Year 12 Physics Module 6 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 17 of 21

Transformers, Theory and Operation

Transformers use mutual induction to change AC voltage and current. This lesson separates the ideal turns-ratio and equal-power model from real efficiency, loss mechanisms and the steady-DC limitation.

Today's hook: The 1895 Niagara Falls power station (Tesla/Westinghouse) used step-up transformers to raise voltage from 2,200 V to 22,000 V for transmission. If it had transmitted at the generator voltage of 2,200 V instead, the current would have been 10× higher. Since power loss = I²R, how many times greater would the transmission losses have been, and why does this make the transformer the most important device in M6?

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Learn0 of 6 steps complete

Retrieve changing flux and separate the ideal transformer from the real device.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Use the available foundations worksheet for guided practice on this lesson.

Build Foundations & guided practice

Before you read, predict

A transformer has a primary coil with 100 turns and a secondary coil with 200 turns. The primary is connected to 10 V AC.

Do you think the secondary voltage will be higher or lower than 10 V?
By what factor do you think it will change? (Hint: consider the ratio of turns.)
Why do you think transformers only work with AC, not DC?

Warm-up, a transformer works by using electromagnetic induction. For induction to occur, what must happen to the magnetic flux through the secondary coil?

Learning Intentions

goals

Know, Transformer Equations

Voltage ratio equals turns ratio: $V_p/V_s = N_p/N_s$
For an ideal transformer: $P_p = P_s$, so $V_p I_p = V_s I_s$
Step-up: $N_s > N_p$, $V_s > V_p$, $I_s < I_p$

Understand, Why AC Only

AC creates a changing magnetic field in the core
Changing flux induces emf in the secondary coil
Steady DC produces constant flux after switching → no sustained secondary emf

Can Do, Calculate and Design

Calculate secondary voltage and current given turns ratio
Determine whether a transformer is step-up or step-down
Explain energy losses and why transformers are not 100% efficient

Scan these before reading

vocab

TransformerA device that changes AC voltage using electromagnetic induction between two coils on a shared core.

Primary coilThe input coil connected to the AC power source.

Secondary coilThe output coil where the induced voltage is delivered.

Step-up transformer$N_s > N_p$, increases voltage, decreases current.

Step-down transformer$N_s < N_p$, decreases voltage, increases current.

Laminated iron coreLayers of insulated iron that concentrate magnetic flux and reduce eddy current losses.

Cross-lesson links: L14 established induction direction and conservation. L15 applies changing flux to two coupled coils; L20 owns the broader grid-loss and AC/HVDC system trade-off.

Core Content

Explain mutual induction, turns ratio and the steady-DC switching condition.

How Transformers Work

+5 XP

Induction between coupled coils

Connect a steady 12 V DC supply to a transformer's primary: the secondary may show a brief switching transient, but after the current settles its emf returns to zero. With 12 V AC, the changing current continuously changes the core flux and induces a sustained alternating emf in the secondary. A transformer therefore requires changing flux; AC supplies it continuously, while switched or pulsed DC can also induce emf during changes.

The primary coil is connected to an AC voltage source. The alternating current creates a continuously changing magnetic field.
The iron core channels this magnetic field through the secondary coil. Because the field is changing, the flux through the secondary coil changes too.
By Faraday's Law, the changing flux induces an alternating emf in the secondary coil.

The ratio of voltages equals the ratio of turns because both coils share the same changing flux:

Transformer equations

$\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$ , voltage–turns ratio

$\dfrac{I_p}{I_s} = \dfrac{N_s}{N_p}$ , current–turns ratio (inverse)

$P_p = P_s \;\Rightarrow\; V_p I_p = V_s I_s$ , power conserved (ideal)

Where $V_p$, $V_s$ = primary and secondary voltage (V); $I_p$, $I_s$ = primary and secondary current (A); $N_p$, $N_s$ = primary and secondary turns.

