--- title: "Transformer at resistive, inductive, capacitive loads" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:transformer-loading-conditions-phasor" source: https://www.bctengineer.com/q/semester_4/electrical_machines/transformer-loading-conditions-phasor site: BCT Engineering --- # Transformer at resistive, inductive, capacitive loads Electrical Machines · Chapter 2: Transformer - Asked in: 2071 Magh (old course), 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-loading-conditions-phasor - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question Explain the operation of transformer at different loading conditions (resistive, inductive, capacitive) showing their corresponding circuit diagram and phasor diagram. ## Short answer With resistive load, I2 is in phase with V2: E2 = V2 + I2R02 (small drop). Inductive load: I2 lags V2; E2 = V2 + I2(R02 cosφ + X02 sinφ) yields a larger drop and poorer regulation. Capacitive load: I2 leads V2; the reactive drop subtracts, making E2 possibly less than V2 and giving a voltage rise. ## Answer The transformer is treated as a practical one with winding resistances $R_1,R_2$ and leakage reactances $X_1,X_2$. For any load, with the turns ratio $a = N_1/N_2$: $$ V_2 = E_2 - I_2(R_2 + jX_2),\qquad I_1 = I_0 + \frac{I_2}{a},\qquad V_1 = -E_1 + I_1(R_1 + jX_1) $$ #### Circuit diagram ```text I1 R1 X1 R2 X2 I2 o--/\/\--UUU--+ +--/\/\--UUU--o | | | V1 E1 ( ) ( ) E2 Z (load) | | | o-------------+--+-------------o ideal transformer N1:N2 ``` #### Phasor diagrams (drawn with $V_2$ as the reference) For each load, add to $V_2$ the drop $I_2R_2$ (parallel to $I_2$) and then $I_2X_2$ (90° ahead of $I_2$) to obtain $E_2$. | Load | $I_2$ w.r.t. $V_2$ | Resulting $E_2$ | Effect on secondary voltage | |---|---|---|---| | Resistive ($\cos\phi_2 = 1$) | In phase | $E_2 > V_2$ slightly | Small fall | | Inductive (lagging) | Lags by $\phi_2$ | $E_2$ much larger than $V_2$ | Largest fall | | Capacitive (leading) | Leads by $\phi_2$ | $E_2$ may be smaller than $V_2$ | Rise (negative regulation) | ```text Inductive Resistive Capacitive (I2 lags V2) (I2 = V2) (I2 leads V2) E2 (big) E2 I2 / /| /| / / | I2X2 / | I2X2 E2 /__ V2 (> E2) /__| /__| V2 I2R2 V2 I2R2 (I2 below V2) (I2 along V2) ``` **Primary side:** in each case, the load component $I_1' = I_2/a$ has the same phase angle $\phi_2$ relative to $-E_1$ as $I_2$ has relative to $E_2$. It is added to the no-load current $I_0$ (which lags $-E_1$ by about 78°) to give $I_1$. Then $V_1$ is the phasor sum of $-E_1$, $I_1R_1$ and $jI_1X_1$. **Result:** the inductive load gives the poorest regulation, the resistive load an intermediate value, and a capacitive load can give a voltage rise. The primary power factor follows the load: lagging for an inductive load, nearly unity for a resistive load, and leading for a strongly capacitive load. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On load, the secondary terminal voltage is: A. V2 = E2 + I2(R2 + jX2) B. V2 = E2 C. V2 = E2 - I2(R2 + jX2) D. V2 = E2 - I2 R2 only Answer: C. The winding drops are subtracted from the induced emf. (Key point: Voltage equations) 2. Which load gives the largest fall in secondary voltage? A. Resistive B. Capacitive (leading) C. All equal D. Inductive (lagging) Answer: D. The reactive drop adds directly in phase with V2, so E2 is much larger than V2. (Key point: Inductive load) 3. For a capacitive load the secondary voltage: A. May rise, since E2 can be smaller than V2 B. Always falls C. Is exactly E2 D. Falls the most Answer: A. Leading current reverses the effect of the reactive drop. (Key point: Capacitive load) 4. The primary power factor of the transformer follows: A. Always lagging by 78 degrees B. The load: lagging for inductive, nearly unity for resistive, leading for strongly capacitive C. Always unity D. Always leading Answer: B. I1 = I0 + I1', with I1' having the same phase angle as I2. 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