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Inquiry Question 4: How are electric and magnetic fields applied in electrical generation, transmission and use?

Analyse the operation of ideal and real transformers, including the turns ratios V_s/V_p = N_s/N_p and I_p/I_s = N_s/N_p, energy losses, and the role of step-up and step-down transformers in AC power transmission

A focused answer to the HSC Physics Module 6 dot point on transformers. Ideal voltage and current ratios, power conservation V_p I_p = V_s I_s, the four energy losses in real transformers, and why high-voltage AC transmission minimises line losses.

Generated by Claude Opus 4.814 min answerUpdated 2026-07-01

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

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What this dot point is asking
The answer
Examples in context

What this dot point is asking

NESA wants you to derive and apply the ideal-transformer voltage and current ratios from Faraday's law, recognise that an ideal transformer conserves power, identify the four main loss mechanisms in real transformers and how the design mitigates them, and explain why AC transmission relies on stepping voltage up for transmission and down for distribution.

The answer

How a transformer works

A transformer is two coils (the primary and secondary) wound on the same ferromagnetic core. An alternating current in the primary produces a changing 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):

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

Dividing:

\boxed{\frac{V_s}{V_p} = \frac{N_s}{N_p}}

The voltage ratio equals the turns ratio.

The must-know equations

Voltage (turns) ratio: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ .

Current ratio (ideal transformer, power conserved): $\dfrac{I_p}{I_s} = \dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}$ , so $V_p I_p = V_s I_s$ .

Transmission-line loss: $P_{\text{loss}} = I^2 R_{\text{line}}$ - it depends on current squared, which is why raising $V$ (and so lowering $I$ ) at a fixed transmitted power $P = VI$ cuts losses.

A step-up transformer has $N_s > N_p$ (raises $V$ , lowers $I$ ); a step-down transformer has $N_s < N_p$ (lowers $V$ , raises $I$ ). Transformers only work on AC, because they need $d\Phi/dt \neq 0$ .

Power conservation and the current ratio

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

V_p I_p = V_s I_s

Combining with the voltage ratio:

\boxed{\frac{I_p}{I_s} = \frac{N_s}{N_p} = \frac{V_s}{V_p}}

Currents are in the inverse ratio to voltages. A step-up transformer ( $N_s > N_p$ ) raises the voltage and lowers the 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 DC primary current produces a constant flux, so the secondary EMF is zero (except during the brief switch-on transient). AC, by changing direction many times per second, produces the continuous flux change required.

The four losses in a real transformer

Loss	Cause	Mitigation
Resistive (copper, $I^2 R$ )	Resistance of the windings	Thick, low-resistance wire; oil cooling for large units
Eddy currents	Induced currents circulating in the iron core	Laminated core (insulated thin sheets)
Hysteresis	Energy dissipated re-magnetising the core each cycle	Soft-magnetic alloys (silicon steel) with a narrow B-H loop
Flux leakage	Some primary flux fails to link the secondary	Closed-loop laminated core; interleaved windings

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

Step-up and step-down in AC transmission

Transmitting electrical power $P = VI$ over a long line of resistance $R_{\text{line}}$ wastes power as $P_{\text{loss}} = I^2 R_{\text{line}}$ . 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:

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

Model 1 - transmission line saving. A small power station delivers $1.0$ MW to a town through a transmission line of total resistance $5.0\,\Omega$ . 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, so $P_{\text{loss}} = I^2 R = (1000)^2 \times 5.0 = 5.0 \times 10^6\ \text{W} = 5.0$ MW. The line cannot even deliver $1.0$ MW: the losses exceed the transmitted power.

At $V = 100$ kV: $I = 10^6 / 10^5 = 10$ A, so $P_{\text{loss}} = (10)^2 \times 5.0 = 500\ \text{W}$ .

The loss drops from $5.0$ MW to $500$ W, a factor of $10\,000$ - exactly the square of the voltage ratio ( $100^2 = 10\,000$ ), illustrating the $V^2$ dependence of transmission efficiency.

Marker's note: show the current calculation at each voltage explicitly before squaring it into $P_{\text{loss}} = I^2R$ ; jumping straight to a "loss ratio" without stating both currents loses method marks.

Model 2 - turns ratio and input power. 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: $N_s/N_p = V_s/V_p = 5000/240 = 20.8$ (to 3 s.f.).

