Apply the ideal-transformer relationships V_s / V_p = N_s / N_p and I_p / I_s = N_s / N_p, and the role of step-up and step-down transformers in minimising I^2 R losses in AC power transmission

What this dot point is asking

QCAA wants you to derive the ideal-transformer voltage and current ratios from Faraday's law, apply them to step-up and step-down problems, recognise that an ideal transformer conserves power ($V_p I_p = V_s I_s$), identify the four main loss mechanisms in real transformers, and explain why AC transmission relies on stepping voltage up for transmission and down for distribution. The dot point closes Topic 2 and is a frequent EA Paper 2 question.

The answer

How a transformer works

A transformer is two coils (the primary and the secondary) wound on the same ferromagnetic core. An alternating current in the primary produces an alternating flux in the core. The same changing flux links the secondary, inducing an EMF in it (Faraday's law).

Because both coils share the same flux $\Phi$ (in the ideal case where all primary flux links the secondary):

$V_p = N_p \, d\Phi / dt$ and $V_s = N_s \, d\Phi / dt$.

Dividing:

The voltage ratio equals the turns ratio.

Power conservation and the current ratio

An ideal transformer has no losses. Power in equals power out:

Combining with the voltage ratio:

Currents are in the inverse ratio to voltages. A step-up transformer ($N_s > N_p$) raises voltage and lowers current; a step-down transformer ($N_s < N_p$) does the reverse.

Why transformers need AC

Faraday's law requires $d\Phi / dt \neq 0$ to induce a secondary EMF. A steady DC primary current produces a constant flux, so the secondary EMF is zero (except during the brief switch-on transient). AC, by reversing direction many times per second, produces the continuous flux change required.

A transformer connected to a DC source also risks overheating or magnetic saturation, because the primary winding's low resistance allows a large steady current with no back-EMF to limit it.

The four losses in a real transformer

Typical efficiencies: about 95 percent for small transformers, above 99 percent for large grid transformers.

Try it: Transformer calculator. Enter turns and either primary or secondary voltage to get the other side, plus the corresponding currents under power conservation.

Step-up and step-down in AC transmission

Transmitting electrical power $P = V I$ over a long line of resistance $R_{\text{line}}$ wastes power as:

The loss depends on the current squared, not the voltage. So for the same transmitted power $P$, a higher transmission voltage means a smaller current and dramatically smaller line losses.

Typical Australian grid:

1. Generation at a power station: about 11 to 25 kV from the generator.
2. Step-up transformer at the station raises this to 132, 220, 330, 500 kV or higher for long-distance transmission.
3. Transmission lines carry power at high voltage (low current, low $I^2 R$ loss).
4. Step-down transformer at a sub-transmission substation drops to 33 or 66 kV.
5. Distribution transformer at street level drops to 11 kV, then a final transformer drops to 415 V (three-phase) or 240 V (single-phase) for delivery to homes and businesses.

Without transformers, this voltage manipulation would not be possible, and long-distance AC transmission would lose most of the generated power as heat in the wires. This is why Tesla's AC system, with its easy transformer-based voltage conversion, won out over Edison's DC system in the 1890s. (Modern HVDC links exist today, but they require expensive electronic converters at each end.)

Worked example: transmission line saving

A power station delivers $1.0$ MW to a town through a transmission line of total resistance $5.0$ ohms. Compare the line losses at $1000$ V transmission versus $100$ kV transmission.

At $V = 1000$ V:

$I = P / V = 10^6 / 10^3 = 1000$ A.
$P_{\text{loss}} = I^2 R = (1000)^2 \times 5.0 = 5.0 \times 10^6$ W = 5 MW.

The line cannot even deliver 1 MW: the losses exceed the input.

At $V = 100$ kV:

$I = 10^6 / 10^5 = 10$ A.
$P_{\text{loss}} = (10)^2 \times 5.0 = 500$ W.

