--- title: "Practical vs ideal transformer, phasor diagram" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:practical-vs-ideal-transformer-phasor" source: https://www.bctengineer.com/q/semester_4/electrical_machines/practical-vs-ideal-transformer-phasor site: BCT Engineering --- # Practical vs ideal transformer, phasor diagram Electrical Machines · Chapter 2: Transformer - Asked in: 2078 Poush, 2+6 marks - Video: [Ideal Transformer and Practical Transformer (Basic Electrical Engineering) BTech 1st year](https://www.youtube.com/watch?v=VaVOmLKCZ3c) — Gautam Varde - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/practical-vs-ideal-transformer-phasor - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question How practical transformer is different from ideal one? Explain with phasor diagram the operation of practical transformer when secondary is connected to load. ## Short answer An ideal transformer has no resistance, no leakage, no core loss, and zero no-load current. A practical transformer has winding resistances, leakage reactances, core loss, and draws an exciting current I0. On load, V2 = E2 – I2(R2 + jX2) and I1 = I0 + I2/a, where a = N1/N2. ## Answer #### Practical vs ideal transformer | Point | Ideal transformer | Practical transformer | |---|---|---| | Winding resistance | Zero | $R_1$, $R_2$ present (copper loss) | | Leakage flux | None (unity coupling) | Leakage reactances $X_1$, $X_2$ | | Core permeability | Infinite | Finite, needs magnetising current $I_\mu$ | | Core loss | Zero | Hysteresis and eddy loss, current $I_c$ | | No-load current | Zero | Small $I_0 = I_c + I_\mu$ | | Efficiency | 100% | 95 to 99% | | Voltage regulation | Zero | Non-zero | | $V_1/V_2$ | Exactly $N_1/N_2$ | Slightly different on load | #### Operation of a practical transformer on load When load current $I_2$ flows, the secondary terminal voltage and primary current are given by $$ V_2 = E_2 - I_2(R_2 + jX_2),\qquad I_1 = I_0 + I_1',\ \ I_1' = \frac{N_2}{N_1}I_2,\qquad V_1 = -E_1 + I_1(R_1 + jX_1) $$ **Phasor diagram (inductive load, lagging $\phi_2$):** ```text V1 (applied) /| I1X1/ | Steps: / | I1R1 1. Take flux phi as reference. -E1 /___| 2. E1, E2 lag phi by 90 deg. | 3. I2 lags V2 by phi2. | E2 4. E2 = V2 + I2R2 + jI2X2. | \ I2X2 5. I1' = I2/a opposes the I2 mmf. phi | \ I2R2 6. I1 = I0 + I1'. v \V2 7. V1 = -E1 + I1R1 + jI1X1. ``` Steps in words: 1. Draw the flux $\phi$ as reference. The emfs $E_1$ and $E_2$ lag it by 90°, and $-E_1$ is drawn opposite to $E_1$. 2. Draw $V_2$, then $I_2$ lagging it by $\phi_2$. Add $I_2R_2$ (parallel to $I_2$) and $I_2X_2$ (90° ahead) to reach $E_2$. 3. The load component $I_1'$ is drawn opposite to $I_2$ (scaled by $N_2/N_1$). The no-load current $I_0$ is drawn leading the flux by the hysteresis angle, with $I_\mu$ along $\phi$ and $I_c$ along $-E_1$. 4. $I_1 = I_0 + I_1'$. Add $I_1R_1$ and $I_1X_1$ to $-E_1$ to get $V_1$. The angle between $V_1$ and $I_1$ is the primary power factor angle $\phi_1$, slightly different from $\phi_2$ because of $I_0$ and the drops. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Compared with an ideal transformer, a practical transformer has: A. Zero regulation and 100% efficiency B. Infinite permeability C. Winding resistance, leakage reactance, finite permeability and core loss D. No-load current of zero Answer: C. These give copper loss, leakage drops, magnetising current and iron loss. (Key point: Differences) 2. In a practical transformer I0 equals: A. Ic - Im B. Zero C. I2/a D. Ic + Im Answer: D. A core-loss component and a magnetising component. (Key point: No-load current) 3. The load component I1' is drawn: A. Opposite to I2 (scaled by N2/N1) B. In phase with I2 C. At 90 degrees to I2 D. Along the flux Answer: A. It balances the secondary mmf. (Key point: Load component) 4. In the phasor construction, flux is the reference and E1, E2 lag it by: A. 0 degrees B. 90 degrees C. 180 degrees D. 45 degrees Answer: B. e = -N dphi/dt lags a sinusoidal flux by 90 degrees. 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