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 $P_{loss} = I^2 R$

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 $\omega$:

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:

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

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):

and energy conservation $P_1 = P_2$ gives:

So the full ideal-transformer relationship is:

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:

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:

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.

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. Hornsdale battery inverter and step-down to distribution. Hornsdale Power Reserve 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.