--- title: "Transformer operation; flux constant at any load" subject: "Electrical Machine" chapter: 2 key: "semester_4/electrical_machine:transformer-load-noload-flux-constant" source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-load-noload-flux-constant site: BCT Engineering --- # Transformer operation; flux constant at any load Electrical Machine · Chapter 2: Transformer - Asked in: 2072 Magh; 2069 Bhadra; 2070 Bhadra, 4 marks (asked 3 times) - Video: [Iron losses | Constant flux | Transformers | Lec - 21](https://www.youtube.com/watch?v=H_g6hy8P2uM) — Education 4u - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-load-noload-flux-constant - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2 ## Question Explain the no-load and loaded operation of a transformer. Prove that the net magnetic flux in the core remains constant at any load. ## Short answer At no-load, primary mmf N₁I₀ establishes flux Φ. On load, secondary mmf N₂I₂ opposes flux, so primary draws extra current I₂′ with N₁I₂′ = N₂I₂, making net mmf N₁I₀ unchanged. Since V₁ ≈ E₁ = 4.44 f N₁ Φₘ and V₁, f are constant, core flux remains the same at all loads. ## Answer A transformer works on mutual induction. Its operation is explained at no-load and on load, and the same net flux is present in both. #### No-load operation ```text I0 <-- Primary Secondary (open) o--+ +-------------+ +--o V1 |N1| core Φ |N2| E2 = V2 (no current) o--+ +-------------+ +--o ``` - The primary is connected to the supply V1 and the secondary is open. - A small no-load current I0 flows. It has a magnetising component Iμ (in phase with Φ, produces the flux) and a core-loss component Iw (in phase with V1, supplies the iron loss). So I0 = √(Iμ² + Iw²) = 2 to 5 % of full-load current. - The alternating flux Φ links both windings: E1 = 4.44 f N1 Φm, E2 = 4.44 f N2 Φm. Since E1 ≈ V1, the flux is fixed by V1. - Primary current lags V1 by nearly 90°, so the no-load power factor is very low. #### Loaded operation - A load is connected to the secondary. A secondary current I2 flows and produces an mmf N2 I2 which, by Lenz's law, opposes the flux Φ. - The flux tends to fall, so E1 falls slightly and the primary draws a larger current. The extra current I2′ gives an mmf N1 I2′ that cancels N2 I2: N1 I2′ = N2 I2 or I2′ = (N2/N1) I2. - Total primary current: I1 = I0 + I2′ (phasor sum). #### Proof that the net flux is constant ```text Net mmf acting on the core = N1 I1 − N2 I2 = N1 (I0 + I2′) − N2 I2 = N1 I0 + N1 I2′ − N2 I2 = N1 I0 + 0 (because N1 I2′ = N2 I2) ``` So the net mmf is N1 I0, the same as at no-load. Since the flux produced by this mmf is Φ, the core flux is the same at all loads. This is also clear from V1 ≈ E1 = 4.44 f N1 Φm. V1 and f are constant, so Φm is constant. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the components of no-load current I0? A. Load and magnetising B. Both in phase with flux C. Both in phase with V1 D. Magnetising Iu (in phase with flux) and core-loss Iw (in phase with V1) Answer: D. I0 = sqrt(Iu^2 + Iw^2); Iw supplies the iron loss, Iu produces the flux. (Key point: No-load current components) 2. When the secondary is loaded, why does the primary draw more current? A. Because frequency falls B. Because the supply voltage rises C. To cancel the demagnetising mmf N2 I2 so that the flux is maintained D. Because the core saturates Answer: C. The extra current I2' satisfies N1 I2' = N2 I2, keeping flux unchanged. (Key point: Load effect) 3. What is the net mmf acting on the core on load? A. N1 I0, same as at no-load B. N1 I1 + N2 I2 C. Zero D. N2 I2 Answer: A. N1(I0 + I2') - N2 I2 = N1 I0 because N1 I2' = N2 I2. (Key point: Net mmf proof) 4. Which relation shows the flux is set by supply? A. V2 = I2 R B. P = V I C. V1 = E1 = 4.44 f N1 Phi_m with constant V1 and f D. E = B l v Answer: C. V1 and f are constant, so Phi_m is constant at any load. 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