AC and DC generators, RMS values and AC transmission: HSC Physics Module 6

An AC generator uses slip rings (two continuous, separate rings, one connection per end of the coil), which let each end of the coil stay connected to the same external terminal throughout the rotation. This produces a smooth sinusoidal output that alternates sign every half-turn.

A DC generator uses a split-ring commutator (a single ring split into two halves), which swaps which end of the coil connects to which external terminal every half-turn, at the same instant the induced emf changes sign. This produces a output that is always the same sign (unipolar) but still varies in size - a "humped" rectified waveform, not steady DC.

Marks: one for correctly naming both components (slip rings vs split-ring commutator), one for correctly linking each to its output waveform (smooth sinusoid vs rectified humps).

The peak emf of a rotating coil generator is $\varepsilon_0 = NBA\omega$.

$\varepsilon_0 = NBA\omega = 200 \times 0.25 \times 0.040 \times 50$

$\varepsilon_0 = 200 \times 0.25 \times 0.040 \times 50 = 100\ \text{V}$.

Marks: one for correctly quoting $\varepsilon_0 = NBA\omega$, one for substituting all four values correctly, one for the final answer $100\ \text{V}$ with the unit.

$V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}} = \dfrac{340}{\sqrt{2}} = 240\ \text{V}$.

RMS voltage is quoted because it is the value of a steady DC supply that would deliver the same average power to a resistive load as the alternating supply, so appliance ratings can be compared directly with DC-style figures.

Marks: one for $V_{\text{rms}} = 240\ \text{V}$ correctly computed from $V_0/\sqrt{2}$, one for the equivalent-average-power explanation of why RMS is used.

(a) Angular velocity

$\omega = 2\pi f = 2\pi \times 50 = 314\ \text{rad s}^{-1}$ (to 3 s.f.).

(b) Peak emf

$\varepsilon_0 = NBA\omega = 150 \times 0.40 \times 0.025 \times 314 = 471\ \text{V}$.

(c) RMS emf

$\varepsilon_{\text{rms}} = \dfrac{\varepsilon_0}{\sqrt{2}} = \dfrac{471}{\sqrt{2}} = 333\ \text{V}$.

Marks: one for $\omega = 2\pi f = 314\ \text{rad s}^{-1}$, one for the peak-emf formula $\varepsilon_0 = NBA\omega$ with correct substitution, one for $\varepsilon_0 = 471\ \text{V}$, one for correctly converting to $\varepsilon_{\text{rms}} = 333\ \text{V}$ using $\varepsilon_0/\sqrt{2}$.

(a) Reading the graph, the peak emf is $\varepsilon_0 = 20\ \text{V}$ and the curve completes one full cycle in a period of $T = 0.020\ \text{s}$ ($20\ \text{ms}$).

(b) Frequency. $f = \dfrac{1}{T} = \dfrac{1}{0.020} = 50\ \text{Hz}$.

(c) RMS emf. $\varepsilon_{\text{rms}} = \dfrac{\varepsilon_0}{\sqrt{2}} = \dfrac{20}{\sqrt{2}} = 14\ \text{V}$.

Marks: one for correctly reading $\varepsilon_0 = 20\ \text{V}$ and $T = 0.020\ \text{s}$ from the graph, one for $f = 1/T = 50\ \text{Hz}$, one for the RMS formula $\varepsilon_0/\sqrt{2}$, one for $\varepsilon_{\text{rms}} = 14\ \text{V}$.

Power lost in the line is $P_{\text{loss}} = I^2 R$.

(a) $P_{\text{loss}} = (1000)^2 \times 5.0 = 1.0 \times 10^6 \times 5.0 = 5.0 \times 10^6\ \text{W}$ (five times the power being delivered - all of it and more is wasted).

(b) $P_{\text{loss}} = (100)^2 \times 5.0 = 1.0 \times 10^4 \times 5.0 = 5.0 \times 10^4\ \text{W}$.

Principle. Reducing the current by a factor of $10$ (by stepping up the voltage with a transformer, since $P = VI$ is fixed) reduces the resistive loss by a factor of $10^2 = 100$. This is why transmission lines run at very high voltage and low current.

Marks: one for the formula $P_{\text{loss}} = I^2R$, one for each correctly computed loss (parts a and b), one for stating the $I^2$ (quadratic) scaling as the reason low current is essential.

Band-6 plan. (1) Describe the shared induction principle (Faraday's law, a rotating coil in a field). (2) Contrast the slip-ring and split-ring-commutator structures and derive why each produces its waveform. (3) State the transformer-transmission argument ($P_{\text{loss}} = I^2R$, step up $V$ to cut $I$). (4) Weigh AC's advantage against DC's limitation (no easy AC-style voltage step-up for DC without electronics) and reach a judgement.

