--- title: "Effect of excitation on armature current and pf" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:excitation-armature-current-pf" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/excitation-armature-current-pf site: BCT Engineering --- # Effect of excitation on armature current and pf Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2079 Bhadra, 6 marks; 2072 Kartik, 8 marks; 2070 Asar, 8 marks (asked 3 times) - Video: [Electrical Machines | Lecture - 9G | Synchronous Motor | Effect of Excitation on Current & PF](https://www.youtube.com/watch?v=tHKvK6tjB1A) — Engineering Devotion - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/excitation-armature-current-pf - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question Explain the variation of armature current and power factor for different values of excitation current (normal excitation, under-excitation and over-excitation) in a synchronous motor with suitable figures and phasor diagrams. ## Short answer With constant load, varying DC excitation changes Eb. Under-excitation (EbV) yields leading pf and increased current. Armature current follows V-curves; pf follows inverted V-curves. ## Answer In a synchronous motor with **constant load** (constant power $P = 3VI\cos\phi$), varying the DC field current changes the back emf $E_b$. Since $V$ and $P$ are fixed, $I\cos\phi$ stays constant, so the tip of the current phasor moves along a line perpendicular to $V$, while $E_b\sin\delta$ also stays constant. #### (a) Normal excitation - Field current gives $E_b$ such that the current is in phase with $V$: **unity pf**. - Armature current is **minimum** for that load. #### (b) Under-excitation - $E_b$ is smaller; the resultant voltage $E_r = V - E_b$ swings so that the current **lags** $V$. - Armature current is larger than at normal excitation; the motor draws lagging reactive power (magnetising current) from the supply. #### (c) Over-excitation - $E_b$ is larger; the current **leads** $V$. - Armature current again increases; the motor supplies reactive power to the supply (synchronous condenser action). ```text Phasors at constant load (V reference, Rs = 0) I3 (lead, over-excited) \ \ I1 (upf, normal) O----+------> V I tip moves on the / vertical line I cos(phi) / = constant I2 (lag, under-excited) ``` #### V-curves and inverted V-curves ```text Ia pf | \ / full load | _ | \ / | / \ (inverted V) | \ \ / / | / \ | \_\_/_/ no load | / \ +------------> If +--------------> If lag | upf | lead lag upf lead ``` - **V-curves**: $I_a$ vs $I_f$ at constant load, shaped like a "V". The minimum point of each curve is at unity pf. The left side (under-excited) is lagging, the right side (over-excited) is leading. Higher load gives a higher curve. - **Inverted V-curves**: pf vs $I_f$; the peak is unity pf. The locus of minimum points (unity pf) shifts to the right as load increases, since more excitation is needed at higher load. | Excitation | pf | Armature current | |---|---|---| | Under | Lagging | High | | Normal | Unity | Minimum | | Over | Leading | High | This property lets an over-excited synchronous motor drive its load and also correct the plant power factor. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At constant load and constant V, as the field current varies, what stays constant? A. Armature current magnitude B. Power factor C. Ia cos(phi), the active current component D. Back emf Eb Answer: C. P = 3VIcos(phi) is fixed, so Ia cos(phi) is constant and the tip of the current phasor moves on a line perpendicular to V. (Key point: Constant-load condition) 2. At normal excitation for a given load, the armature current is: A. Maximum with lagging pf B. Minimum with unity pf C. Maximum with leading pf D. Zero Answer: B. At unity pf the current is wholly active, so it has its minimum value for that load. (Key point: Normal excitation) 3. On a V-curve (Ia vs If at constant load), which side is under-excited and what does the minimum point represent? A. Left side is under-excited (lagging); minimum is at unity pf B. Left side is over-excited (leading); minimum is at zero pf C. Right side is under-excited; minimum is at lagging pf D. Both sides are leading; minimum is at unity pf Answer: A. The left side has low If (lagging pf), the right side high If (leading pf), and the minimum Ia is at unity pf. (Key point: V-curve shape) 4. What happens to the V-curve and the unity-pf minimum as load increases? A. Curve shifts down and minimum moves left B. Curve shifts up and the unity-pf point moves to higher excitation (right) C. Curve is unchanged D. Curve becomes flat Answer: B. Higher load gives a higher curve, and more excitation is needed for unity pf, so the minimum shifts right. 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