Why can a transformer change voltage but never give you free power?

How to use this simulation

1. Press play. The two waves under the core show the primary and secondary voltages. With the defaults, Vp = 220 V, Np = 1,000 and Ns = 100, so the secondary wave is 10 times smaller: Vs = 22.0 V. The label says step-down.
2. Drag Ns upwards. The secondary coil gains turns, the green wave grows, and at Ns = 1,000 it equals the primary. Go past that and the label flips to step-up. Notice that the bigger wave always fills the panel and the other is drawn to scale.
3. Watch the readings while you change Ns. When Vs rises, the secondary current Is rises too, because the load is fixed, and the source has to supply more power. Compare Ip and Is: their ratio is the inverse of the turns ratio.
4. Slide the efficiency down from 100%. The power from the source Pp climbs above the power in the load Ps, and the difference shows up as “wasted in the transformer”. That is the heat your charger gives off.
5. Change the load resistance RL. A smaller RL draws a larger current and more power. The primary current follows, because the primary has to feed whatever the secondary takes.
6. Read the last two lines in the readings. They compare the loss in a 0.5 Ω line if the same power travels at the secondary voltage or at the primary voltage. Make Vs bigger and watch the loss collapse.

Why we need to change voltage at all

Look at the voltages in a single day. Your phone charger delivers about 5 V. The socket on your wall gives 220 V. A power station generates at around 11 kV. The lines that carry power across the country run at 132 kV and more. No single voltage suits all of these jobs: low voltage is safe in the home, high voltage is efficient on the road.

We need something that changes AC voltage up and down cheaply, with no moving parts, and wasting very little. That something is the transformer. It is a lump of iron and copper that sits quietly on a pole outside your house, hums a little, and does one of the most important jobs in the power system.

What is inside

A basic transformer has just three parts, all drawn in the simulation.

• The iron core: a closed loop of soft iron. It guides almost all the magnetic flux from one coil to the other. Real cores are built from thin insulated sheets (laminations), not a solid block.
• The primary coil: Np turns of wire, connected to the AC source. It makes the changing flux.
• The secondary coil: Ns turns of wire, connected to the load. The changing flux induces the output voltage in it.

How it works: mutual induction

The alternating voltage on the primary drives an alternating current, which magnetises the core first one way, then the other, 50 times a second on the Bangladesh and India mains. The moving dots in the core show this surge of flux, and they glow brightest when the flux is changing fastest.

Because the core links the two coils, the same flux passes through every turn of both. Faraday’s law says each turn has the same emf induced in it, equal to the rate of change of flux. Call that emf per turn ε. Then the primary, with Np turns, has a voltage Np·ε across it, and the secondary, with Ns turns, has Ns·ε. Divide one by the other and ε cancels.

Step-up and step-down

If Ns is greater than Np the secondary voltage is higher than the primary: a step-up transformer. If Ns is smaller, the voltage falls: a step-down transformer. If they are equal the voltage is unchanged, and the transformer is used only to isolate two circuits electrically.

With 220 V on the primary of 1,000 turns, every turn carries 0.22 V. The default secondary of 100 turns therefore gives 22.0 V. Here is what other secondaries would do, with the same primary:

No free lunch: power in equals power out

A step-up transformer makes a bigger voltage, so it must be getting something for it, right? No. The power on the secondary cannot exceed the power fed to the primary. For an ideal transformer, with no losses at all, the two powers are equal.

With the defaults, the secondary has 22.0 V across a 10 Ω load, so Is = 22.0 ÷ 10 = 2.20 A and Ps = 22.0 × 2.20 = 48.4 W. For the primary to supply 48.4 W at 220 V it must draw Ip = 48.4 ÷ 220 = 0.220 A. The secondary current is 10 times the primary current, and the secondary voltage is 10 times smaller. Voltage and current trade places.

A real transformer is not perfect

Efficiency is the useful power out divided by the power in: η = Ps ÷ Pp. An ideal transformer has η = 100%, and large power transformers come remarkably close, at 95% to 99%. Small chargers are less efficient. The missing power goes to heat in four ways.

• Copper loss: the wire has resistance, so I²R heating happens in both coils. Thicker wire reduces it.
• Eddy-current loss: the changing flux induces swirling currents in the iron core itself. Laminated, insulated sheets break these currents up.
• Hysteresis loss: the iron is magnetised back and forth, and some energy is lost each time. Soft iron or silicon steel, which magnetises easily, keeps it small.
• Flux leakage: not every line of flux links both coils. Winding the secondary on top of the primary, or on the same limb, helps.

