--- title: "Equivalent ABCD of lines in parallel" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:parallel-lines-abcd-derivation" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/parallel-lines-abcd-derivation site: BCT Engineering --- # Equivalent ABCD of lines in parallel Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2068 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-derivation - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question Two non-identical transmission lines represented by their respective ABCD parameters are connected in parallel. Derive the expression for the equivalent ABCD parameters of the transmission system. ## Short answer Expressing each line in admittance form and summing currents yields equivalent ABCD: A = (A₁B₂ + A₂B₁)/(B₁+B₂), B = B₁B₂/(B₁+B₂), D = (D₁B₂ + D₂B₁)/(B₁+B₂), C = C₁ + C₂ + ((A₁ − A₂)(D₂ − D₁))/(B₁+B₂). ## Answer Two lines with constants $(A_1, B_1, C_1, D_1)$ and $(A_2, B_2, C_2, D_2)$ are connected in parallel: same $V_S$ and $V_R$ at both ends; the currents add. ```text IS1 +------------+ IR1 IS +--->---| A1 B1 C1 D1|--->---+ IR o-->--+ +------------+ +-->--o VS | IS2 +------------+ IR2 | VR +--->---| A2 B2 C2 D2|--->---+ +------------+ o------------------------------------o ``` #### Step 1: Write each line in "admittance" form From $V_S = A_kV_R + B_kI_{Rk}$: $$ I_{Rk} = \frac{V_S - A_kV_R}{B_k}, \quad k = 1, 2 $$ Total receiving current: $$ I_R = I_{R1} + I_{R2} = V_S\left(\frac{1}{B_1} + \frac{1}{B_2}\right) - V_R\left(\frac{A_1}{B_1} + \frac{A_2}{B_2}\right) $$ #### Step 2: Solve for $V_S$ (gives A and B) $$ V_S\,\frac{B_1 + B_2}{B_1B_2} = I_R + V_R\,\frac{A_1B_2 + A_2B_1}{B_1B_2} $$ $$ V_S = \frac{A_1B_2 + A_2B_1}{B_1 + B_2}V_R + \frac{B_1B_2}{B_1 + B_2}I_R $$ $$ A = \frac{A_1B_2 + A_2B_1}{B_1 + B_2}, \qquad B = \frac{B_1B_2}{B_1 + B_2} $$ #### Step 3: Sending current (gives C and D) $I_{Sk} = C_kV_R + D_kI_{Rk}$, and $I_S = I_{S1} + I_{S2}$: $$ I_S = (C_1 + C_2)V_R + D_1\frac{V_S - A_1V_R}{B_1} + D_2\frac{V_S - A_2V_R}{B_2} $$ Substituting $V_S$ from Step 2 and simplifying with $A_kD_k - B_kC_k = 1$: $$ D = \frac{D_1B_2 + D_2B_1}{B_1 + B_2}, \qquad C = C_1 + C_2 + \frac{(A_1 - A_2)(D_2 - D_1)}{B_1 + B_2} $$ #### Result $$ \begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} \frac{A_1B_2 + A_2B_1}{B_1 + B_2} & \frac{B_1B_2}{B_1 + B_2} \\ C_1 + C_2 + \frac{(A_1 - A_2)(D_2 - D_1)}{B_1 + B_2} & \frac{D_1B_2 + D_2B_1}{B_1 + B_2} \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix} $$ Special case: for two identical lines, $A = A_1$, $B = B_1/2$, $C = 2C_1$, $D = D_1$. The result still satisfies $AD - BC = 1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For two lines connected in parallel, which quantities are common and which add? A. Sending and receiving voltages are common; the currents add B. Currents are common; the voltages add C. Sending voltage is common; receiving voltages add D. Voltages and currents both add Answer: A. Both lines share the same V_S and V_R across their terminals, so I_R = I_R1 + I_R2 and I_S = I_S1 + I_S2. (Key point: Parallel connection conditions) 2. Starting point of the derivation: from V_S = A_k V_R + B_k I_Rk, the receiving current of line k is A. I_Rk = (V_S + A_k V_R)/B_k B. I_Rk = (V_S - A_k V_R)/B_k C. I_Rk = (V_S - B_k V_R)/A_k D. I_Rk = A_k(V_S - V_R)/B_k Answer: B. Rearranging V_S = A_k V_R + B_k I_Rk for I_Rk gives (V_S - A_k V_R)/B_k. (Key point: Admittance form) 3. What is the equivalent B constant of two lines in parallel? A. B1 + B2 B. (B1 + B2)/2 C. B1B2/(B1 + B2) D. (A1B2 + A2B1)/(B1 + B2) Answer: C. Series impedances B1 and B2 behave like parallel impedances, so B = B1B2/(B1+B2). (Key point: Equivalent B) 4. Which expression gives the equivalent A of the parallel combination? A. A1 + A2 B. A1A2 + B1C2 C. (A1B1 + A2B2)/(B1 + B2) D. (A1B2 + A2B1)/(B1 + B2) Answer: D. Solving for V_S after adding the receiving currents gives A = (A1B2 + A2B1)/(B1 + B2); D has the same form with D1, D2. 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