Primary current: $I_p = (V_s/V_p)I_s = 20.8 \times 0.050 = 1.04$ 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, confirming power is conserved for an ideal transformer).

Marker's note: checking $V_p I_p = V_s I_s$ at the end is a fast self-check that the current calculation is correct - use it whenever time allows.

Common traps

Inverting the turns ratio: $V_s / V_p = N_s / N_p$ , but $I_p / I_s = N_s / N_p$ . The voltage ratio and the current ratio are inverses of each other.
Trying to use a transformer on DC: A common error in design questions; the device gives no output (no induced EMF) and may overheat. Always state the AC requirement when explaining transformer operation.
Confusing the four losses: Eddy-current losses and hysteresis losses are both in the core; resistive losses are in the windings; flux leakage is a geometry issue. Markers expect you to distinguish them.
Forgetting why we step up voltage for transmission: It is to reduce $I^2 R$ loss, which depends on $I^2$ , not on $V$ . Higher $V$ at the same $P$ means lower $I$ .

Saying "transformers create energy." A step-up transformer increases voltage at the cost of decreasing current; total power is conserved (or reduced slightly by losses in a real transformer).

Using the wrong direction for "primary" and "secondary." The primary is whichever coil is connected to the source; the secondary is whichever is connected to the load. A step-up transformer used backwards becomes a step-down transformer.

Exam technique

State the relationship before you substitute: write " $V_s/V_p = N_s/N_p$ " or " $V_p I_p = V_s I_s$ " as a line on its own before putting numbers in - a stated relationship earns method marks even if the arithmetic slips.

Watch which ratio you need. $V_s/V_p = N_s/N_p$ but $I_p/I_s = N_s/N_p$ (not $I_s/I_p$ ): the current ratio is the inverse of the voltage ratio, so check your answer's sign makes physical sense (step-up transformers draw more current on the primary side than they deliver on the secondary side).

For a graph question, read the axis units carefully (turns vs volts, or amps vs volts) before computing a gradient, and always give the gradient a unit - it usually corresponds to a named physical quantity ( $V_p/N_p$ , or a resistance in a $V$ - $I$ graph).

For "outline/describe" questions on losses, name the loss AND its physical cause; for "analyse/evaluate" extended response, do not just list the losses - link them back to how closely the device approaches $V_pI_p = V_sI_s$ .

In plain English

A transformer is like a set of gears on a bicycle, but for electricity instead of pedalling force. Gears trade speed for torque: a small gear driving a big one turns slower but pushes harder, while the total pedalling effort you put in still has to match the work the wheel does (ignoring friction). A transformer trades voltage for current in exactly the same way: more turns of wire on one side means more "voltage push" but less current, and fewer turns means less "voltage push" but more current - and the total electrical power going in still has to (almost) equal the power coming out, because energy cannot appear from nowhere. The two coils of wire, wrapped around the same iron core, only "talk" to each other through a changing magnetic field, which is why transformers only work on the constantly alternating current, AC, from the wall or the grid - a steady, unchanging DC current produces no changing field and so no output at all. This is exactly why the electricity grid uses AC: it lets power stations push electricity down the wires at extremely high voltage and low current (which wastes very little energy as heat), then use a chain of transformers to gear it back down, step by step, to the safe 240 V that reaches your wall socket.

Examples in context

Example 1. TransGrid Liddell step-up transformer. Snowy Hydro's generator output of $V_p = 20 \text{ kV}$ feeds a step-up transformer with $N_p = 200$ primary turns. To reach the TransGrid backbone at $V_s = 330 \text{ kV}$ , the secondary needs $N_s = N_p \times V_s/V_p = 200 \times 330/20 = 3300$ turns. Power throughput is $P = 600 \text{ MW}$ , so primary current is $I_p = P/V_p = 6.0 \times 10^8 / 2.0 \times 10^4 = 30{,}000 \text{ A}$ , while secondary current is only $I_s = I_p N_p/N_s = 30{,}000 \times 200/3300 = 1818 \text{ A}$ . The high-voltage low-current secondary feeds the transmission line.

