Transformers, their efficiency, and sending power over long distances
A transformer changes AC voltage by mutual induction, with V_s/V_p = N_s/N_p. Power is sent at high voltage to cut the I²R loss in the lines.
Construction and working
A transformer raises or lowers an alternating voltage. It has two coils of insulated copper wire, wound on the same soft iron core. The primary is joined to the supply and the secondary to the load. The coils are not joined electrically; only the changing flux in the core links them. The core is made of thin laminated sheets, insulated from each other, to cut eddy currents. An alternating current in the primary sets up an alternating flux in the core. This flux passes through the secondary and induces an emf in it by mutual induction. A step-up transformer has more turns in the secondary and raises the voltage. A step-down transformer has fewer turns in the secondary and lowers it. A steady DC gives a constant flux, so a transformer does not work on DC.
Deriving the turns ratio for an ideal transformer
An **ideal transformer** has coils of zero resistance, no flux leakage (every turn of both coils has the same flux) and no energy loss in the core.
- 1Let the flux in the core change at the rate dφ/dt. It is the same through every turn of both coils.
- 2Back emf in the primary of N_p turns: E_p = −N_p dφ/dt. With zero resistance, this equals the applied voltage V_p.
- 3emf in the secondary of N_s turns: E_s = −N_s dφ/dt, which equals the output voltage V_s.
- 4★Dividing: V_s/V_p = N_s/N_p.
- 5No energy is lost, so input power = output power: V_pI_p = V_sI_s.
- 6★So, I_p/I_s = V_s/V_p = N_s/N_p. Raising the voltage lowers the current in the same ratio.
★ key step
Energy losses in a transformer and how they are reduced
| Loss | Cause | How it is reduced |
|---|---|---|
| Copper loss | The current heats the copper windings (I²R) | Use thick wire of good conductivity |
| Eddy current loss | The changing flux induces eddy currents that heat the iron core | Make the core of thin laminated sheets, insulated from each other |
| Hysteresis loss | The core is magnetised and demagnetised every cycle, which heats it | Use soft iron, which magnetises and demagnetises easily |
| Flux leakage | Some flux of the primary does not pass through the secondary | Wind the two coils close together, one over the other |
Sending power to your home at high voltage
Power stations are often far from towns. Suppose a power P is sent at a voltage V through lines of total resistance R. The current is I = P/V, so the power lost as heat in the lines is I²R = P²R/V². So a higher voltage means a smaller current and a much smaller loss. Doubling V cuts the loss to a quarter. Large power-station generators give about 11 kV to 25 kV (your NIOS lesson says about 15 kV). Step-up transformers raise this to hundreds of kilovolts. In India the long lines run at 220 kV, 400 kV and 765 kV (your NIOS lesson gives about 330 kV as an example). Step-down transformers at sub-stations then lower it in stages, until it reaches homes at 230 V. Very high voltages need tall towers and good insulation, which cost more. Transformers work easily only with AC, which is one main reason the world uses AC for its power supply.
Worked example: a step-down transformer and a transmission line
(a) A transformer lowers the 230 V mains to 12 V for a 24 W lamp. Its primary has 1150 turns. Taking it as ideal, find the turns in the secondary and the currents in both coils. (b) If the transformer is 90% efficient, what current does its primary draw? (c) A station sends 200 kW through lines of total resistance 4.0 Ω. Compare the loss when it sends at 2000 V and at 20 000 V.
- (a) N_s = N_p × V_s/V_p = 1150 × 12/230 = 60 turns.
- Secondary current I_s = P/V_s = 24 W ÷ 12 V = 2.0 A. Primary current I_p = 24 W ÷ 230 V ≈ 0.10 A.
- (b) Input power = 24 W ÷ 0.90 ≈ 26.7 W, so I_p = 26.7 W ÷ 230 V ≈ 0.12 A.
- (c) At 2000 V: I = 200 000 W ÷ 2000 V = 100 A, so the loss = I²R = 100² × 4.0 W = 40 000 W = 40 kW, a fifth of the power sent.
- At 20 000 V: I = 10 A, so the loss = 10² × 4.0 W = 400 W.
- Raising the voltage 10 times cuts the current 10 times and the loss 100 times.
(a) 60 turns, 2.0 A and about 0.10 A; (b) about 0.12 A; (c) 40 kW at 2000 V, but only 0.4 kW at 20 000 V
How to draw a transformer
- Draw the soft iron core as a thick square frame. Draw a few parallel lines across it to show the laminations.
- Wind the primary on the left side of the frame and label N_p, V_p and I_p. Join it to an AC source, a circle with a ~ sign.
- Wind the secondary on the right side: more loops for step-up, fewer for step-down. Label N_s, V_s and I_s, and join it to a load R.
- Draw arrows round the core to show the alternating flux φ.
- Write the title: step-up (or step-down) transformer.
The transformer on the electric pole or in the sub-station near your home steps the voltage down to 230 V. Doorbells, battery chargers and welding machines also use transformers.
3 marks: “Explain the working of a transformer. Show that V_s/V_p = N_s/N_p for an ideal transformer, and name two energy losses.” Primary and secondary on a laminated soft iron core; AC in the primary makes a changing flux that induces an emf in the secondary by mutual induction; E = −N dφ/dt for each, so V_s/V_p = N_s/N_p; losses: copper, eddy current, hysteresis, flux leakage.
2 marks: “Why is electric power sent over long distances at high voltage?” Loss = I²R = P²R/V². At a high V the current is small, so the heating loss in the lines is much smaller; the voltage is stepped down near homes.
2 marks: “A transformer takes 1500 W from the mains and gives 1350 W to its load. Find its efficiency.” η = (1350 ÷ 1500) × 100% = 90%.
Turns ratio = voltage ratio = inverse current ratio: V_s/V_p = N_s/N_p = I_p/I_s. Step up the volts, step down the amps.