--- title: "Transformer on load and no load, phasor diagrams" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:transformer-load-phasor-resistive-capacitive" source: https://www.bctengineer.com/q/semester_4/electrical_machines/transformer-load-phasor-resistive-capacitive site: BCT Engineering --- # Transformer on load and no load, phasor diagrams Electrical Machines · Chapter 2: Transformer - Asked in: 2068 Magh, 8 marks - Video: [Transformer On Load - Operation and Phasor Diagram | Electrical Deck](https://www.youtube.com/watch?v=7CrKC36BLZw) — Electrical Deck - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/transformer-load-phasor-resistive-capacitive - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question Explain the transformer on load and no load with the phasor diagram of resistive and capacitive load. ## Short answer At no load, primary draws a small lagging exciting current I0 with core-loss and magnetizing components. On resistive load, I2 is in phase with V2, and E2 > V2 giving a slight voltage drop. On capacitive load, I2 leads V2; the reactance drop partially opposes V2, so E2 may be smaller than V2, causing a voltage rise (negative regulation). ## Answer A practical transformer has winding resistances $R_1, R_2$, leakage reactances $X_1, X_2$ and a core that draws an exciting current $I_0$. #### No-load operation The secondary is open ($I_2 = 0$). The primary draws only $I_0$, which has two components: $$ I_0 = I_c + I_\mu,\qquad I_c = I_0\cos\phi_0\ (\text{core loss}),\qquad I_\mu = I_0\sin\phi_0\ (\text{magnetising}) $$ ```text V1 (= -E1 approx) ^ | Ic | . I0 (along V1)| . | . phi0 = angle between V1 and I0 |. (about 75-80 deg) +-------------------> Im, flux phi | v E1, E2 (lag phi by 90 deg) ``` - $I_c$ is in phase with $V_1$ and supplies the core loss. - $I_\mu$ lags $V_1$ by 90°, and the flux $\phi$ is in phase with $I_\mu$. - $E_1$ and $E_2$ lag $\phi$ by 90°, so $E_1$ is opposite to $V_1$. - $I_0$ is small (2 to 5% of full-load current) and lags $V_1$ by $\phi_0$, so the no-load power factor is low (about 0.1 to 0.25). #### On load The secondary current $I_2$ flows, and the primary takes the additional load component $I_1' = I_2/a$, so $I_1 = I_0 + I_1'$ (phasor sum). The voltage equations are $$ V_1 = -E_1 + I_1(R_1 + jX_1),\qquad V_2 = E_2 - I_2(R_2 + jX_2) $$ **Resistive load.** $I_2$ is in phase with $V_2$. $E_2$ is found by adding to $V_2$ the drop $I_2R_2$ (in phase with $I_2$) and $I_2X_2$ (90° ahead of $I_2$). So $E_2 > V_2$ and the terminal voltage falls on load. $I_1'$ is in phase with $-I_2$ (it is drawn in phase with $-E_1$), and $I_1$ is close to $I_1'$. The primary power factor is nearly unity. ```text Resistive load (phasors referred to secondary) E2 /| / | I2X2 / | /___|______ V2 I2R2 I2 along V2 ``` **Capacitive load.** $I_2$ leads $V_2$ by $\phi_2$. The drops $I_2R_2$ and $I_2X_2$ are drawn along $I_2$ and 90° ahead of it. Because $I_2$ is now ahead of $V_2$, the drop $I_2X_2$ points partly opposite to $V_2$, and $E_2$ becomes *smaller* than $V_2$: approximately $E_2 \approx V_2 + I_2R_2\cos\phi_2 - I_2X_2\sin\phi_2$. So the secondary terminal voltage rises on load (negative regulation) when the leading power factor is low enough. ```text Capacitive load (V2 as reference, I2 leads V2 by phi2) E2 tip <-----------+ .' I2X2 / .' / I2R2 (along I2) O ----------------------------+ V2 E2 runs from O to E2 tip and is shorter than V2 ``` In each case, $I_1$ is obtained as $I_0 + I_1'$, and $V_1 = -E_1 + I_1(R_1 + jX_1)$ completes the primary diagram. For the capacitive load $I_1'$ leads $-E_1$, so the primary power factor is leading when the load current is large compared with $I_0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The no-load current I0 has components: A. Ic lagging V1 by 90 degrees and Im in phase B. Both in phase with V1 C. Ic in phase with V1 (core loss) and Im lagging by 90 degrees (magnetising) D. Ic and Im both lead V1 Answer: C. I0 = Ic + Im, with Ic = I0 cosphi0 and Im = I0 sinphi0. (Key point: No-load current) 2. No-load power factor is: A. Near unity B. Exactly zero C. Leading and about 0.9 D. Low, about 0.1 to 0.2 Answer: D. I0 is mostly magnetising, so phi0 is about 75 to 80 degrees. (Key point: No-load power factor) 3. The primary current on load is: A. I1 = I0 + I2/a (phasor sum) B. I1 = I2 only C. I1 = I0 - I2/a D. I1 = I0 + a I2 Answer: A. The load component I1' = I2/a is added to I0. (Key point: Primary current on load) 4. For a low leading-pf capacitive load, the secondary terminal voltage: A. Always falls more than for resistive load B. Can rise on load (negative regulation) because E2 < V2 C. Is unchanged D. Falls to zero Answer: B. The I2X2 drop points partly opposite to V2, so V2 exceeds E2. 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