Transformers

Mutual Induction, Engineered

A transformer changes an alternating voltage from one value to another (greater or smaller) using mutual induction (Chapter 6).

Construction: two coils, insulated from each other, wound on a soft-iron core — either one on top of the other, or on separate limbs of the core. The primary coil ($N_p$ turns) is the input; the secondary ($N_s$ turns) is the output.

Working: an alternating voltage across the primary drives an alternating current, producing an alternating flux $\phi$ in the core that links both coils. By Faraday's law the secondary acquires an induced emf

$$\varepsilon_s = -N_s\frac{d\phi}{dt}$$

while the same changing flux produces a back emf $\varepsilon_p = -N_p\,d\phi/dt$ in the primary (which equals the applied $v_p$ for a negligible-resistance primary).

The Transformer Equations

Dividing the two emf relations (with $\varepsilon_s = v_s$, $\varepsilon_p = v_p$ for an ideal transformer):

$$\boxed{\frac{v_s}{v_p} = \frac{N_s}{N_p}}$$

The same flux change per turn in both coils — so voltage divides in proportion to turns. NCERT's three small print assumptions: (i) negligible primary resistance, (ii) the same flux links both coils (very little leakage), (iii) small secondary current.

If the transformer is 100 percent efficient (no losses), power in = power out:

$$i_pv_p = i_sv_s \quad\Rightarrow\quad \frac{i_p}{i_s} = \frac{v_s}{v_p} = \frac{N_s}{N_p}$$

The same ratios hold for amplitudes and rms values.

• Step-up ($N_s > N_p$): voltage up, current down — e.g. $N_s/N_p = 2$ turns 220 V at 10 A into 440 V at 5 A (NCERT's own example).
• Step-down ($N_s < N_p$): voltage down, current up.

Key Point: a transformer trades voltage for current at (nearly) constant power. It is not a power amplifier — and it does nothing at all on steady DC (no changing flux, no induction!).

Real Transformers: The Four Losses, and the Grid

Well-designed transformers exceed 95 percent efficiency, but small losses persist (NCERT's four, with remedies):

1. Flux leakage — not all primary flux links the secondary (air gaps, poor design). Remedy: wind one coil over the other.
2. Winding resistance — wires heat up ($I^2R$). Remedy: thick wires for high-current windings.
3. Eddy currents — the alternating flux induces swirling currents in the iron core itself, heating it. Remedy: a laminated core (thin insulated sheets) that breaks up the current loops.
4. Hysteresis — repeated magnetisation reversal of the core dissipates energy. Remedy: core material with low hysteresis loss (soft iron).

The large-scale payoff: at the generating station the voltage is stepped up (reducing the current and hence the $I^2R$ line loss), transmitted cross-country, then stepped down in stages — at area sub-stations, then utility poles — to the 240 V that reaches homes.

[NEET Important] The four losses + remedies are a standing 3-mark Board question and a NEET favourite. Learn them as pairs (loss → fix).

Solved Examples

Example 1: NCERT's own numbers [NEET Numerical]

A transformer has 100 primary turns and 200 secondary turns. The input is 220 V at 10 A. Find the output voltage and current (ideal transformer).

Solution:

1. $N_s/N_p = 2$: $v_s = 2 \times 220 = 440$ V — stepped up.
2. Current steps down by the same ratio: $i_s = i_p \times \frac{N_p}{N_s} = 10/2 = 5$ A.
3. Check: power in $= 220 \times 10 = 2200$ W; power out $= 440 \times 5 = 2200$ W. Conserved.

Example 2: A step-down at the sub-station [JEE Numerical]

A 4400 V line feeds a step-down transformer delivering 220 V to a locality drawing 2.2 kW. Find the turns ratio, the secondary current and the primary current (ideal).

Solution:

1. $\frac{N_s}{N_p} = \frac{220}{4400} = \frac{1}{20}$.
2. $i_s = P/v_s = 2200/220 = 10$ A.
3. $i_p = P/v_p = 2200/4400 = 0.5$ A — twenty times smaller, exactly the inverse of the voltage ratio.

