Transformers and power distribution: WACE Year 12 Physics Unit 3

Syllabus dot point

“Apply the transformer equation and explain high-voltage transmission of electrical power”

8 min readUpdated 2 Jun 2026

Quick answer

A transformer transfers energy between two coils on a shared iron core by mutual induction, with voltages in the turns ratio $\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}$ . For an ideal transformer power is conserved, so $V_pI_p=V_sI_s$ and stepping voltage up steps current down. The grid sends power at very high voltage because line loss is $P_{\text{loss}}=I^2R$ , so a smaller current means far smaller losses in the transmission wires.

What this dot point is asking

WACE wants you to use the transformer equation, apply power conservation, and explain quantitatively why high-voltage transmission is efficient. This is the systems-level payoff of electromagnetic induction.

The transformer equation

A changing current in the primary coil sets up a changing flux that the iron core guides to the secondary, inducing an emf there. For an ideal transformer the induced emf per turn is the same in both coils, so

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

More secondary turns than primary gives a step-up transformer; fewer gives a step-down. The iron core, often laminated to reduce eddy-current heating, links the flux efficiently between the coils.

Power conservation and current

An ideal transformer wastes no power, so the power into the primary equals the power out of the secondary:

V_pI_p=V_sI_s.

This means voltage and current trade off: a transformer that doubles the voltage halves the current. You cannot get more energy out than you put in; transformers change the voltage at which energy is delivered, not the amount.

Why transformers need AC

A transformer relies on a continually changing flux to induce a secondary emf. Direct current produces a steady flux once established, so $\Delta\Phi/\Delta t=0$ and no emf is induced after the initial switch-on. This is the central reason mains power is distributed as alternating current.

High-voltage transmission

Power lines have resistance $R$ , and the power lost heating them is

P_{\text{loss}}=I^2 R.

Because the loss depends on the square of the current, halving the current cuts the loss to a quarter. To deliver a given power $P=VI$ with a small current, the voltage must be large. So a step-up transformer raises the voltage to hundreds of kilovolts for long-distance transmission, and step-down transformers near homes reduce it to safe levels. This is why the grid uses transformers at both ends.

Transmission loss before and after stepping up

Comparing line loss at two voltages

A power station delivers $1.0\ \text{MW}$ along a line of total resistance $5.0\ \Omega$ . Compare the loss when transmitting at $10\ \text{kV}$ and at $100\ \text{kV}$ .

Step 1: Find the line current at $10\ \text{kV}$ and calculate the power loss.

The power $P = VI$ gives the current once the voltage is known. Because the line loss is $P_{\text{loss}} = I^2 R$ , a higher current wastes far more power as heat in the resistance of the wires:

I=\frac{P}{V}=\frac{1.0\times10^6}{1.0\times10^4}=100\ \text{A},\qquad P_{\text{loss}}=I^2R=(100)^2(5.0)=5.0\times10^4\ \text{W}.

Step 2: Find the line current at $100\ \text{kV}$ and calculate the power loss.

Raising the transmission voltage tenfold reduces the current tenfold for the same delivered power. Because the loss depends on the square of the current, the loss falls by a factor of $100$ :

I=\frac{1.0\times10^6}{1.0\times10^5}=10\ \text{A},\qquad P_{\text{loss}}=(10)^2(5.0)=5.0\times10^2\ \text{W}.

Step 3: Compare the two losses.

The tenfold increase in voltage has cut the loss a hundredfold, from $50\ \text{kW}$ to $0.5\ \text{kW}$ . This is why the grid steps voltage up to very high levels for long-distance transmission.

Raising the voltage tenfold cuts the current tenfold and the loss a hundredfold, from $50\ \text{kW}$ to $0.5\ \text{kW}$ .

Real transformer losses

Real transformers are not quite ideal, and WACE expects you to name the loss mechanisms and how engineers reduce them. Eddy currents are loops of current induced in the solid iron core by the changing flux; they dissipate energy as heat, so cores are laminated (built from thin insulated sheets) to break up these loops. Resistive heating in the copper windings is reduced by using thick, low-resistance wire. Hysteresis loss, the energy used repeatedly re-magnetising the core, is reduced by choosing a soft magnetic material that magnetises and demagnetises easily. Despite these effects, large grid transformers reach efficiencies above $99\%$ , so the ideal equations are an excellent approximation in exam calculations.

Applying the equations together

Many questions chain the transformer equation with $P=VI$ and $P_{\text{loss}}=I^2R$ . Work in order: use the turns ratio for the transmission voltage, find the line current from $P=VI$ , then the loss from $I^2R$ . Keep primary and secondary quantities clearly labelled.

Thinking a step-up transformer creates energy

A step-up transformer raises voltage but lowers current in the same proportion, so the power is unchanged (and slightly reduced by real losses). It does not generate energy. Likewise, students sometimes use the line voltage in $P_{\text{loss}}=I^2R$ instead of the current through the line; the loss depends on the transmission current and the wire resistance, not the supply voltage.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks

An ideal step-up transformer has

240

turns on its primary and

9600

turns on its secondary. The primary is connected to a

230\ \text{V}

AC supply and draws a current of

5.0\ \text{A}

. (a) Calculate the secondary voltage. (b) Calculate the secondary current. (c) Explain why this transformer would not work on a DC supply.

Show worked answer →

A 7 mark calculation rewards the turns-ratio voltage, power conservation for current and a flux argument.

(a) Secondary voltage: $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$ , so
$V_s=V_p\frac{N_s}{N_p}=230\times\frac{9600}{240}=9200\ \text{V}.$
(b) Secondary current: For an ideal transformer $V_pI_p=V_sI_s$ , so
$I_s=\frac{V_pI_p}{V_s}=\frac{230\times5.0}{9200}=0.125\ \text{A}.$
(c) Why not DC: A transformer needs a continually changing flux to induce a secondary emf. DC produces a steady flux once established, so $\Delta\Phi/\Delta t=0$ and no emf is induced in the secondary after switch-on.

Markers reward the turns ratio giving $9200\ \text{V}$ , power conservation giving $0.125\ \text{A}$ and the changing-flux requirement for AC.

WACE 2021-style6 marks

A town requires

2.0\ \text{MW}

of power delivered through transmission lines of total resistance

8.0\ \Omega

. Compare the power lost in the lines when the power is transmitted at

20\ \text{kV}

with the loss when it is transmitted at

200\ \text{kV}

, and explain the significance of the result for the design of the grid.