--- title: "Nominal-T phasor diagram and ABCD parameters" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:nominal-t-phasor-abcd" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-t-phasor-abcd site: BCT Engineering --- # Nominal-T phasor diagram and ABCD parameters Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2081 Chaitra (new course), 4 marks; 2075 Baisakh, 5 marks (asked 2 times) - Video: [ABCD parameters of medium transmission line by Nominal T method explained by Irfan Mujawar](https://www.youtube.com/watch?v=HzHGlzKCBPI) — Dr. Irfan Mujawar - 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/nominal-t-phasor-abcd - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question Draw the phasor diagram for a nominal 'Tee' circuit of a transmission line for lagging condition and find its A, B, C, D parameters. ## Short answer The phasor diagram takes V_R as reference, then Vc = V_R + I_R Z/2, Ic leading Vc, Is = I_R + Ic, and Vs = Vc + Is Z/2. ABCD constants: A = D = 1+YZ/2, B = Z(1+YZ/4), C = Y. ## Answer In the **nominal-T** model of a medium transmission line, the total series impedance $Z = R + jX$ is split into two halves $Z/2$ at each end, and the total shunt admittance $Y = j\omega C$ is placed at the middle. #### Circuit ```text Is Z/2 Vc Z/2 Ir o---[R/2 X/2]---+---[R/2 X/2]---o + | + Vs Ic| Y Vr - | - o---------------+---------------o ``` #### Phasor diagram (lagging load) Take $V_R$ as reference; $I_R$ lags it by $\phi_R$. Currents: ```text Ic (leads Vc by ~90 deg) ^ | Is = Ir + Ic | / | / O----+-+--------------> Vr (reference) \ \ phi_R (lag) v Ir ``` Voltages (drops added head to tail): ```text Vs /| / | Is.X/2 (90 deg / | ahead of Is) /---+ Is.R/2 (along Is) Vc /| / | Ir.X/2 (90 deg / | ahead of Ir) /---+ Ir.R/2 (along Ir) / O-----------------+ Vr Vs leads Vr by the load angle delta ``` Construction steps: 1. Draw $V_R$ along the reference axis; draw $I_R$ lagging by $\phi_R$. 2. Add $I_R R/2$ parallel to $I_R$ and $I_R X/2$ perpendicular (leading $I_R$ by 90°) to get $V_C$. 3. Draw $I_C$ leading $V_C$ by 90°; $I_S = I_R + I_C$. 4. Add $I_S R/2$ parallel to $I_S$ and $I_S X/2$ perpendicular to it to get $V_S$. #### Derivation of ABCD constants Voltage across the shunt branch: $$ V_C = V_R + I_R\frac{Z}{2} $$ Shunt current and sending-end current: $$ \begin{aligned} I_C &= YV_C = YV_R + \frac{YZ}{2}I_R \\ I_S &= I_R + I_C = YV_R + \left(1 + \frac{YZ}{2}\right)I_R \end{aligned} $$ Sending-end voltage: $$ \begin{aligned} V_S &= V_C + I_S\frac{Z}{2} \\ &= V_R + I_R\frac{Z}{2} + \frac{Z}{2}\left[YV_R + \left(1 + \frac{YZ}{2}\right)I_R\right] \\ &= \left(1 + \frac{YZ}{2}\right)V_R + Z\left(1 + \frac{YZ}{4}\right)I_R \end{aligned} $$ Comparing with $V_S = AV_R + BI_R$ and $I_S = CV_R + DI_R$: | Constant | Nominal-T value | Unit | |---|---|---| | $A$ | $1 + \dfrac{YZ}{2}$ | none | | $B$ | $Z\left(1 + \dfrac{YZ}{4}\right)$ | Ω | | $C$ | $Y$ | S | | $D$ | $1 + \dfrac{YZ}{2}$ | none | **Check:** $AD - BC = \left(1 + \frac{YZ}{2}\right)^2 - YZ\left(1 + \frac{YZ}{4}\right) = 1$, and $A = D$ since the circuit is symmetrical. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the nominal-T model, the shunt admittance is placed: A. At the midpoint, with Z/2 on each side B. At the sending end C. Split Y/2 at each end D. At the receiving end only Answer: A. T model has Z/2 - Y - Z/2. (Key point: Nominal-T) 2. Starting from V_R, to get V_C you add: A. I_S Z B. I_R R/2 along I_R and I_R X/2 90° ahead of I_R C. I_C Z/2 D. I_R X along V_R Answer: B. V_C = V_R + I_R(Z/2), whose drops are in phase and in quadrature with I_R. (Key point: Phasor construction) 3. In nominal-T, A = D equals: A. 1 + YZ/4 B. 1 C. 1 + YZ/2 D. Z(1 + YZ/4) Answer: C. V_S coefficient of V_R is 1 + YZ/2. (Key point: A and D) 4. The B and C constants of nominal-T are: A. B = Z, C = Y(1+YZ/4) B. B = Y, C = Z C. B = ZY, C = 1 D. B = Z(1 + YZ/4), C = Y Answer: D. Substitution gives B = Z(1+YZ/4) and C = Y; AD - BC = 1. 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