Why power lines use high voltage

Here is the question every exam asks. Suppose a power plant must send 50 kW along a line whose total resistance is 0.5 Ω. The power is P = VI, so the current needed is I = P ÷ V. The line wastes heat at the rate I²R.

Send it at 220 V and the current is 227.3 A. The loss is 227.3² × 0.5 = 25.8 kW, which is 51.7% of what you sent: the wire eats half the power. Send the same 50 kW at 11 kV instead and the current falls to 4.55 A. The loss is now only 10.3 W. Raising the voltage 50 times cut the loss 2,500 times, because the loss goes as the square of the current.

The journey from power station to socket

A generator produces power at around 11 kV. A step-up transformer raises it to 132 kV or more for the long trip across the country, a ratio of 12. Near towns, step-down transformers at substations bring it down in stages. Finally, a transformer on a pole or in a small house on your street reduces about 11 kV to 400 V, a ratio of 27.5, and the 220 V single-phase supply reaches your meter.

So the same electricity may pass through four or five transformers on its way to your phone. At each stage the voltage is chosen for the best balance between safety, cost and loss.

Worked example 1: the default transformer

A transformer has Np = 1,000, Ns = 100, a 220 V primary and a 10 Ω load. Assume it is ideal. Find Vs, Is, Ps and Ip.

Worked example 2: how many turns for 6 V?

You want 6 V from the 220 V supply using the same 1,000-turn primary. Ns = Np × Vs ÷ Vp = 1,000 × 6 ÷ 220 = 27.27. You cannot wind a fraction of a turn, so use 27 turns, which gives 5.94 V. A small mismatch like this is normal and the load copes with it.

Worked example 3: the same transformer, 90% efficient

Set the efficiency to 90% in the simulation. The load still gets 48.4 W, but the source must supply Pp = Ps ÷ η = 48.4 ÷ 0.90 = 53.78 W. The primary current rises to 0.2444 A, and 5.38 W is lost as heat in the core and coils. Note that the secondary side does not change at all: the load never notices the losses, only the source does.

Worked example 4: a phone charger

A charger gives 12 V at 1.5 A, so the load power is 18 W. If it is 90% efficient, it draws 20.0 W from the 220 V socket, which means Ip = 20.0 ÷ 220 = 0.091 A. The 2.0 W of difference warms the charger. With a 1,000-turn primary the secondary would need about 54.5 turns.

Worked example 5: a power-station step-up

A power station produces 11 kV and the line needs 132 kV. The turns ratio is Ns ÷ Np = 12. With 500 turns on the primary, the secondary needs 6,000. The current on the high-voltage side is 12 times smaller than the generator current, which is what makes the thin line cables possible.

Common mistakes

These four lose marks every year.

• Thinking a step-up transformer gives more power. It gives more voltage and less current, with the product equal (or a little smaller).
• Writing the turns ratio upside down. It is Vs/Vp = Ns/Np, and for currents Is/Ip = Np/Ns.
• Expecting a transformer to work on DC. It needs a changing flux.
• Forgetting that efficiency is below 100% in real problems: Pp = Ps ÷ η, not Ps × η.

Exam corner

Typical questions: explain the principle and working of a transformer with a diagram, derive Vs/Vp = Ns/Np, list the energy losses and how each is reduced, or explain why electrical power is transmitted at high voltage. For the last one write the chain: P = VI, so for a fixed P a higher V needs a smaller I, and the power wasted is I²R.

Numericals ask for Vs, Ip from Is, the number of turns, or the efficiency. Use the readings panel to check each answer with the matching slider values.

Revision list

Check yourself without looking.

• Principle: mutual induction between two coils on a shared iron core.
• Vs/Vp = Ns/Np. Ns > Np is step-up and Ns < Np is step-down.
• Ideal: VpIp = VsIs, so Is/Ip = Np/Ns.
• η = Ps/Pp; real efficiencies are 95% to 99% for large units.
• Losses: copper, eddy current, hysteresis, flux leakage.
• Transmission is at high voltage because loss = I²R and I = P/V.
• A transformer does not work on DC.

Frequently asked questions