.552\cos 71.608^\circ = 9.955\ \Omega \\
X_L &= 31.552\sin 71.608^\circ = 29.94\ \Omega
\end{aligned}
$$

#### Shunt admittance
From $A = 1 + \frac{ZY}{2}$:

$$
\begin{aligned}
A - 1 &= (0.995499 + j0.001494) - 1 = -0.004501 + j0.001494 = 4.743\times10^{-3}\angle 161.63^\circ \\
Y &= \frac{2(A - 1)}{Z} = \frac{2\times4.743\times10^{-3}\angle 161.63^\circ}{31.552\angle 71.608^\circ} \\
&= 3.006\times10^{-4}\angle 90.03^\circ \approx j3.0\times10^{-4}\ \text{S}
\end{aligned}
$$

Check with the given $C$: $Y(1 + ZY/4) = 2.999\times10^{-4}\angle 90.07^\circ$ S $\approx 3\times10^{-4}\angle 90^\circ$ S, which agrees. (A nominal-T reading, $C = Y$, gives nearly the same numbers: $Z = 9.96 + j30.01\ \Omega$.)

**Answer:** Series resistance $R \approx 9.96\ \Omega$, series inductive reactance $X_L \approx 29.94\ \Omega$, shunt admittance $Y \approx j3.0\times10^{-4}$ S (all per phase).

## Check yourself

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

1. In the nominal-Π assumption, how are R and XL found from the given constants?
   A. Z = B, so R = |B| cos θ and XL = |B| sin θ
   B. Z = A, so R = |A| cos θ
   C. Z = 1/C
   D. Z = B/A
   Answer: A. For nominal-Π B = Z, so R = 31.552 cos 71.608° = 9.955 Ω and XL = 31.552 sin 71.608° = 29.94 Ω. (Key point: Series impedance from B)

2. How is Y obtained after Z is known?
   A. Y = (A - 1)/Z
   B. Y = 2(A - 1)/Z
   C. Y = C/2
   D. Y = 2(1 - A) Z
   Answer: B. From A = 1 + ZY/2, Y = 2(A - 1)/Z, giving about j3.0e-4 S. (Key point: Shunt admittance from A)

3. How is the result checked against the given C?
   A. Compute Y(1 + ZY/4) and compare with C
   B. Check that A = D
   C. Check that B = 1/C
   D. Check that Z = Y
   Answer: A. For nominal-Π C = Y(1 + ZY/4); it comes to 2.999e-4∠90.07° S, matching the given 3e-4∠90° S. (Key point: Consistency check)

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