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How do generators produce alternating voltage, and how do transformers change voltage for efficient transmission?

Explain the operation of an AC generator and apply the transformer equation to step voltage up or down.

How a rotating coil produces alternating EMF in a generator, how transformers use mutual induction to change voltage, the turns-ratio equation, and the role in power transmission.

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The AC generator
  3. The transformer
  4. Power transmission
  5. How SACE assesses this

What this dot point is asking

You need to explain how an AC generator produces alternating voltage and apply the transformer turns-ratio equation.

The AC generator

A generator is the reverse of a motor: instead of using a current to produce motion, it uses motion to produce a current.

A coil is rotated in a magnetic field. As it turns, the flux through it changes continuously (because the angle θ\theta in Φ=BAcosθ\Phi = BA\cos\theta changes). By Faraday's law this changing flux induces an EMF.

A generator uses slip rings (not a split-ring commutator), which keep the connection to each end of the coil fixed, allowing the output to alternate naturally.

The transformer

A transformer changes the size of an alternating voltage using mutual induction. Two coils - primary and secondary - are wound on a shared iron core.

  • AC in the primary creates a continuously changing magnetic flux in the core.
  • The core channels this flux through the secondary coil.
  • The changing flux induces an alternating EMF in the secondary (Faraday's law).

If the secondary has more turns it is a step-up transformer (higher voltage, lower current); fewer turns makes a step-down transformer (lower voltage, higher current).

A transformer only works with AC. A steady DC current produces no changing flux, so no EMF is induced in the secondary.

Power transmission

Power lost in transmission lines is Ploss=I2RP_{\text{loss}} = I^2 R. To minimise this loss, power is sent at very high voltage and therefore low current (since P=VIP = VI). Halving the current quarters the line loss. Step-up transformers raise the voltage at the power station for transmission, and step-down transformers reduce it to safe levels for homes. This is the main reason the grid uses AC: it can be transformed, whereas DC cannot be transformed simply.

How SACE assesses this

SACE Stage 2 transformer questions are direct substitutions into VpVs=NpNs\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s} to find a missing voltage or turns count, often followed by a current calculation using VpIp=VsIsV_p I_p = V_s I_s for an ideal transformer. The reliable sanity check is the direction of the change: a step-up transformer (more secondary turns, higher voltage) draws a higher primary current, while a step-down transformer does the reverse, because power is conserved. Generator questions are usually qualitative, asking you to explain why the output is alternating and where in the rotation the EMF peaks. State that the EMF is greatest when the coil is parallel to the field (flux changing fastest) and zero when perpendicular, and name slip rings as the component that delivers an alternating output.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20232 marksA laptop charger contains a transformer that converts 220 V220\ \text{V} to 11 V11\ \text{V}. The input coil has 30003000 turns. Determine the number of turns in the output coil.
Show worked answer →

Use the transformer equation, VpVs=NpNs\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}.

Rearrange for the output turns: Ns=NpVsVpN_s = N_p\dfrac{V_s}{V_p}.

Ns=3000×11220=3000×0.0500=150 turns.N_s = 3000\times\dfrac{11}{220} = 3000\times0.0500 = 150\ \text{turns}.

1 mark for the rearrangement, 1 mark for the answer of 150150 turns. The voltage is stepped down, so the output coil must have fewer turns, a useful sanity check.

SACE 20243 marksA transformer in a microwave oven converts 220 V220\ \text{V} to 2400 V2400\ \text{V}. The output coil contains 720720 loops and draws an output current of 0.30 A0.30\ \text{A}. (a) Determine the number of loops in the input coil. (b) Assuming the transformer is ideal, calculate the input current.
Show worked answer →

(a) Apply VpVs=NpNs\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}, rearranged for the input turns Np=NsVpVsN_p = N_s\dfrac{V_p}{V_s}:

Np=720×2202400=66 turns.N_p = 720\times\dfrac{220}{2400} = 66\ \text{turns}.
(2 marks)

(b) For an ideal transformer power is conserved, VpIp=VsIsV_p I_p = V_s I_s, so

Ip=VsIsVp=(2400)(0.30)220=3.3 A.I_p = \dfrac{V_s I_s}{V_p} = \dfrac{(2400)(0.30)}{220} = 3.3\ \text{A}.
(1 mark) This is a step-up transformer (more output turns, higher output voltage), so the input current exceeds the output current, consistent with conserved power.

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