working principle

working principle

1. How a Single-Phase Transformer Works (Faraday + Mutual Induction)

Working principle: AC in primary creates time-varying flux $φ$ in iron core, which links secondary ⇒ induces emf in secondary via Faraday's law ($e = -N dφ/dt$). No conductor between windings — energy transferred magnetically.

Steps:

1. Primary connected to AC supply $V_1$ ⇒ current $I_1$ flows ⇒ magnetomotive force $N_1 I_1$ drives alternating flux $φ$ in core ($φ = (N_1 I_1)/R$, $R$ reluctance).
2. Flux $φ$ is common to both windings (ideally, $k≈1$, leakage small).
3. Secondary experiences same $φ$ ⇒ induced emf

(and primary back emf

). For ideal (no loss, no leakage),

Current ratio inverse (power in ≈ power out for ideal):

and phase (dot convention): $V_2$ in phase with $V_1$ if dots on same side.

2. Ideal Transformer Equations (no loss)

Step-up ($N_2>N_1$): $V_2>V_1$, $I_2<I_1$; Step-down opposite.

3. Practical Aspects

• No-load: secondary open, primary draws magnetizing current $I_0$ (small, $2$–$5$% of full load, lags $V_1$ ~90° due to $L_m$) to create $φ$.
• On load: secondary $I_2$ creates opposing $MMF$ $N_2 I_2$, primary automatically draws extra $I_2' = (N_2/N_1)I_2$ to maintain $φ$ (ampere-turn balance $N_1 I_1 + N_2 I_2 = N_1 I_0$).

4. Worked Checks

(a) N1=100, N2=1000, V1=230V ⇒ V2=2300V (step-up 10×) ✓

(b) Load 2300V, 10Ω ⇒ I2=230A, I1=2300A? Wait power 2300·230=529kW, I1=529k/230=2300A correct (10×).

(c) Why not DC? DC I1 constant ⇒ dφ/dt=0 ⇒ e2=0 ⇒ no transfer, primary draws huge DC current limited only by R (short).

5. Exam Points

• State working principle as Faraday mutual induction with $e=-N dφ/dt$ — 5 marks, very frequent (almost certain to ask principle).
• Show $V_1/V_2=N_1/N_2$, $I_2/I_1=N_1/N_2$ — 5 marks, very frequent.
• Explain no-load $I_0$ and ampere-turn balance — 5 marks, frequent.
• Mention why DC fails — 3 marks.