Model answer. Both generator types work by the same principle: a coil of $N$ turns and area $A$ rotates at angular velocity $\omega$ in a uniform magnetic field $B$, so the flux through it varies as $\Phi = BA\cos(\omega t)$ and Faraday's law gives an induced emf $\varepsilon = NBA\omega\sin(\omega t)$, which alternates sign every half-rotation regardless of generator type.

The two designs differ only in how this alternating emf is delivered to the external circuit. An AC generator uses slip rings: two separate, continuous conducting rings, each permanently connected to one end of the coil and each in constant sliding contact with its own brush. Because the connection never changes, the external circuit sees exactly the coil's alternating emf, and the output is a smooth sine wave, $\varepsilon = \varepsilon_0\sin(\omega t)$. A DC generator instead uses a split-ring commutator: a single ring split into two half-ring segments, each wired to one end of the coil. As the coil turns, the segments swap which brush they touch at the exact instant the coil's emf reverses sign, so the external circuit terminal that was positive stays positive. The output is therefore always one sign, but its size still rises and falls sinusoidally between rotations - a rectified, "humped" waveform, not steady DC.

This structural difference is why AC (not DC) generation paired naturally with transformers to win electricity transmission. A transformer works only on a changing flux, so it requires an alternating supply; it can step an AC voltage up or down via $V_s/V_p = N_s/N_p$ with very little energy loss. Since transmission-line heating is $P_{\text{loss}} = I^2R$, and the power delivered is fixed by the load ($P = VI$), stepping the voltage up by a step-up transformer at the power station reduces the transmission current by the same factor, cutting $I^2R$ losses by the square of that factor, then a step-down transformer restores a safe voltage near consumers. A DC system generated at a safe, usable voltage has no equivalently simple, low-loss way to step that voltage up for transmission and back down again (this only became practical much later, with power electronics), so historically it could not exploit the $I^2R$ trick at all.

Weighing this, AC's compatibility with transformers gives it a decisive practical and economic advantage for large-scale transmission over distance, which is why national grids, including Australia's, are built on AC generation despite DC's simpler, non-reversing output being attractive for some local applications (e.g. battery charging, some electronics).

Marker's note: the top band explains the physical origin of each waveform from the ring/commutator structure (not just "AC is a sine wave, DC is not"), correctly states and uses $P_{\text{loss}} = I^2R$ with the transformer voltage-current trade-off, and reaches an explicit evaluative judgement about why AC became standard. A response that lists facts about both generator types without the transformer-and-transmission argument caps in the middle band.

Band-6 plan. (1) Calculate the transmission current and $I^2R$ loss at $22\ \text{kV}$. (2) Calculate the transmission current and $I^2R$ loss at $220\ \text{kV}$. (3) Compare the loss as a proportion of power sent. (4) Evaluate: recommend the higher voltage, noting the trade-off (transformer cost, insulation, safety) that is outweighed by the loss saving.

Model answer. At $22\ \text{kV}$, the transmission current is $I = \dfrac{P}{V} = \dfrac{4.0 \times 10^7}{2.2 \times 10^4} = 1.8 \times 10^3\ \text{A}$. The line loss is $P_{\text{loss}} = I^2R = (1.8 \times 10^3)^2 \times 8.0 = 2.6 \times 10^7\ \text{W}$, which is $65\%$ of the generated power - more than half the electricity would be wasted as heat in the wires before reaching the city.

Stepping up to $220\ \text{kV}$ (a factor of $10$) with a transformer, the current becomes $I = \dfrac{4.0 \times 10^7}{2.2 \times 10^5} = 1.8 \times 10^2\ \text{A}$, one-tenth of the previous current. The loss is now $P_{\text{loss}} = I^2R = (1.8 \times 10^2)^2 \times 8.0 = 2.6 \times 10^5\ \text{W}$, only $0.65\%$ of the generated power - a reduction in loss by a factor of $100$, consistent with loss scaling as $I^2$ for a $10$-fold drop in current.

This is only possible because the supply is AC: a step-up transformer at the station raises $V$ and lowers $I$ for the transmission line, and a step-down transformer near the city restores a safe local voltage, both with very little energy loss in the transformers themselves. Although high-voltage transmission requires more expensive transformers, taller towers and better insulation to manage the higher voltage safely, this cost is far outweighed by recovering roughly $2.6 \times 10^7\ \text{W}$ of power that would otherwise be lost as heat, in addition to reducing the required conductor thickness for a given loss target.

Marker's note: the top band computes both currents and both losses correctly, expresses the loss as a proportion of power sent (making the $65\%$ versus $0.65\%$ contrast explicit), and gives a genuine evaluation (a recommendation weighed against the real cost of high-voltage infrastructure), not just "higher voltage is better."