--- title: "Synchronous motor adjusting to increasing load" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:motor-adjust-increasing-load" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-adjust-increasing-load site: BCT Engineering --- # Synchronous motor adjusting to increasing load Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2074 Chaitra, 6 marks - Video: [Electrical Machines | Lecture - 9C | Synchronous Motor | Load with Constant & Different Excitations](https://www.youtube.com/watch?v=TCTMH1ZpjWg) — Engineering Devotion - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-adjust-increasing-load - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question In what manner does a synchronous motor adjust itself to an increasing shaft load? ## Short answer Increasing shaft load momentarily slows the rotor, increasing the load angle δ between rotor and stator fields. This raises armature current and developed power (P ∝ sinδ) until torque balances, maintaining synchronous speed. The limit is the pull-out torque at δ=90°. ## Answer A synchronous motor meets an increasing shaft load **without changing speed**. It does so by increasing its **load (torque) angle** $\delta$, the angle by which the rotor poles (and $E_f$) fall behind the stator field (and $V$). #### Step-by-step action 1. **At no load**, the rotor poles are almost exactly under the stator poles; $\delta$ is very small and the motor draws only a small current for losses. 2. **Load is increased.** Load torque becomes greater than the developed torque, so the rotor decelerates momentarily. 3. **Rotor falls back.** Because the stator field keeps rotating at $N_s$, the rotor poles slip back by a larger angle $\delta$. The magnetic "spring" linking the poles is stretched more. 4. **More current.** $E_f$ keeps the same magnitude (field current unchanged) but swings further behind $V$. The resultant voltage $E_r = V - E_f$ increases, so $I_a = E_r/X_s$ increases. 5. **More power.** Developed power rises: $$ P = \frac{3VE_f}{X_s}\sin\delta $$ 6. **New balance.** When the developed torque equals load torque plus losses, the rotor again runs at exactly $N_s$, but with a larger fixed $\delta$. ```text V ------------------------------> \ \ \ \ delta2 (more load) \ \ \ '----> Ef2 (same length) delta1 '---------> Ef1 (light load) Er2 > Er1 -> Ia2 > Ia1 ``` #### Limit of loading - Power rises with $\delta$ only up to $\delta = 90^\circ$ (cylindrical rotor), where $P_{max} = 3VE_f/X_s$. - If load is increased beyond this **pull-out torque**, the rotor cannot stay locked; it falls out of step, and the motor stalls with heavy current (the protection then trips). - The pull-out limit can be raised by **increasing excitation** ($E_f$). #### Other effects - With fixed excitation, the power factor changes with load; for a given $E_f$ the pf moves towards lagging as load increases. - A sudden load change makes the rotor overshoot and oscillate about the new $\delta$ (**hunting**); damper windings reduce this. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does a synchronous motor meet an increase in shaft load? A. Its speed falls until torque balances B. It increases its load (torque) angle while speed stays at Ns C. It increases its field current automatically D. It increases the supply frequency Answer: B. The rotor slips back by a larger angle delta, develops more torque, and keeps running at synchronous speed. (Key point: Speed stays constant) 2. As delta increases at constant excitation, why does Ia increase? A. Ef becomes larger in magnitude B. V drops C. Ef swings further behind V, so Er = V - Ef grows and Ia = Er/Xs grows D. The speed increases Answer: C. Ef keeps its magnitude but its phase lags more, which increases the resultant voltage Er and hence the current. (Key point: Why current rises) 3. For a cylindrical-rotor motor, what happens if the load goes beyond the maximum (pull-out) torque at delta = 90 degrees? A. The motor speeds up B. The pf becomes unity C. The rotor falls out of step and stalls D. Delta settles at 120 degrees Answer: C. Power P = 3VEf sin(delta)/Xs peaks at 90 degrees; beyond that the rotor cannot stay locked and the motor stalls. (Key point: Pull-out limit) 4. How can the pull-out power be raised? A. Decrease excitation Ef B. Increase excitation Ef C. Reduce the supply frequency D. Remove the damper winding Answer: B. Pmax = 3VEf/Xs, so a larger Ef gives a larger pull-out torque. 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