Example 3: Counting turns [NEET Numerical]

A transformer must convert 220 V mains to 11 V for a doorbell. Its primary has 400 turns. How many turns does the secondary need?

Solution:

1. $\frac{N_s}{N_p} = \frac{v_s}{v_p} = \frac{11}{220} = \frac{1}{20}$.
2. $N_s = 400/20 = 20$ turns.
3. Small voltage, few turns — the flux change per turn is identical in both coils.

Example 4: A 90 percent efficient transformer [JEE Numerical]

A transformer draws 1000 W from the mains but delivers only 90 percent of it to the 220 V secondary circuit. Find the output power and the secondary current.

Solution:

1. $P_{out} = 0.90 \times 1000 = 900$ W.
2. $i_s = P_{out}/v_s = 900/220 \approx 4.1$ A.
3. The missing 100 W heats the transformer — flux leakage, winding resistance, eddy currents, hysteresis.

Example 5: The transmission-loss miracle [JEE Numerical]

A plant must send 22 kW through a line of total resistance 10 $\Omega$. Compare the line losses when transmitting (a) at 220 V, (b) stepped up to 22 kV.

Solution:

1. (a) $I = P/V = 22000/220 = 100$ A: loss $= I^2R = 100^2 \times 10 = 100$ kW — more than the power being sent! Absurd.
2. (b) $I = 22000/22000 = 1$ A: loss $= 1^2 \times 10 = 10$ W — utterly negligible.
3. Stepping the voltage up 100x cut the loss by $100^2 = 10^4$. This single calculation is why the grid exists.

Example 6: Flux change per turn [NEET Numerical]

The primary (300 turns) of a transformer is fed such that the flux in the core changes at $2 \times 10^{-2}$ Wb/s. Find the emf per turn and the secondary emf if $N_s = 750$.

Solution:

1. Emf per turn $= d\phi/dt = 2 \times 10^{-2}$ V — the same for every turn of both coils.
2. Primary back emf: $300 \times 0.02 = 6$ V; secondary: $\varepsilon_s = 750 \times 0.02 = 15$ V.
3. Ratio $15/6 = 750/300$: the turns ratio, of course.

Example 7: Why no transformer on DC?

A student connects a transformer's primary to a battery, expecting a stepped-up DC output. What actually happens and why?

Solution:

1. A steady current makes a steady flux — and induction needs $d\phi/dt \ne 0$.
2. Apart from a momentary kick at switch-on/off, the secondary shows zero voltage.
3. (Worse: with no back emf to limit it, the primary current grows dangerously — transformers on DC tend to burn.) Transformers are an AC-only technology, which is precisely why the grid is AC.

Example 8: The four losses, with remedies

List the energy losses in a real transformer and one remedy for each.

Solution:

1. Flux leakage → wind the coils one over the other.
2. Winding resistance ($I^2R$ heat) → use thick copper wire.
3. Eddy currents in the core → laminate the core (thin insulated sheets).
4. Hysteresis (repeated magnetisation reversal) → use a core material with low hysteresis loss, like soft iron. Even so, good transformers already exceed 95 percent efficiency.

Example 9: Ideal-transformer assumptions

State the three assumptions behind $v_s/v_p = N_s/N_p$.

Solution:

1. The primary resistance (and current) is small — applied voltage equals back emf.
2. The same flux links both coils — negligible leakage from the core.
3. The secondary current is small (light loading).
4. Real, well-designed transformers approximate all three closely — hence efficiencies above 95 percent.

Example 10: From power plant to your wall

Trace the voltage of electrical power from generation to a home, naming the transformer at each stage.

Solution:

1. At the plant: generated at several kV, immediately stepped UP to very high voltage for transmission (small current, tiny $I^2R$ loss).
2. Area sub-station: stepped down to distribution levels.
3. Utility pole / local transformer: stepped down again to the 240 V (rms) that enters homes. Up once for the journey, down in stages for safety.