--- title: "Motor adjustment to load and excitation effects" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:motor-adjust-load-excitation" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-adjust-load-excitation site: BCT Engineering --- # Motor adjustment to load and excitation effects Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2082 Baisakh, 6 marks - Video: [Synchronous motor on Load](https://www.youtube.com/watch?v=E6gTAWI39h0) — B.R verma - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-adjust-load-excitation - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question Explain how synchronous motor adjust itself for the change in load. Elaborate the effects of change in excitation of synchronous motor on armature current and power factor with proper diagram. ## Short answer When load increases, rotor angle δ grows while speed stays synchronous, increasing armature current to match load; beyond δ=90° it pulls out. Changing excitation at constant load varies armature current and pf: under-excited gives lagging pf, normal gives minimum current unity pf, over-excited gives leading pf (synchronous condenser). ## Answer A synchronous motor always runs at synchronous speed $N_s = 120f/P$. When the shaft load changes it cannot slow down (as an induction motor does); instead the rotor shifts backwards in space by a larger **load angle** $\delta$, and the armature current changes to supply the new power. #### Adjustment to change in load 1. At no load the rotor poles are almost in line with the stator field ($\delta \approx 0$), and the motor draws only enough current to meet losses. 2. When load torque increases, the rotor momentarily slows down. The rotor poles fall back with respect to the rotating stator field, so $\delta$ increases. 3. Back emf $E_f$ stays the same in magnitude (excitation unchanged) but moves further behind $V$. The resultant voltage $E_r = V - E_f$ grows, so armature current $I_a = E_r / jX_s$ increases. 4. Power drawn $P = \dfrac{3VE_f}{X_s}\sin\delta$ rises until it equals the new load plus losses. The rotor then runs again at $N_s$ with a larger fixed $\delta$. 5. If load exceeds $P_{max}$ (at $\delta = 90^\circ$ for a cylindrical rotor), the motor **pulls out of step** and stops. ```text V (fixed) ------------------> \ delta1 \--------> Ef (light load) \ \ delta2 > delta1 \------> Ef (heavy load, same length) ``` #### Effect of change in excitation (constant load) With load (and hence $P = VI_a\cos\phi$) constant, $I_a\cos\phi$ is constant, so the tip of $I_a$ moves on a line perpendicular to $V$, and $E_f\sin\delta$ is also constant. | Excitation | $E_f$ vs $V$ | Armature current | Power factor | |---|---|---|---| | Under-excited | $E_f < V$ | Large | Lagging | | Normal | $E_f \approx V$ | Minimum | Unity | | Over-excited | $E_f > V$ | Large | Leading | ```text I(lead) I(unity) \ | \ | ------------ +-------+--------> V \ | I(lag) locus of I tip (I cos(phi) const) ``` - Raising the field current from a low value first **reduces** $I_a$ (lagging pf improves) until $I_a$ is minimum at unity pf. - Further increase makes $I_a$ rise again with a **leading** pf. - Plotting $I_a$ against field current gives the **V-curves**; the pf vs field current curve is an inverted V. This is why an over-excited synchronous motor can act as a **synchronous condenser** to improve plant power factor. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When shaft load on a synchronous motor increases, how does it adjust? A. Speed drops permanently B. The load angle delta increases and armature current rises C. Back emf Ef increases in magnitude D. Supply frequency falls Answer: B. The rotor falls back in space, so delta increases, Er grows and Ia increases, while the speed returns to Ns. (Key point: Load adjustment) 2. With excitation unchanged, what happens to Ef as load increases? A. Its magnitude is the same but it lags V by a larger angle B. It grows in magnitude C. It falls in magnitude D. It leads V Answer: A. Ef depends on field current, which is unchanged, so only its angle behind V changes. (Key point: Ef behaviour) 3. For a cylindrical-rotor motor, at what load angle is the maximum power and what happens beyond it? A. 45 degrees; the motor speeds up B. 90 degrees; the motor pulls out of step C. 180 degrees; nothing happens D. 0 degrees; the motor stalls Answer: B. P = (3VEf/Xs) sin(delta) is maximum at 90 degrees; a load above that makes the motor lose synchronism. (Key point: Pull-out) 4. At constant load, raising the field current from a low value does what to Ia? A. Increases it continuously B. Decreases it to a minimum at unity pf, then increases it with leading pf C. Keeps it constant D. Decreases it forever Answer: B. This gives the V-curve: lagging current falls to a minimum at unity pf, then rises again as leading current. 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