Type	Turns	Voltage	Current	Example use
Step-up	$N_s > N_p$	$V_s > V_p$	$I_s < I_p$	Power transmission
Step-down	$N_s < N_p$	$V_s < V_p$	$I_s > I_p$	Phone chargers

Stop & Check

A transformer has 400 primary turns and 800 secondary turns. The primary voltage is 230 V AC. Calculate the secondary voltage and explain whether this is a step-up or step-down transformer.

Transformer: $V_p/V_s = N_p/N_s$ (voltage–turns ratio); $I_p/I_s = N_s/N_p$ (current ratio, inverse); $V_p I_p = V_s I_s$ (ideal power conservation). Step-up: $N_s > N_p$ → higher $V_s$, lower $I_s$. Step-down: $N_s < N_p$ → lower $V_s$, higher $I_s$.

Pause, copy the highlighted transformer equations and step-up/down rules into your book before moving on.

A transformer has $N_p = 100$ and $N_s = 500$ connected to $V_p = 20$ V. The secondary voltage $V_s$ is:

Why Transformers Need AC

+5 XP

The critical role of changing flux

We just saw the transformer equations $V_p/V_s = N_p/N_s$. That raises a question: why can't we use DC in a transformer, wouldn't DC still create a magnetic field in the core? This card answers it → DC creates constant (not changing) flux, so $\Delta\Phi/\Delta t = 0$ and no emf is induced.

Transformers only work with alternating current. AC in the primary coil produces a continuously changing magnetic field in the core. This changing field produces changing flux through the secondary coil, and by Faraday's Law, changing flux induces an emf.

If you connected DC to the primary:

After the switching transient, steady DC produces a constant current and therefore constant magnetic flux.
Constant field means constant flux through the secondary.
Constant flux means zero rate of change → zero induced emf.

After the transient, the secondary has no sustained induced emf. Steady DC must be switched or converted so that the flux changes before transformer action can continue.

Key insight

The transformer equation $V_p/V_s = N_p/N_s$ is derived from the fact that both coils experience the same rate of change of flux. Each turn contributes the same induced emf, so the total emf is proportional to the number of turns. With DC, the rate of change is zero, so the equation produces zero on both sides.

AC: continuously changing $B$ → changing $\Phi$ → sustained induced emf. Steady DC after switching: constant $B$ → constant $\Phi$ → $\Delta\Phi/\Delta t = 0$ → zero sustained emf. Switching or pulsed DC can induce emf while flux changes. In the ideal coupled-core model, each turn links the same flux, so emf is proportional to turns.

Add the highlighted AC vs DC transformer argument to your notes before the check below.

Connecting DC to a transformer's primary coil will produce a larger DC voltage in the secondary.

A transformer works because changing flux in the iron core induces an emf in the secondary coil.

In a step-up transformer, both voltage and current increase on the secondary side.

Use the ideal voltage, turns, current and equal-power relationships with units.

Interactive Tool, Transformers & Power Transmission Open fullscreen ↗

Use the transformer tool with 100 primary turns and 500 secondary turns. A 230 V input gives an output of:

Worked Example, Transformer Calculations

+5 XP

Apply the transformer equations to real scenarios

We just saw why transformers need AC. That raises a question: given $N_p$, $N_s$, $V_p$ and $I_p$ for a step-down transformer, how do we calculate $V_s$, $I_s$ and power? This card answers it → $V_s = V_p(N_s/N_p)$, then $I_s = V_p I_p / V_s$, then check $P_p = P_s$.

A step-down transformer has 1200 primary turns and 60 secondary turns. The primary is connected to 230 V AC and draws 0.50 A. Calculate (a) secondary voltage, (b) secondary current, and (c) power delivered.

Part (a), Secondary voltage

Given. $N_p = 1200$, $N_s = 60$, $V_p = 240$ V.
Method. Use $V_p/V_s = N_p/N_s$, rearrange for $V_s$.
Solve. $V_s = V_p \times \dfrac{N_s}{N_p} = 240 \times \dfrac{60}{1200} = 12$ V.

Part (b), Secondary current

Given. $V_p = 240$ V, $I_p = 0.50$ A, $V_s = 12$ V (from part a).
Method. For an ideal transformer: $V_p I_p = V_s I_s$.
Solve. $I_s = \dfrac{V_p I_p}{V_s} = \dfrac{240 \times 0.50}{12} = 10$ A.