Example 2. $I^2 R$ loss saving by stepping up to 330 kV. Suppose the Liddell-to-Sydney transmission line is $300 \text{ km}$ of aluminium conductor with total resistance $R = 6.0 \Omega$ . At $20 \text{ kV}$ , delivering $600 \text{ MW}$ requires $I = 30{,}000 \text{ A}$ and the line dissipates $I^2 R = (3.0 \times 10^4)^2 \times 6.0 = 5.4 \times 10^9 \text{ W} = 5400 \text{ MW}$ - more than the power being sent. At $330 \text{ kV}$ , $I = 1818 \text{ A}$ , so $I^2 R = 1818^2 \times 6.0 = 1.98 \times 10^7 \text{ W} = 19.8 \text{ MW}$ , or just $3.3\%$ loss. Stepping up the voltage by a factor of $16.5$ cuts the loss by $16.5^2 = 272$ .

Exam-style practice questions

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

2022 HSC4 marksAn ideal transformer steps 240 V AC down to 12 V to power a 36 W lamp. Calculate the turns ratio, the primary and secondary currents, and explain why the transformer cannot be used with a DC supply.

Show worked answer →

Turns ratio:

$\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{240}{12} = 20$ , that is, $N_p : N_s = 20 : 1$ .

Secondary current (from the lamp's rating):

$I_s = P / V_s = 36 / 12 = 3.0$ A.

Primary current (ideal transformer, power in = power out):

$V_p I_p = V_s I_s \Rightarrow I_p = (V_s / V_p) I_s = (12/240) \times 3.0 = 0.15$ A.

Equivalently, $I_p / I_s = N_s / N_p = 1/20$ , giving $I_p = 0.15$ A.

A transformer needs a changing magnetic flux in the core to induce a secondary EMF. A steady DC primary current produces a steady flux, so $d\Phi / dt = 0$ in the secondary and no EMF is induced. A DC supply would also potentially overheat or saturate the core because the primary winding's low resistance allows a large continuous current.

Markers reward both ratios with correct orientation, the power-conservation step, and a clear "needs a changing flux" reason for the AC requirement.

2018 HSC5 marksOutline the four main energy losses in a real transformer and describe one design feature used to minimise each.

Show worked answer →

Four losses and their mitigations:

Resistive (copper, I^2 R) losses in the windings. The current in each coil dissipates energy as heat in the wire's resistance. Mitigation: use thick, low-resistance copper or aluminium wire; for very large transformers, oil cooling carries heat away.
Eddy-current losses in the core. The changing flux induces circulating currents in the iron, dissipating energy as heat. Mitigation: laminate the core (thin iron sheets insulated from each other by varnish), which breaks the eddy-current paths and dramatically reduces the loss.
Hysteresis losses in the core. Each AC cycle re-magnetises the core, and energy is dissipated against the internal magnetic friction of the ferromagnet (area of the B-H loop). Mitigation: use a soft-magnetic alloy (silicon steel or grain-oriented steel) with a narrow hysteresis loop.
Flux leakage. Not all flux from the primary links the secondary; some escapes through the air. Mitigation: a closed-loop laminated iron core that guides flux around the magnetic circuit, and concentric or interleaved windings so the secondary surrounds the primary closely.

Markers reward each loss correctly identified with both cause and a specific mitigation; full marks require all four pairs.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksAn ideal transformer has

N_p = 1000

primary turns and is connected to a

240

V AC supply. It must supply

6.0

V to a low-voltage doorbell. Calculate the number of secondary turns required.

Show worked solution →

Use the turns ratio $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ , so $N_s = N_p \times \dfrac{V_s}{V_p}$ .

$N_s = 1000 \times \dfrac{6.0}{240} = 25$ turns.

Marks: one for the correctly rearranged ratio with values substituted, one for the answer $N_s = 25$ turns (an integer, with no unit needed for a turns count).

foundation2 marksState whether each of the following is a step-up or a step-down transformer, and justify each answer using the turns ratio: (a)

N_p = 300

N_s = 6000

. (b)

N_p = 6000

N_s = 300

Show worked solution →

(a) Step-up. $N_s > N_p$ ( $6000 > 300$ ), so by $V_s/V_p = N_s/N_p$ the secondary voltage is greater than the primary voltage.

(b) Step-down. $N_s < N_p$ ( $300 < 6000$ ), so the secondary voltage is less than the primary voltage.

Marks: one for correctly classifying both transformers, one for justifying each using the turns-ratio inequality (not just asserting the answer).

foundation3 marksA step-up transformer is connected to a

240

2.0

A AC supply and is ideal (lossless). Its secondary delivers

1200

V. Calculate the secondary current

I_s

Show worked solution →

An ideal transformer conserves power: $V_p I_p = V_s I_s$ .

