--- title: "Effect of excitation on synchronous motor" subject: "Electrical Machine" chapter: 6 key: "semester_4/electrical_machine:sync-motor-excitation-effect" source: https://www.bctengineer.com/q/semester_4/electrical_machine/sync-motor-excitation-effect site: BCT Engineering --- # Effect of excitation on synchronous motor Electrical Machine · Chapter 6: Three Phase Synchronous Machines - Asked in: 2074 Bhadra, 6 marks; 2073 Bhadra, 8 marks; 2072 Asoj; 2070 Bhadra, 8 marks; 2068 Bhadra, 8 marks; 2069 Poush (asked 6 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 Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/sync-motor-excitation-effect - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-6 ## Question With the help of phasor diagrams, explain the effect of excitation on armature current and power factor of a 3-phase synchronous motor, and explain how it can be operated in leading and lagging pf mode. ## Short answer In a synchronous motor with constant load, varying field excitation changes the back emf and armature current. Under-excitation gives lagging power factor; over-excitation gives leading power factor. Unity pf occurs at minimum armature current. Leading pf operation is obtained by over-exciting the field. ## Answer In a synchronous motor, the field excitation (field current I_f) controls the back emf E_b. At constant supply voltage V and constant load (constant power input), changing I_f changes the armature current I_a and the power factor, but not the speed. #### Basis Per phase, E_r = V − E_b (phasor difference) and I_a = E_r / Z_s, where Z_s = R_a + jX_s. For constant load, the input power per phase V·I_a·cos φ is constant, so the in-phase component I_a cos φ is constant. Only the quadrature component changes with excitation. R_a is neglected in the diagrams, so E_r = jX_s·I_a and I_a lags E_r by 90°. E_b lags V by the load angle δ. E_r is drawn from the tip of E_b to the tip of V. #### Phasor diagrams (V as reference, motor mode) ```text (a) Under-excited: Eb < V, Ia LAGS V O-------------------------------------->V \ ------ // \ ----- // Er \ ----- /// \ Eb ------ // \ Ia -->/ \ \ v (b) Normal excitation: Ia IN PHASE with V (unity pf, minimum Ia) O-------------------------------------->V Ia along V -------- | -------- | Er -------- | Eb -------- | ---> (c) Over-excited: Eb > V, Ia LEADS V ^ Ia / / // / / O-------------------------------------->V ----------- \\ --------- \\ Er ----------- \\ Eb --------- \\ -----> In every case Er = V - Eb = jXs·Ia, so Ia is 90° behind Er; Ia cosφ stays the same for constant load. ``` | Excitation | Back emf E_b | I_a | Power factor | Machine acts as | |---|---|---|---|---| | Under-excited | small | lags V | lagging | inductor (R-L load) | | Normal excited | medium (I_a minimum) | in phase with V | unity | resistor | | Over-excited | large | leads V | leading | capacitor (supplies lagging kVAR) | #### Explanation - With reduced I_f, E_b is small (less than V). The resultant emf E_r = V − E_b is large and points up and to the right of the E_b tip, so I_a, which is 90° behind E_r, lags V. The motor draws magnetising (lagging reactive) current from the supply, because part of the main flux must be supplied by the stator. This is the **lagging pf** mode. - As I_f is increased, E_b rises, E_r shrinks and I_a falls. At one value of I_f, E_r is at right angles to V, and I_a is minimum and in phase with V: unity pf. - Further increase of I_f (over-excitation) makes E_r larger again, but it now points up and to the left, so I_a leads V. The motor delivers lagging reactive power to the line, i.e. it acts as a capacitor. This is the **leading pf** mode. #### Operating in leading or lagging mode - To run at leading pf, raise the d.c. field current above the value for unity pf (over-excite). This is used for power-factor correction; a machine run this way without a mechanical load is called a synchronous condenser. - To run at lagging pf, reduce the field current below that value (under-excite). This is rarely done, as the stability margin is poorer. - The plot of I_a against I_f for a constant load is the V-curve. Its minimum point is at unity pf. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At constant load, what stays constant when excitation is changed? A. Ia B. In-phase component Ia cos(phi) C. Back emf Eb D. Power factor Answer: B. Input power 3VIa cos(phi) is constant so only the quadrature component of current changes. (Key point: What stays constant) 2. What happens in an under-excited synchronous motor? A. Ia leads V, leading pf B. Ia is in phase, unity pf C. Ia lags V, lagging pf D. Ia is zero Answer: C. With small Eb, the motor draws magnetising current and behaves as an inductor. (Key point: Under-excited mode) 3. How can the motor operate at leading pf? A. Reduce field current below the unity-pf value B. Raise field current above the unity-pf value C. Reduce supply frequency D. Increase mechanical load only Answer: B. Over-excitation makes Ia lead V; the motor acts like a capacitor and supplies lagging kVAR. (Key point: Over-excited mode) 4. At which excitation is armature current minimum? A. Lowest excitation B. Highest excitation C. Normal excitation at unity pf D. At zero load only Answer: C. At unity pf all current is in-phase, so Ia is minimum. 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