What is applying the equations together?

WAPhysicsSyllabus dot point

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 4.77 min answerUpdated 2026-05-29

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 $100\ \text{kV}$ .

At $10\ \text{kV}$ , the current is

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}.

At $100\ \text{kV}$ , the current is

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}.

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

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.

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