--- title: "Loaded transformer and phasor diagram for inductive load" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:transformer-loaded-operation-phasor-inductive" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/transformer-loaded-operation-phasor-inductive site: BCT Engineering --- # Loaded transformer and phasor diagram for inductive load Electrical Machines I · Chapter 2: Transformer - Asked in: 2071 Bhadra, 4+3 marks - Video: [PHASOR DIAGRAM ( INDUCTIVE LOAD) FOR A SINGLE PHASE TRANSFORMER](https://www.youtube.com/watch?v=IKx5yHPn3V8) — Xplanator - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/transformer-loaded-operation-phasor-inductive - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Explain the loaded operation of transformer. Draw the phasor diagram of a power transformer for inductive load. ## Short answer When loaded, secondary MMF opposes main flux, so primary draws an extra current I2' to cancel it, keeping core flux nearly constant. The primary current is I0 + I2'. For an inductive load, I2 lags V2, V2 is less than E2 due to impedance drops, and I1 lags V1. ## Answer #### Loaded operation of a transformer When a load is connected to the secondary, a current $I_2$ flows, set by the load: $I_2 = V_2/Z_L$. 1. $I_2$ in $N_2$ turns produces an MMF $N_2I_2$ that, by Lenz's law, **opposes** the main flux $\phi$. 2. The flux tends to fall, so the back EMF $E_1$ falls slightly and the primary draws an extra current $I_2'$ from the supply. 3. $I_2'$ is just large enough for its MMF to cancel the secondary MMF: $N_1I_2' = N_2I_2$, so $I_2' = KI_2$ (with $K = N_2/N_1$). 4. The total primary current is the phasor sum $\bar I_1 = \bar I_0 + \bar I_2'$. 5. Because the net MMF is still $N_1I_0$, the **core flux stays almost constant** from no load to full load, so the iron loss is constant. In a real transformer the windings have resistance and leakage reactance, so: $$ \begin{aligned} \bar V_1 &= -\bar E_1 + \bar I_1(R_1 + jX_1) \\ \bar V_2 &= \bar E_2 - \bar I_2(R_2 + jX_2) \end{aligned} $$ As the load increases, these drops increase, so $V_2$ falls (voltage regulation). #### Phasor diagram for inductive (lagging) load Steps: draw $\phi$ as the reference; $E_1$ and $E_2$ lag $\phi$ by 90°; $I_0$ leads $\phi$ by a small angle (because of core loss); $I_2$ lags $V_2$ by $\phi_2$; $V_2 = E_2 - I_2R_2 - jI_2X_2$; $I_2'$ is drawn equal and opposite to $I_2$; $I_1 = I_0 + I_2'$; $V_1 = -E_1 + I_1R_1 + jI_1X_1$. ```text V1 ^ /| I1.X1 (perpendicular to I1) / | -E1 ^ /--+ I1.R1 (parallel to I1) | / | / ^ I2' (equal and opposite to I2) | / / |/ / ^ I1 = I0 + I2' | / / | / / | / / ___...---> I0 (leads phi slightly) |/ /..--' ---------O-------------------------> phi (main flux, reference) /| / | / | v | I2 v V2 = E2 - I2(R2 + jX2) (lags V2 | by phi2) v E2 (E1 also lies along this line) phi1 = angle between V1 and I1; phi2 = angle between V2 and I2 ``` (For clarity many textbooks take $K = 1$, so $E_1 = E_2$ and $I_2' = I_2$ in magnitude.) Key features for an inductive load: - $I_2$ lags $V_2$ by $\phi_2$, so it points down and to the left of $V_2$; $I_2'$ is exactly opposite to it, lagging $-E_1$. - $I_1$ lags $V_1$ by $\phi_1$, with $\phi_1$ slightly greater than $\phi_2$ because of $I_0$ and the drops. - $V_2 < E_2$ and $V_1 > E_1$; the drop is largest for lagging loads. - On full load $I_0$ is small compared with $I_2'$, so $I_1 \approx I_2'$ and $\frac{I_1}{I_2} \approx K$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When load current I2 flows, its MMF N2 I2 tends to: A. Increase the main flux B. Oppose the main flux C. Have no effect on the flux D. Reverse the supply voltage Answer: B. By Lenz's law it opposes the flux, so the primary draws extra current. (Key point: Secondary MMF effect) 2. The extra primary current I2' satisfies: A. N1 I2' = N2 I2 B. N1 I2' = I0 C. I2' = I2 / K D. I2' = I2 K^2 Answer: A. I2' cancels the secondary MMF, giving I2' = K I2. (Key point: Primary balance) 3. From no load to full load, the core flux and hence iron loss: A. Doubles B. Falls to zero C. Stays almost constant D. Rises as load squared Answer: C. The net MMF stays equal to N1 I0, so flux is almost constant. (Key point: Core flux) 4. The total primary current is: A. I0 - I2' B. The phasor sum I0 + I2' C. I2' only D. I0 x I2' Answer: B. I1 is the phasor sum of the no-load current and the load component I2'. 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