$I_s = \dfrac{V_p I_p}{V_s} = \dfrac{240 \times 2.0}{1200} = 0.40$ A.

Marks: one for stating power conservation $V_p I_p = V_s I_s$ , one for correctly substituting and rearranging, one for the answer $I_s = 0.40$ A with the unit.

core4 marksA sub-transmission step-down transformer converts

V_p = 66\,000

V to

V_s = 11\,000

V. The secondary supplies a substation load drawing

I_s = 240

A. Calculate **(a)** the turns ratio

N_p : N_s

and **(b)** the primary current

I_p

, assuming an ideal transformer.

Show worked solution →

(a) Turns ratio. $\dfrac{N_p}{N_s} = \dfrac{V_p}{V_s} = \dfrac{66\,000}{11\,000} = 6.0$ , so $N_p : N_s = 6.0 : 1$ .

(b) Primary current. From power conservation, $I_p = \dfrac{V_s}{V_p} I_s = \dfrac{11\,000}{66\,000} \times 240 = 40.0$ A.

Marks: one for the turns-ratio formula with correct orientation, one for $N_p : N_s = 6.0 : 1$ , one for the current relation $I_p = (V_s/V_p)I_s$ , one for $I_p = 40.0$ A with the unit.

core4 marksThe figure shows secondary voltage

V_s

plotted against the number of secondary turns

N_s

for an ideal transformer with a fixed primary of

V_p = 240

V and

N_p = 500

turns. **(a)** Describe the relationship shown. **(b)** Using the points

(N_s = 250,\ V_s = 120\ \text{V})

and

(N_s = 1000,\ V_s = 480\ \text{V})

, calculate the gradient. **(c)** Show that this gradient equals

V_p / N_p

for this transformer.

Show worked solution →

(a) The graph is a straight line through the origin, so $V_s$ is directly proportional to $N_s$ ( $V_s \propto N_s$ ), consistent with $V_s = (V_p/N_p) N_s$ for fixed $V_p$ and $N_p$ .

(b) Gradient $= \dfrac{\Delta V_s}{\Delta N_s} = \dfrac{480 - 120}{1000 - 250} = \dfrac{360}{750} = 0.48\ \text{V per turn}$ .

(c) $\dfrac{V_p}{N_p} = \dfrac{240}{500} = 0.48\ \text{V per turn}$ , which matches the measured gradient, confirming $V_s = (V_p/N_p)N_s$ .

Marks: one for identifying direct proportionality through the origin, one for a correctly computed gradient with the unit V per turn, one for computing $V_p/N_p$ , one for the explicit comparison showing the two values agree.

core3 marksA transformer core is re-designed from solid iron to thin, varnish-insulated laminated sheets, while keeping the windings unchanged. Identify which energy loss this change targets, explain the physical mechanism it prevents, and state one loss this change does NOT reduce.

Show worked solution →

Loss targeted: Eddy-current losses in the core.
Mechanism prevented: The changing magnetic flux in a solid iron core induces circulating (eddy) currents within the iron itself, which dissipate energy as heat through the core's own resistance. Laminating the core into thin sheets, each insulated from its neighbours, breaks up these current loops into much smaller, higher-resistance paths, greatly reducing the induced eddy currents and the heat they generate.
Loss not reduced: Resistive ( $I^2R$ ) losses in the windings, because lamination changes only the core, not the wire; hysteresis losses would also be unchanged unless the core material itself is also changed to a soft-magnetic alloy.

Marks: one for correctly naming eddy-current loss, one for explaining the induced circulating-current mechanism and how thin insulated sheets break the current paths, one for correctly identifying a loss the change does not address (winding resistance or hysteresis).

exam6 marksAnalyse the role of transformers in minimising energy loss during AC electricity transmission from a power station to a town, referring to relevant equations.

Show worked solution →

Band-6 plan. (1) State the transmission-loss equation $P_{\text{loss}} = I^2 R_{\text{line}}$ and identify that loss depends on current, not voltage. (2) Explain how a step-up transformer reduces current for a given transmitted power via $V_p I_p = V_s I_s$ . (3) Explain the step-down stage is needed for safe delivery. (4) Quantify with a worked comparison to show the scale of the saving. (5) Conclude on why transformers (not any other device) make this possible.

