---
title: "Two lines in parallel: ABCD constants"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:parallel-lines-abcd-numerical"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/parallel-lines-abcd-numerical
site: BCT Engineering
---

# Two lines in parallel: ABCD constants

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2069 Bhadra, 6 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/parallel-lines-abcd-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

Two lines are represented by their respective ABCD constants and are connected in parallel. The line constants are as follows: A1 = 0.980∠0.1° = D1, B1 = 45∠73.2° Ohm; A2 = 0.975∠0.2° = D2, B2 = 55∠75.1° Ohm. [The paper does not state what is to be found; presumably the equivalent ABCD constants of the combination.]

## Short answer

Using parallel two-port formulas with C_i = (A_i²-1)/B_i: equivalent parameters are A = D = 0.9778 angle 0.147°, B = 24.75 angle 74.06° Ω, C = 1.788×10⁻³ angle 99.57° S. The method combines A1B2 + A2B1 over B1+B2 for A, and B1B2/(B1+B2) for B.

## Answer

Assumption: the equivalent ABCD constants of the two lines in parallel are required.

**Given:** $A_1 = D_1 = 0.980\angle 0.1^\circ$, $B_1 = 45\angle 73.2^\circ\ \Omega$; $A_2 = D_2 = 0.975\angle 0.2^\circ$, $B_2 = 55\angle 75.1^\circ\ \Omega$.

#### Formulas for two networks in parallel

$$
\begin{aligned}
A &= \frac{A_1B_2 + A_2B_1}{B_1 + B_2}, \quad B = \frac{B_1B_2}{B_1 + B_2}, \quad D = \frac{D_1B_2 + D_2B_1}{B_1 + B_2} \\
C &= C_1 + C_2 + \frac{(A_1 - A_2)(D_2 - D_1)}{B_1 + B_2}
\end{aligned}
$$

$C_1$ and $C_2$ are not given; since each line is symmetrical and reciprocal, $C = (AD - 1)/B = (A^2 - 1)/B$.

#### Step 1: C of each line

$$
\begin{aligned}
C_1 &= \frac{A_1^2 - 1}{B_1} = 8.833\times10^{-4}\angle 101.96^\circ\ \text{S} \\
C_2 &= \frac{A_2^2 - 1}{B_2} = 9.062\times10^{-4}\angle 97.25^\circ\ \text{S}
\end{aligned}
$$

#### Step 2: Common terms

$$
\begin{aligned}
B_1 + B_2 &= (13.01 + j43.08) + (14.14 + j53.15) = 27.15 + j96.23 = 99.99\angle 74.25^\circ\ \Omega \\
A_1B_2 &= 53.90\angle 75.20^\circ, \quad A_2B_1 = 43.875\angle 73.40^\circ \\
A_1B_2 + A_2B_1 &= (13.77 + j52.11) + (12.54 + j42.05) = 26.30 + j94.16
\end{aligned}
$$

#### Step 3: Equivalent constants

$$
\begin{aligned}
A = D &= \frac{26.30 + j94.16}{27.15 + j96.23} = 0.9778\angle 0.147^\circ \\
B &= \frac{45\times55\angle 148.3^\circ}{99.99\angle 74.25^\circ} = \frac{2475\angle 148.3^\circ}{99.99\angle 74.25^\circ} = 24.75\angle 74.06^\circ\ \Omega \\
C &= C_1 + C_2 - \frac{(A_1 - A_2)^2}{B_1 + B_2} \\
&= (-1.831 + j8.641)\times10^{-4} + (-1.143 + j8.990)\times10^{-4} - 2.8\times10^{-7}\angle -111.6^\circ \\
&= 1.788\times10^{-3}\angle 99.57^\circ\ \text{S}
\end{aligned}
$$

Check: $AD - BC = 1.000$.

**Answer:** $A = D = 0.9778\angle 0.147^\circ$, $B = 24.75\angle 74.06^\circ\ \Omega$, $C = 1.788\times10^{-3}\angle 99.57^\circ$ S.

(The C angles above 90° come from the given A and B values; real lines have C close to 90°.)

## Check yourself

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

1. What is the equivalent B of two lines in parallel?
   A. B1 + B2
   B. B1 B2 / (B1 + B2)
   C. (B1 + B2)/2
   D. B1 / B2
   Answer: B. Like parallel impedances, B = B1B2/(B1 + B2), here 24.75∠74.06° Ω. (Key point: Equivalent B of parallel lines)

2. What is the equivalent A of two parallel lines?
   A. A = (A1 + A2)/2
   B. A = (A1B2 + A2B1)/(B1 + B2)
   C. A = A1 A2
   D. A = (A1B1 + A2B2)/(B1 + B2)
   Answer: B. The parallel formulas weight each A by the other line's B: A = (A1B2 + A2B1)/(B1 + B2). (Key point: Equivalent A formula)

3. C1 and C2 are not given. How are they found for symmetrical, reciprocal lines?
   A. C = (AD - 1)/B = (A^2 - 1)/B
   B. C = B/(A - 1)
   C. C = A/B
   D. C = 1/B
   Answer: A. Reciprocity gives AD - BC = 1, so C = (AD - 1)/B and with A = D, C = (A^2 - 1)/B. (Key point: Finding missing C)

4. Which formula gives C of the parallel combination?
   A. C = C1 + C2
   B. C = C1 + C2 + (A1 - A2)(D2 - D1)/(B1 + B2)
   C. C = C1 C2/(C1 + C2)
   D. C = C1 + C2 - B1 B2
   Answer: B. Besides adding C1 and C2, there is a cross term from the difference in A and D of the two lines; with A = D it equals -(A1 - A2)^2/(B1 + B2). (Key point: Equivalent C term)

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