Transformers — Step-Up vs Step-Down

Why Transformers Matter

Almost every gadget in the country is plugged into a transformer of some kind. The mobile charger on the wall, the substation at the end of the lane, the giant towers on the highway carrying power from a hydroelectric dam — all transformers, scaling voltage up or down to suit the job. Class 12 board exams and JEE-NEET both test the working principle, and AC circuits chapter is incomplete without it.

The core idea is simple: a changing magnetic flux in an iron core links two coils, and the ratio of voltages equals the ratio of turns. We’ll derive this, work through ideal and real-world examples, and look at how transformers achieve their famous efficiency above 95%.

Key Terms & Definitions

Primary coil: the coil connected to the input AC source. Has $N_p$ turns.

Secondary coil: the coil connected to the load. Has $N_s$ turns.

Turns ratio ( $k$ ): $k = N_s/N_p$ . If $k > 1$ , step-up. If $k < 1$ , step-down.

Mutual inductance: the property that links the changing flux in one coil to the EMF induced in the other.

Soft iron core: a high-permeability material that channels almost all the flux from primary to secondary, minimising leakage.

Eddy currents: parasitic loops of current induced in the core, dissipating energy as heat. Suppressed by laminating the core.

The Working Principle

When AC voltage $V_p$ drives current through the primary, the resulting alternating flux $\Phi$ threads the iron core. The same flux passes through the secondary, inducing an EMF in each turn equal to $-d\Phi/dt$ .

Since both coils see the same flux per turn:

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

For an ideal (lossless) transformer, power input equals power output:

$V_p I_p = V_s I_s \implies \frac{I_s}{I_p} = \frac{N_p}{N_s}$

So a step-up transformer raises voltage and lowers current; a step-down does the reverse. Power is conserved.

Step-Up vs Step-Down — At a Glance

Property	Step-Up	Step-Down
$N_s$ vs $N_p$	$N_s > N_p$	$N_s < N_p$
Voltage	Increases	Decreases
Current	Decreases	Increases
Wire thickness	Secondary thin, primary thick	Secondary thick, primary thin
Use	Power transmission, X-ray	Mobile charger, doorbell

The wire thickness rule is a memory aid — high-current side needs thick wire to avoid heat loss.

Solved Examples

Example 1 (Easy, CBSE) — Voltage Calculation

A transformer has $200$ primary turns and $1000$ secondary turns. If $V_p = 220\,\text{V}$ , find $V_s$ .

$V_s = V_p \frac{N_s}{N_p} = 220 \times 5 = 1100\,\text{V}$

This is a step-up transformer.

Example 2 (Easy, CBSE) — Current Calculation

The same transformer drives a $50\,\Omega$ resistor. Find the primary current. Assume the transformer is ideal.

Secondary current $I_s = V_s/R = 1100/50 = 22\,\text{A}$ . Primary current $I_p = I_s \times (N_s/N_p) = 22 \times 5 = 110\,\text{A}$ .

Example 3 (Medium, JEE Main) — Real Transformer Efficiency

A transformer is fed $V_p = 200\,\text{V}$ , $I_p = 5\,\text{A}$ . Output is $V_s = 1000\,\text{V}$ , $I_s = 0.95\,\text{A}$ . Find efficiency.

$\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{1000 \times 0.95}{200 \times 5} = \frac{950}{1000} = 95\%$

The $5\%$ loss goes into heat (resistive losses), eddy currents, and hysteresis.

Example 4 (Medium, JEE Main) — Reflected Impedance

A step-down transformer with turns ratio $10:1$ has a $4\,\Omega$ load on the secondary. Find the impedance seen at the primary.

Reflected impedance scales as the square of the turns ratio:

$Z_p = Z_s \left(\frac{N_p}{N_s}\right)^2 = 4 \times 100 = 400\,\Omega$

This is the principle behind impedance matching in audio systems.

Example 5 (Hard, JEE Advanced) — Power Loss in Transmission

A power station generates $100\,\text{kW}$ at $250\,\text{V}$ . The transmission line has resistance $1\,\Omega$ . Compare power lost (a) without a transformer and (b) with a step-up to $25\,\text{kV}$ .

(a) Current $= 100000/250 = 400\,\text{A}$ . Loss $= I^2 R = 160\,\text{kW}$ . More than the generated power — impossible.

(b) Current $= 100000/25000 = 4\,\text{A}$ . Loss $= 16\,\text{W}$ . Negligible.

This is precisely why long-distance transmission uses high-voltage lines.

Exam-Specific Tips

CBSE Class 12 Boards: Direct formula $V_s/V_p = N_s/N_p$ and one-step efficiency questions. Three to five marks per year.

JEE Main: Transformer efficiency, reflected impedance, power loss in transmission. One question typical.

JEE Advanced: Combined with AC circuits — phase, power factor, and transformer in the same problem.

NEET: Definition-level questions and one numerical on turns ratio.

Common Mistakes to Avoid

Mistake 1: Assuming transformers work with DC. They need changing flux, which DC cannot provide. A DC transformer is a heating coil at best, a fire hazard at worst.

Mistake 2: Confusing step-up with high current. Step-up means high voltage, low current.

Mistake 3: Ignoring the squared factor in reflected impedance. Turns ratio appears squared because both $V$ and $I$ transform.

Mistake 4: Forgetting that real transformers have efficiency below 100%. Always check whether the problem says “ideal”.

Mistake 5: Confusing primary/secondary. Primary is always the input side connected to the source.

Practice Questions

Q1. A transformer steps up $220\,\text{V}$ to $11{,}000\,\text{V}$ . If primary has $50$ turns, how many turns on secondary?

$N_s = N_p \times V_s/V_p = 50 \times 50 = 2500$ turns.

Q2. An ideal transformer with turns ratio $1:20$ has $V_p = 220\,\text{V}$ . The secondary drives a $440\,\Omega$ load. Find $I_p$ .

$V_s = 4400\,\text{V}$ , $I_s = 10\,\text{A}$ , $I_p = 200\,\text{A}$ .

Q3. Why is the transformer core laminated?

To break up paths for eddy currents and reduce core losses.

Q4. A transformer has $1000$ primary and $100$ secondary turns. Input is $V_p = 230\,\text{V}, I_p = 0.5\,\text{A}$ . If efficiency is $90\%$ , find $I_s$ .

$V_s = 23\,\text{V}$ . Output power $= 0.9 \times 230 \times 0.5 = 103.5\,\text{W}$ . $I_s = 103.5/23 = 4.5\,\text{A}$ .

Q5. Why do we use AC for power transmission instead of DC?

AC voltage can be easily stepped up by transformers for low-loss transmission and stepped down for safe domestic use. DC transformation is technically possible but historically much more expensive.

FAQs

Why is iron used for the core? High magnetic permeability concentrates the flux in the core, ensuring almost all of it links both coils. Soft iron has low hysteresis loss compared to steel.

Can a transformer work in vacuum? Yes — the core just needs to be magnetisable. Air-core transformers exist, used in radio frequency circuits where iron losses dominate.

What is the buzzing sound from a transformer? Magnetostriction — the iron core slightly changes shape with the alternating flux, vibrating at $100\,\text{Hz}$ (twice the supply frequency).

Are mobile chargers transformers? Modern chargers use switching power supplies that include a small high-frequency transformer. Older “wall warts” used a classic 50 Hz iron-core transformer plus a rectifier.

What’s the difference between auto-transformer and ordinary transformer? An auto-transformer uses a single tapped winding for both primary and secondary. Cheaper and lighter but doesn’t isolate the two circuits.