Generators, transformers and AC power transmission: VCE Physics Unit 3

Syllabus dot point

“Explain the operation of AC and DC generators, distinguish between peak and RMS values of voltage and current using $V_{RMS} = V_{peak} / \sqrt{2}$ and $I_{RMS} = I_{peak} / \sqrt{2}$ , and apply the ideal transformer relationship $V_1 / V_2 = N_1 / N_2 = I_2 / I_1$ to AC power transmission, including resistive losses $P_{loss} = I^2 R$ ”

12 min readUpdated 21 May 2026

Quick answer

A generator induces a sinusoidal EMF $\varepsilon = NBA\omega \sin(\omega t)$ in a rotating coil (slip rings for AC, split-ring commutator for DC), with RMS values $V_{RMS} = V_{peak}/\sqrt{2}$ used for power, while an ideal transformer obeys $V_1/V_2 = N_1/N_2 = I_2/I_1$ and enables high-voltage AC transmission to slash the $I^2 R$ cable losses.

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What this dot point is asking
The answer
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Exam-style questions

What this dot point is asking

VCAA wants you to explain how AC and DC generators work, distinguish peak and RMS values, and apply the ideal transformer equation to AC power transmission, including the role of $P_{loss} = I^2 R$ in cable losses.

The answer

How a generator works

A generator is a coil rotated in a magnetic field. By Faraday's law, the changing flux through the coil induces an EMF. If the coil has $N$ turns, area $A$ , in field $B$ , rotating at angular frequency $\omega$ :

\varepsilon(t) = N B A \omega \sin(\omega t)

The EMF is sinusoidal, with peak value $\varepsilon_{peak} = N B A \omega$ .

AC generator. Uses slip rings: a continuous ring on each end of the coil, with brushes feeding the external circuit. As the coil rotates, the current reverses every half-turn. The external circuit sees alternating current.

DC generator. Uses a split-ring commutator (the same device as in a DC motor). It swaps the brush connections every half-turn, so the external current always flows in the same direction. The output is a series of half-wave humps; with many coils at different angles, the output approaches a smooth DC voltage.

Peak and RMS values

A sinusoidal AC voltage swings between $+V_{peak}$ and $-V_{peak}$ . The average voltage over a cycle is zero, which is unhelpful for calculating power. Instead we use the root-mean-square (RMS) value:

V_{RMS} = \frac{V_{peak}}{\sqrt{2}}, \quad I_{RMS} = \frac{I_{peak}}{\sqrt{2}}

The RMS value is the equivalent DC that delivers the same average power to a resistor. For AC:

P_{avg} = V_{RMS} I_{RMS} = I_{RMS}^2 R = \frac{V_{RMS}^2}{R}

Australian mains is quoted as $230$ V RMS (peak $\approx 325$ V) at $50$ Hz.

The ideal transformer

A transformer is two coils wound around a common iron core. AC in the primary coil produces a changing magnetic flux in the core. The same changing flux passes through the secondary coil, inducing an EMF.

For an ideal transformer (no flux leakage, no resistive losses):

\frac{V_1}{V_2} = \frac{N_1}{N_2}

and energy conservation $P_1 = P_2$ gives:

\frac{I_1}{I_2} = \frac{N_2}{N_1}

So the full ideal-transformer relationship is:

\frac{V_1}{V_2} = \frac{N_1}{N_2} = \frac{I_2}{I_1}

Step-up ( $N_2 > N_1$ ): voltage rises, current falls. Step-down ( $N_2 < N_1$ ): voltage falls, current rises.

A transformer works only on alternating current. A constant DC current produces no change in flux and no induced EMF in the secondary.

Power transmission losses

The power lost as heat in transmission lines of resistance $R$ depends on the current squared:

P_{loss} = I^2 R

For a given transmitted power $P = V I$ , doubling $V$ halves $I$ and reduces $P_{loss}$ by a factor of $4$ . This is why electricity is transmitted at very high voltage (typically 132 kV to 500 kV in Australia).

The chain is: generator (10-25 kV) -> step-up transformer (up to 500 kV) -> transmission lines -> step-down transformer (substation, 11-66 kV) -> distribution transformer (240 V) -> household.

Only AC can use transformers, which is why AC won the "war of currents" against DC for grid distribution.

Voltage at the load

The transmission cables form a voltage divider with the load. Voltage delivered to the load:

V_{load} = V_{transmitted} - I R_{cable}

At low transmission voltage, $I$ is large and the cable voltage drop ( $IR$ ) becomes a significant fraction of the supply. At high transmission voltage, $I$ is small and the load receives nearly the full voltage.

Worked example

A 1.0 MW generator outputs at 25 kV. A step-up transformer raises this to 250 kV for transmission along cables of total resistance 4.0 ohm. A step-down transformer at the city end converts the voltage to 11 kV. Find the cable current, the cable power loss, and the percentage of power lost. Assume ideal transformers.

Transmission current: $I = P / V = 1.0 \times 10^6 / 2.5 \times 10^5 = 4.0$ A.

Cable loss: $P_{loss} = I^2 R = 4.0^2 \times 4.0 = 64$ W.

Percentage lost: $64 / 10^6 = 6.4 \times 10^{-3}$ percent (essentially zero).

If the same 1.0 MW were transmitted at 25 kV (no step-up), $I = 40$ A and $P_{loss} = 40^2 \times 4.0 = 6400$ W (0.64 percent). Stepping up by a factor of 10 reduced losses by a factor of 100.