The losses drop from 5 MW to 500 W, a factor of $10000 = 100^2$. The $V^2$ dependence of efficiency is the entire reason transmission voltages are so high.

Worked example: turns ratio

A neon sign requires $5.0$ kV from a $240$ V mains supply. Find the turns ratio, the primary current when the sign draws $50$ mA, and the input power.

Turns ratio: $N_s / N_p = V_s / V_p = 5000 / 240 \approx 20.8$.

Primary current (ideal transformer): $I_p = (V_s / V_p) I_s = 20.8 \times 0.050 = 1.04$ A.

Input power: $P = V_p I_p = 240 \times 1.04 = 250$ W (equal to $V_s I_s = 5000 \times 0.050 = 250$ W, as expected for an ideal transformer).

How this appears in IA1 and IA2

IA1 data test. Expect a transformer characteristics table (turns counts, voltage and current measurements) with questions on the inferred turns ratio, the deviation from the ideal model, and the energy dissipated in the windings. Alternatively, a transmission-line stimulus with two transmission voltages and a question on the $I^2 R$ losses.

IA2 student experiment. A standard IA2 design measures secondary voltage against primary voltage (or against $N_s / N_p$) using a low-voltage AC supply and a hand-wound transformer. Strong reports linearise $V_s$ against $V_p$ for fixed turns ratio (slope $= N_s / N_p$) and discuss systematic deviations from the ideal model (flux leakage at low turns counts, $I^2 R$ heating at high primary currents).

Examples in context

Example 1. A Townsville $275 \text{ kV}$ Powerlink transmission line of $\sim 300 \text{ km}$ from Strathmore substation to the Townsville ring carries $300 \text{ MW}$ at $V = 275 \text{ kV}$, giving $I = P/V = 1091 \text{ A}$. Total line resistance is about $25 \text{ ohms}$, so $I^2 R$ loss is $1091^2 \times 25 = 2.98 \times 10^7 \text{ W}$, roughly $10$ per cent. Stepping down at the local substation to $11 \text{ kV}$ via a $275 / 11 \text{ kV}$ transformer with turns ratio $25:1$ converts the high-voltage transmission to local distribution. This dot-point chain is exactly QCAA Unit 3 IA1.

Example 2. A Gladstone industrial transformer at the Boyne Island smelter steps $132 \text{ kV}$ down to $11 \text{ kV}$ for the potline rectifier. Turns ratio is $12:1$. At rated $250 \text{ MVA}$, primary current is $1894 \text{ A}$ and secondary is $22730 \text{ A}$. Eddy currents are limited by lamination ($I^2 R$ in core sheets), hysteresis by silicon-steel core, and resistive loss by copper-bar windings. QCAA EA Unit 3 Paper 2 routinely uses this kind of grid stem.

Try this

Q1. State the ideal transformer voltage and current ratios. [2 marks]

• Cue. $V_s/V_p = N_s/N_p$; $I_p/I_s = N_s/N_p$.

Q2. A step-down transformer has $N_p = 5000$, $N_s = 250$. With $V_p = 11000 \text{ V}$ and secondary delivering $20 \text{ A}$, calculate $V_s$, $I_p$ and the transferred power. [4 marks]

• Cue. $V_s = 550 \text{ V}$; $I_p = 1.0 \text{ A}$; $P = 11 \text{ kW}$.

Q3. A Townsville Powerlink line delivers $300 \text{ MW}$ at $275 \text{ kV}$ over $25 \text{ ohms}$. (a) Calculate the line current and the $I^2 R$ loss. (b) Compare with the same power at $33 \text{ kV}$ on the same line. (c) Justify the use of step-up and step-down transformers in this chain. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $I = 1091 \text{ A}$, loss $29.8 \text{ MW}$ ($10$ per cent); (b) $I = 9091 \text{ A}$, loss $2.07 \text{ GW}$, infeasible; (c) high V lowers $I^2 R$, transformers enable conversion.