SCSA WACE 2022-style practice: An ideal step-up transformer has 240 turns on its primary and 9600 turns on its secondary. The primary is connected to a 230\ V AC supply and draws a current of 5.0\ A. (a) Calculate the secondary voltage. (b) Calculate the secondary current. (c) Explain why this transformer would not work on a DC supply.

A 7 mark calculation rewards the turns-ratio voltage, power conservation for current and a flux argument. **(a) Secondary voltage.** V_sV_p=N_sN_p, so $V_s=V_p(N_s)/(N_p)=230×(9600)/(240)=9200\ V.$ **(b) Secondary current.** For an ideal transformer V_pI_p=V_sI_s, so $I_s=(V_pI_p)/(V_s)=(2305.0)/(9200)=0.125\ A.$ **(c) Why not DC.** A transformer needs a continually changing flux to induce a secondary emf. DC produces a steady flux once established, so / t=0 and no emf is induced in the secondary after switch-on. Markers reward the turns ratio giving 9200\ V, power conservation giving 0.125\ A and the changing-flux requirement for AC.

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Why is electrical energy transmitted at high voltage and stepped down with transformers?

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

A focused answer to the WACE Year 12 Physics Unit 3 content point on transformers and the grid. The turns-ratio equation, ideal power conservation, why transformers need AC, and why transmitting at high voltage cuts resistive line losses.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksAn 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 20216 marksA 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.

Show worked answer →

A 6 mark answer rewards two correct loss calculations and an explanation.

At $20\ \text{kV}$: Line current $I=\dfrac{P}{V}=\dfrac{2.0\times10^6}{2.0\times10^4}=100\ \text{A}$ , so $P_{\text{loss}}=I^2R=(100)^2(8.0)=8.0\times10^4\ \text{W}=80\ \text{kW}$ .
At $200\ \text{kV}$: $I=\dfrac{2.0\times10^6}{2.0\times10^5}=10\ \text{A}$ , so $P_{\text{loss}}=(10)^2(8.0)=8.0\times10^2\ \text{W}=0.80\ \text{kW}$ .
Significance: Raising the voltage tenfold cuts the current tenfold and the loss a hundredfold (from $80\ \text{kW}$ to $0.80\ \text{kW}$ ), because $P_{\text{loss}}=I^2R$ . This is why the grid steps voltage up to hundreds of kilovolts for long-distance transmission and steps it back down near consumers.

Markers reward both currents, both losses via $I^2R$ , the factor-of-one-hundred reduction and the link to high-voltage transmission.