--- title: "Effect of excitation on steady state stability" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:excitation-effect-steady-state" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/excitation-effect-steady-state site: BCT Engineering --- # Effect of excitation on steady state stability Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2071 Shrawan, 4 marks - Video: [Steady state stability with AVR](https://www.youtube.com/watch?v=LiyoVh2eYM8) — Dr. Bipasa Bimalendu Patra - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/excitation-effect-steady-state - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question Explain the effect of change of excitation on the steady state stability of a synchronous generator feeding an infinite bus bar. ## Short answer Excitation determines internal emf E. Increasing E raises the steady-state stability limit Pmax=EV/X and stability margin. Decreasing E reduces Pmax and can cause loss of synchronism if Pmax falls below mechanical input. ## Answer For a generator connected to an infinite bus, the power transferred is $$ P_e = \frac{E\,V}{X}\sin\delta, \qquad P_{max} = \frac{EV}{X} $$ The **steady-state stability limit** is $P_{max}$, reached at $\delta = 90^\circ$. Excitation sets the internal emf $E$, so it changes this limit directly. #### Increase in excitation (over-excitation) - $E$ increases, so $P_{max} = EV/X$ increases and the power-angle curve becomes taller. - For the same mechanical input $P_m$, the operating angle $\delta = \sin^{-1}(P_m X/EV)$ becomes smaller. - The synchronising power coefficient $P_s = (EV/X)\cos\delta$ becomes larger. - The steady-state stability margin $(P_{max}-P_m)/P_{max}$ increases. The machine also supplies lagging reactive power to the bus. #### Decrease in excitation (under-excitation) - $E$ falls, $P_{max}$ falls, and $\delta$ rises toward $90^\circ$ for the same $P_m$. - $P_s$ becomes small, so the machine is weakly held in synchronism. - If excitation is reduced too far, $P_{max}$ drops below $P_m$ and the machine **pulls out of step**. Under-excitation limiters guard against this. ```text Pe | ___ E2 > E1 (higher excitation) | / \ | / __ \ E1 | / / \ \ Pm-+-*---*-----\---------- | /d2 d1 \ +--------90------180-- delta ``` #### Effect of the AVR A fast automatic voltage regulator raises $E$ when the load angle increases. This effectively holds the terminal voltage, and the practical stability limit rises above the value set by a fixed $E$ (**dynamic stability limit**). Too high an AVR gain can, however, reduce damping, which is why power system stabilisers are used. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a generator on an infinite bus, the steady-state stability limit is: A. Pmax = EV/X reached at delta = 90 degrees B. Pmax = EV/X reached at delta = 180 degrees C. Pm = 0 D. EV X at delta = 0 Answer: A. Pe = (EV/X) sin(delta) is maximum at delta = 90 degrees. (Key point: Stability limit) 2. If excitation increases with Pm fixed, the operating angle delta: A. Increases B. Decreases, and Ps increases C. Stays the same D. Goes beyond 90 degrees Answer: B. Higher E raises Pmax, so delta = asin(PmX/EV) falls and Ps = (EV/X)cos(delta) rises. (Key point: Increase in excitation) 3. If excitation is reduced too far: A. Pmax rises above Pm B. Ps becomes very large C. Pmax drops below Pm and the machine pulls out of step D. The AVR is disconnected Answer: C. Lower E reduces Pmax; once Pmax < Pm there is no equilibrium angle. (Key point: Under-excitation) 4. What does a fast AVR do to the practical stability limit? A. Lowers it below EV/X B. Has no effect C. Makes Pe zero D. Raises it above the fixed-E value (dynamic stability limit) Answer: D. The AVR raises E as the angle rises, extending the limit, though too high a gain reduces damping. 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