--- title: "Varying excitation and V curves" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:excitation-v-curves" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/excitation-v-curves site: BCT Engineering --- # Varying excitation and V curves Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2073 Shrawan, 7 marks - Video: [V-Curves And Inverted V-Curves Of Synchronous Motor](https://www.youtube.com/watch?v=SHk49rSvVp8) — Electrical Engineering Idea's - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/excitation-v-curves - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question Explain the effect of varying excitation on armature current and power factor in a synchronous motor. Draw V curves and state their significance. ## Short answer Varying field excitation changes the armature current and power factor at constant load. Under-excitation gives lagging pf and high current; normal excitation yields minimum current at unity pf; over-excitation gives leading pf. V-curves plot armature current vs field current, with minima at unity pf. They show excitation for unity pf, leading kVAR supply as a synchronous condenser, and stability limits. ## Answer At constant load and supply voltage, the field current $I_f$ of a synchronous motor controls the back emf $E_f$; this changes the **armature current** and the **power factor**, but not the speed or the active power. #### Effect of varying excitation Input power per phase $P = VI_a\cos\phi$ is constant, so the **active component $I_a\cos\phi$ is constant**; the tip of the $I_a$ phasor moves along a line perpendicular to $V$. Also $E_f\sin\delta$ is constant, so the tip of $E_f$ moves on a line parallel to $V$. ```text Ia3 (lead) Ia2 (upf) \ | \ | ------------+-------------+--------> V \ | locus of Ia tip Ia1 (lag) | (Ia cos(phi) const) ``` 1. **Under-excitation ($E_f < V$):** $I_a$ lags $V$; it has a large lagging reactive component. Current is high, pf **lagging**. 2. **Increasing $I_f$:** the lagging component falls, so $I_a$ decreases and pf improves. 3. **Normal excitation:** $I_a$ is in phase with $V$; current is **minimum**, pf = **unity**. 4. **Over-excitation ($E_f > V$):** $I_a$ leads $V$ and grows again; pf **leading**. The motor now supplies reactive power to the system. #### V-curves Plotting $I_a$ against $I_f$ for constant power output gives V-shaped curves. ```text Ia |F F F = full load | F F H = half load |H F F H N = no load | H F F F H |N H H N | N H H N | N N N +------------------- If lagging | leading upf (minima) ``` - One V-curve for each load (no load, half load, full load); higher load curves lie higher and to the right. - The minimum point of each curve corresponds to unity pf. The line joining them is the **unity pf compounding curve**; points left of it are lagging, right are leading. - Plotting pf against $I_f$ gives **inverted V-curves**, with peak (pf = 1) at normal excitation. #### Significance of V-curves 1. Show the field current needed for **unity pf** (minimum current, least copper loss) at any load. 2. Show how much **leading kVAR** an over-excited motor can supply, so it can be used as a **synchronous condenser** for power factor correction. 3. Show the limits of operation: the left end of each curve (very low $I_f$) approaches the stability limit, where the motor would pull out of step. 4. Help the operator choose excitation to keep armature current within rating. 5. Used to determine the excitation required for voltage control at substations. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When excitation is varied at constant load and supply voltage, what stays constant? A. The armature current B. The active component Ia cos(phi) and Ef sin(delta) C. The reactive component of Ia D. The power factor Answer: B. Input power VIa cos(phi) is constant, so the tip of Ia moves along a line perpendicular to V. (Key point: What stays constant) 2. Where on a V-curve is the armature current minimum? A. At the lagging end B. At normal excitation where pf is unity C. At the leading end D. At zero field current Answer: B. At unity pf the current is purely active, so it is minimum. (Key point: Minimum current at unity pf) 3. For under-excitation (Ef < V), what is the armature current like? A. Small and leading B. Large and lagging C. Zero D. Minimum and in phase with V Answer: B. A large lagging reactive component makes the current high and the pf lagging. (Key point: Under-excited behaviour) 4. Which is a significance of V-curves? A. They show the speed at various loads B. They show the field current for unity pf and how much leading kVAR an over-excited motor can supply as a synchronous condenser C. They give the starting torque D. They show hunting frequency Answer: B. V-curves help choose excitation for minimum current and show the leading reactive capability for pf correction. 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