--- title: "Transformer no-load and load operation, constant core flux" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:ideal-transformer-noload-load-constant-flux" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/ideal-transformer-noload-load-constant-flux site: BCT Engineering --- # Transformer no-load and load operation, constant core flux Electrical Machines I · Chapter 2: Transformer - Asked in: 2077 Chaitra, 8 marks; 2075 Bhadra, 8 marks; 2073 Bhadra, 8 marks; 2070 Magh, 8 marks; 2069 Poush, 8 marks (asked 5 times) - Video: [Why transformer is known as constant flux machine?|Transformer ko constant flux machine](https://www.youtube.com/watch?v=N_Offk-ZpVM) — Electrical Urja - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/ideal-transformer-noload-load-constant-flux - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Explain the no-load and loaded operation of an ideal transformer. Prove that the net magnetic flux in the core remains constant at any load. ## Short answer No-load: V1 = E1 = 4.44fN1φm; only magnetizing current I0 flows. Loaded: secondary mmf N2I2 tends to reduce φ; primary draws extra load current I2' with N1I2' = N2I2, so net mmf stays N1I0. Since V1 and f are fixed, φm = V1/(4.44fN1) remains constant at any load. ## Answer An **ideal transformer** has no winding resistance, no leakage flux, no core loss and a core of infinite permeability (so it needs negligible magnetising current). Its primary and secondary are linked by the same mutual flux φ. #### No-load operation ```text phi (mutual flux) +---------------------+ I0 | | o-->|] N1 N2 [|---o V1 |] E1 E2 [| open o---|] [ |---o +---------------------+ ``` - The primary is connected to $V_1$; the secondary is open, so $I_2 = 0$. - Since the winding has no resistance, the applied voltage is balanced only by the induced emf: $V_1 = E_1 = 4.44 f N_1\phi_m$. This fixes the flux. - A very small current $I_0$ (the magnetising current $I_m$) flows. It is in phase with φ and lags $V_1$ by 90°, so the input power is zero. - The same flux induces $E_2 = 4.44 f N_2\phi_m$ in the secondary. $E_1$ and $E_2$ are in phase, both lagging φ by 90°, and $$ \frac{E_1}{E_2} = \frac{N_1}{N_2} $$ Phasor diagram (no load): $\phi$ along the reference, $I_0$ in phase with φ, $V_1$ leading φ by 90°, $E_1$ and $E_2$ opposite to $V_1$. #### Loaded operation 1. A load is connected; current $I_2$ flows in the secondary. It produces mmf $N_2 I_2$ which, by Lenz's law, opposes the mutual flux. 2. If φ fell, $E_1$ would fall below $V_1$. The primary therefore draws an extra **load component** of current $I_2'$ whose mmf exactly cancels the secondary mmf: $$ N_1 I_2' = N_2 I_2 \Rightarrow I_2' = \frac{N_2}{N_1}I_2 $$ 3. Total primary current $I_1 = I_0 + I_2'$. Power drawn from the supply rises to match the load: $V_1 I_1 \cos\phi_1 = V_2 I_2\cos\phi_2$. #### Proof that the core flux remains constant at any load Net mmf acting on the core: $$ \begin{aligned} F_{net} &= N_1 I_1 - N_2 I_2 \\ &= N_1 (I_0 + I_2') - N_2 I_2 \\ &= N_1 I_0 + (N_1 I_2' - N_2 I_2) \\ &= N_1 I_0 \end{aligned} $$ So the net mmf, and hence the flux $\phi = N_1 I_0 / S$, is the **same as on no load**. The same result follows from the voltage equation: since $V_1 = E_1 = 4.44 f N_1\phi_m$ at every load and $V_1$, $f$ and $N_1$ are fixed, $$ \phi_m = \frac{V_1}{4.44 f N_1} = \text{constant} $$ Hence the mutual flux of an ideal transformer is independent of load, from no load to full load. In a practical transformer it falls only very slightly because of the small leakage impedance drop. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which set describes an ideal transformer? A. Small resistance, no leakage, small core loss, finite permeability B. No winding resistance, no leakage flux, no core loss, infinite permeability C. No resistance but with leakage flux D. Zero core but nonzero winding resistance Answer: B. These assumptions make V1 = E1 and no magnetising current needed. (Key point: Ideal assumptions) 2. What is the no-load current of an ideal transformer like? A. A current in phase with V1 giving large input power B. A small magnetising current in phase with φ, lagging V1 by 90°, so input power is zero C. Zero current because the secondary is open D. A current leading V1 by 90° Answer: B. Without core loss, I0 is purely magnetising, lagging the voltage by 90°. (Key point: No-load current) 3. What happens when a load current I2 flows? A. The primary current stays the same and flux rises B. The secondary mmf adds to the flux C. The primary draws an extra I2' = (N2/N1) I2 whose mmf cancels the secondary mmf D. The primary draws I2' = (N1/N2) I2 Answer: C. N1 I2' = N2 I2 restores the flux that the secondary mmf would reduce. (Key point: Load component) 4. Why is the core flux the same at all loads? A. Because the load current is always zero B. Because the net mmf equals N1 I2 which is constant C. Because the secondary induces a larger emf at load D. Net mmf = N1(I0+I2') − N2 I2 = N1 I0, and also φm = V1/(4.44 f N1) is fixed Answer: D. Both the mmf balance and the fixed V1, f, N1 show φm does not depend on load. 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