Model answer. The power dissipated as heat in a transmission line is $P_{\text{loss}} = I^2 R_{\text{line}}$ , which depends on the square of the current carried, not on the transmission voltage directly. Since the power to be delivered is $P = VI$ , for a fixed $P$ a higher transmission voltage $V$ means a proportionally smaller current $I$ , and because the loss scales as $I^2$ , even a modest increase in voltage produces a large reduction in loss.

A step-up transformer at the power station raises the generator's output voltage (typically tens of kV) to a much higher transmission voltage (up to $500$ kV on the Australian grid) before the line. Because an ideal transformer conserves power, $V_p I_p = V_s I_s$ , so stepping the voltage up by a large factor steps the current down by the same factor, directly cutting $I^2 R_{\text{line}}$ losses.

At the receiving end, this high voltage is far too dangerous and impractical for factories and homes, so one or more step-down transformers progressively reduce it (through sub-transmission and distribution levels) to the standard $240$ V supply voltage, restoring a safe, usable current at the point of use.

Because transformers only work with a changing flux, this whole scheme requires AC rather than DC: only AC gives the continuously changing primary current needed to induce a secondary EMF, which is why the AC grid (not a DC one) can exploit cheap, efficient voltage transformation.

Marker's note: the top band names and uses both key equations ( $P_{\text{loss}} = I^2 R$ and $V_p I_p = V_s I_s$ ), explicitly links "step up voltage" to "step down current" to "step down loss" as a causal chain, and explains both the step-up and the step-down stage. A response that only says "high voltage reduces loss" without the current/equation link caps in the middle band.

exam7 marksEvaluate the extent to which a real transformer can approach the ideal-transformer relationships

V_s/V_p = N_s/N_p

and

V_p I_p = V_s I_s

, referring to specific loss mechanisms and design features used to minimise them.

Show worked solution →

Band-6 plan. Thesis: real transformers can approach the ideal case very closely but never reach it. (1) State what "ideal" assumes. (2) Go through each real loss mechanism (resistive, eddy, hysteresis, flux leakage) with its physical cause and its design mitigation. (3) Note the practical outcome (efficiencies often above 99% for large transformers). (4) Reach an evaluative judgement rather than just listing losses.

Model answer. The ideal-transformer relationships assume every unit of flux produced by the primary links the secondary, and that no energy is dissipated anywhere in the device, so that $V_s/V_p = N_s/N_p$ exactly and $V_p I_p = V_s I_s$ with no loss term. Real transformers depart from this in four distinct ways.

Resistive ( $I^2R$ ) losses occur because the copper or aluminium windings have non-zero resistance, dissipating heat as current flows; this is reduced, but never eliminated, by using thick, low-resistance conductors and, in large transformers, oil cooling to remove the heat. Eddy-current losses arise because the changing flux induces circulating currents within the iron core itself; laminating the core into thin, varnish-insulated sheets confines these currents to small loops with high resistance, sharply cutting (but not abolishing) this loss. Hysteresis losses occur because each AC cycle re-magnetises the domains of the core material, dissipating energy proportional to the area of the material's $B$ - $H$ loop; using a soft-magnetic alloy such as grain-oriented silicon steel narrows this loop and reduces the loss. Finally, flux leakage means not all of the primary's flux threads the secondary coil; a closed-loop laminated core and tightly interleaved or concentric windings minimise, but do not perfectly eliminate, this leakage.

Because these four mechanisms can each be engineered down to a small fraction of the throughput power, well-designed large grid transformers achieve efficiencies above 99%, meaning $V_p I_p \approx V_s I_s$ is an excellent approximation in practice, even though it is never exactly true. Smaller transformers, with proportionally larger winding resistance and core losses relative to their throughput, typically only reach around 95% efficiency.

Overall, a real transformer can approach the ideal relationships extremely closely at large scale, which is precisely why the ideal-transformer equations remain the standard tool for grid calculations, but it can never satisfy them exactly, because every one of the four loss mechanisms is a physical consequence of using real conductors and real magnetic materials rather than idealised ones.

Marker's note: the top band treats all four losses with cause AND mitigation (not a bare list), connects the cumulative effect back to the ideal equations (how close, not just "some loss occurs"), and closes with an explicit evaluative judgement (very close at scale, never exact). A response that states the four losses without linking back to $V_p I_p = V_s I_s$ caps in the middle band.