--- title: "No-load and loaded operation, constant flux" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:ideal-transformer-no-load-loaded-flux" source: https://www.bctengineer.com/q/semester_4/electrical_machines/ideal-transformer-no-load-loaded-flux site: BCT Engineering --- # No-load and loaded operation, constant flux Electrical Machines · Chapter 2: Transformer - Asked in: 2077 Chaitra, 8 marks; 2073 Bhadra (old course), 8 marks; 2069 Bhadra (asked 3 times) - Video: [Transformer on Load : Ideal transformer on Load](https://www.youtube.com/watch?v=Lwmu4vRF98E) — E Gyan - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/ideal-transformer-no-load-loaded-flux - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/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 irrespective of the change in load. ## Short answer At no-load, supply V1 sets core flux φm = V1/(4.44 f N1). On load, secondary mmf tends to reduce flux; primary draws extra load current so that N1 I1' = N2 I2, cancelling secondary mmf. Net mmf remains N1 Iμ (magnetizing), so flux is unchanged. Since V1 and f are fixed, φm stays constant. ## Answer An **ideal transformer** has no winding resistance, no leakage flux, no core loss and infinite core permeability, so it needs no magnetising current. #### No-load operation The secondary is open. The primary is connected to a sinusoidal supply $V_1$. ```text I1=0 +------+ I2=0 o--->---+ ||||||| +--- o V1 ) core ( E2 open o-------+ ||||||| +--- o N1 N2 ``` - The primary draws only the magnetising current $I_\mu$ needed to set up the core flux $\phi$. Because the core of an ideal transformer has infinite permeability, $I_\mu$ is taken as zero. The flux itself is not zero: it has the finite value $\phi_m = V_1/(4.44fN_1)$ needed to balance the supply voltage. - The flux induces $e_1 = -N_1\,d\phi/dt$ in the primary. For an ideal transformer $E_1$ is equal in magnitude and opposite in direction to $V_1$. - The same flux links the secondary, so $E_2 = V_2 = -N_2\,d\phi/dt$ and $\dfrac{V_1}{V_2} = \dfrac{N_1}{N_2}$. #### Loaded operation A load is connected to the secondary and a current $I_2$ flows. - $I_2$ sets up a secondary mmf $N_2I_2$ which, by Lenz's law, tends to oppose the core flux. - The primary draws an extra current $I_1'$ so that its mmf $N_1I_1'$ cancels the secondary mmf: $$ N_1I_1' = N_2I_2 \Rightarrow \frac{I_1}{I_2} = \frac{N_2}{N_1} $$ - The input power equals the output power, $V_1I_1 = V_2I_2$. #### Proof that the net flux remains constant The primary equation gives $V_1 \approx E_1 = 4.44fN_1\phi_m$. Since $V_1$ and $f$ are fixed by the supply, $\phi_m = V_1/(4.44fN_1)$ is fixed, whatever the load. Also, from the mmf balance, the net mmf acting on the core on load is: $$ \begin{aligned} F_{net} &= N_1I_1 - N_2I_2 \\ &= N_1(I_\mu + I_1') - N_2I_2 \\ &= N_1I_\mu + (N_1I_1' - N_2I_2) = N_1I_\mu \end{aligned} $$ The net mmf is $N_1I_\mu$, the same as at no load. The core flux is $\phi = N_1I_\mu/S$, so the flux is constant. Any load only adds the balancing component $I_1'$ in the primary, which cancels the secondary mmf and leaves the flux unchanged. (In the ideal limit both $I_\mu$ and the core reluctance $S$ tend to zero, but their ratio still gives the same no-load flux $\phi_m$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which feature belongs to an ideal transformer? A. Small winding resistance but no leakage B. No resistance, leakage flux or core loss, and infinite permeability C. Finite permeability with core loss D. Leakage flux but no resistance Answer: B. These idealisations mean no magnetising current is needed. (Key point: Ideal transformer) 2. At no load, V1/V2 equals: A. N2/N1, because the secondary is open B. 1, because no current flows C. N1/N2, because the same flux links both windings D. I2/I1, because power is conserved at no load Answer: C. E1 = -N1 dphi/dt and E2 = -N2 dphi/dt, so the ratio is N1/N2. (Key point: No-load voltage ratio) 3. On load, the extra primary current I1' satisfies: A. N1 I1' = N2 I2 / 2 B. N1 I1' = N2 I2 only after the flux has fallen C. I1' = I2 D. N1 I1' = N2 I2, cancelling the secondary mmf Answer: D. The primary draws I1' so its mmf cancels the secondary mmf that tries to reduce the flux. (Key point: Loaded mmf balance) 4. Why does the core flux remain constant with load? A. V1 ~ E1 = 4.44 f N1 phi_m fixes phi_m, and net mmf stays N1*I_mu B. Because the load current is always constant C. Because the secondary mmf adds to the primary mmf D. Because core reluctance rises with load Answer: A. With V1 and f fixed, phi_m = V1/(4.44 f N1); also N1I1 - N2I2 = N1 I_mu as at no load. 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