Try it: Transformer calculator - enter primary/secondary turns and one of $V$ or $I$ and get the other side, plus power-loss comparisons.

Common traps

Confusing slip rings and split-ring commutators: Slip rings (continuous) give AC. A split-ring commutator (two halves) gives DC.
Plugging peak values into AC power formulas: $P = V I$ uses RMS values. Using peak values overestimates power by a factor of 2.
Trying to step up DC with a transformer: Transformers need a changing flux. Constant DC produces no induced EMF in the secondary.
Forgetting the inverse relationship of current to turns: When voltage steps up, current steps down (energy is conserved). $V_1 / V_2 = I_2 / I_1$ .
Calculating losses with $V^2 / R$ using the transmitted voltage: $P_{loss} = I^2 R$ uses the cable current and cable resistance, not the transmitted voltage. The transmitted voltage drops mostly across the load, not the cables.

Examples in context

Example 1. Loy Yang to Melbourne CBD transmission via step-up transformer. Loy Yang A generators produce electricity at $23$ kV, then step up to $500$ kV via on-site transformers for transmission to Melbourne. Turns ratio is $N_2/N_1 = 500/23 = 21.7$ , so a generator-side current of $10\,000$ A becomes $10\,000/21.7 = 461$ A on the line. Power loss in a $200$ km, $0.05$ $\Omega$ km $^{-1}$ line is $I^2 R = 461^2 \times 10 = 2.13$ MW per circuit. At $23$ kV the same line current would be $1.96 \times 10^7$ W, illustrating why transmission uses very high voltage.

Example 2. Grid battery inverter and step-down to distribution (simplified model). Take a grid battery such as the Hornsdale Power Reserve (owned by Neoen) and, for this made-up model, suppose it outputs $1500$ V DC, inverted to $33$ kV three-phase AC, then stepped down for distribution at $11$ kV via transformers. Turns ratio for the step-down stage is $N_1/N_2 = 33/11 = 3.0$ . Peak voltage in the $11$ kV system is $V_{\rm peak} = V_{\rm RMS} \times \sqrt{2} = 11\,000 \times 1.414 = 15\,560$ V. With distribution-line resistance of $0.1$ $\Omega$ km $^{-1}$ over $5$ km feeders, and currents around $300$ A, losses are $I^2 R = 9 \times 10^4 \times 0.5 = 45$ kW, or about $0.15\%$ of the $30$ MW load.

Try this

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

Cue. $V_1/V_2 = N_1/N_2$ ; $I_1/I_2 = N_2/N_1$ . Power $V_1 I_1 = V_2 I_2$ in ideal case.

Q2. A transformer steps $240$ V down to $12$ V. (a) Calculate the turns ratio. (b) Find the primary current when the secondary delivers $5.0$ A. Assume ideal. [4 marks]

Cue. (a) $N_1/N_2 = 240/12 = 20$ . (b) $I_1 = I_2 (N_2/N_1) = 5 \times 1/20 = 0.25$ A.

Q3. Refer to Loy Yang transmission. (a) Calculate the turns ratio for stepping $23$ kV to $500$ kV. (b) Determine the current on the line if generator output is $10\,000$ A. (c) Explain why transmitting at $500$ kV reduces resistive losses. [2+2+3 marks]

Cue. (a) $21.7$ . (b) $461$ A. (c) For fixed power, raising $V$ lowers $I$ ; losses $I^2 R$ fall as the square of current.

Exam-style questions

Questions in the style of VCAA 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.

Practice4 marks

A power station produces 500 kW at 4000 V. The power is to be transmitted along cables of total resistance 8.0 ohm. (a) Calculate the power lost in the cables if the power is transmitted at 4000 V. (b) Calculate the power lost if a step-up transformer first raises the voltage to 80 000 V.

Show worked answer →

(a) At 4000 V the current is $I = P/V = 500\,000 / 4000 = 125$ A.

Power lost in the cables: $P_{loss} = I^2 R = 125^2 \times 8.0 = 15\,625 \times 8.0 = 1.25 \times 10^5$ W = 125 kW.

That is 25 percent of the generated power lost as heat.

(b) At 80 000 V the current is $I = 500\,000 / 80\,000 = 6.25$ A.

Power lost in the cables: $P_{loss} = 6.25^2 \times 8.0 = 39.06 \times 8.0 = 313$ W.

Stepping up the voltage by a factor of 20 reduces the current by a factor of 20 and the power loss by a factor of $20^2 = 400$ .

Markers reward the $P = VI$ step, the $I^2 R$ loss formula, and the explicit factor-of-400 reduction.

Practice3 marks

An AC voltage has a peak value of 340 V. Calculate the RMS voltage. Explain why the RMS value, rather than the peak value, is quoted for household electricity.

Show worked answer →

$V_{RMS} = \frac{V_{peak}}{\sqrt{2}} = \frac{340}{\sqrt{2}} = 240$ V. (This is the standard Australian mains voltage.)

The RMS value is the equivalent DC voltage that would deliver the same average power to a resistor. Power in a resistor scales as $V^2$ , and the average of $V^2$ over a sinusoidal cycle is $V_{peak}^2 / 2$ . Quoting RMS lets engineers use the same $P = V I$ and $P = V^2 / R$ relationships for AC as for DC without correction factors.

Markers reward the numerical conversion, the link between RMS and average power, and the equivalent-DC interpretation.