Part (c), Power

Solve. $P_s = V_s I_s = 12 \times 10 = 120$ W.
Check. $P_p = V_p I_p = 240 \times 0.50 = 120$ W. Power is conserved. ✓

Step order: $V_s = V_p(N_s/N_p)$; $I_s = V_p I_p / V_s$; check $P_p = P_s$. Worked result: $V_s = 12$ V (step-down 20:1), $I_s = 10$ A, $P = 120$ W. Power must be equal on both sides for an ideal transformer.

Pause, write the highlighted step order and worked results into your book before moving on.

An ideal transformer has $V_p = 240$ V, $V_s = 12$ V, and $I_s = 5.0$ A. The primary current $I_p$ is:

Distinguish ideal equal power from real efficiency and identify each loss pathway.

Real Transformers, Why They Are Not 100% Efficient

+5 XP

Energy losses and the strategies that reduce them

We just saw that an ideal transformer conserves power ($P_p = P_s$). That raises a question: real transformers don't achieve 100% efficiency, where does the energy go? This card answers it → copper losses ($I^2R$), eddy currents, flux leakage, and hysteresis losses, each with a specific engineering fix.

An ideal transformer assumes all magnetic flux from the primary passes through the secondary and no energy is lost as heat. Real transformers deviate from this ideal. Three main energy losses reduce efficiency.

1. Resistive Heat Production (Copper Losses)

The primary and secondary coils are made of copper wire, which has resistance. When current flows, energy is dissipated as heat according to $P = I^2 R$. Thicker wire reduces resistance, but increases cost and weight.

2. Eddy Currents in the Core (Iron Losses)

The changing magnetic flux in the iron core induces swirling currents, eddy currents within the core itself. These currents dissipate energy as heat. To reduce this, the core is laminated: built from thin sheets of iron insulated from each other. This breaks up large current loops and increases resistance, cutting eddy current losses dramatically.

3. Incomplete Flux Linkage

Not all magnetic flux produced by the primary coil passes through the secondary. Some leaks into the surrounding air. Using a soft iron core channels most flux through both coils, but some leakage is inevitable. Better core design and closed magnetic circuits minimise this.

4. Hysteresis Losses

The alternating current repeatedly magnetises and demagnetises the iron core. The energy required to realign magnetic domains during each cycle is lost as heat. Using soft magnetic materials (which magnetise and demagnetise easily) reduces this loss.

HSC Tip

In extended response questions, name at least two energy losses and two strategies to reduce them. Always link the strategy to the specific loss it addresses.

Real transformer losses: (1) Copper losses: $P = I^2R$ in windings → thicker wire. (2) Eddy currents: induced loops in iron core → laminated core. (3) Flux leakage → closed soft iron core. (4) Hysteresis: repeated magnetisation/demagnetisation → soft magnetic materials. Name 2 losses + 2 fixes in the exam.

Add the highlighted four losses and their fixes to your notes before the check below.

Three of these statements about transformer energy losses are correct. Pick the odd one out.

Compare strong core coupling with the weaker air-core extension and state model limits.

Extension, Mutual Induction Without a Shared Core

+5 XP

Two separate solenoids can still couple, just weakly

We just saw that even inside a transformer, some flux leaks out of the iron core instead of reaching the secondary. That raises a question: what if there is no shared core at all, just two separate solenoids sitting near each other? This card answers it → mutual induction still happens, the changing field simply has to extend through open space to reach the second coil, which makes the coupling far weaker than a transformer's.

Take two ordinary solenoids, wind one length of wire into a coil (the primary), and place a second, completely separate coil (the secondary) a short distance away in air, no iron core joining them, no shared former, nothing connecting them electrically. Connect the primary to an AC source. A sensitive galvanometer on the secondary registers a small deflection, an EMF is being induced in a coil that is not touching, not sharing a core with, and not wired to the primary. This is mutual induction between two discrete air-core solenoids, the same underlying phenomenon as a transformer, but without anything to concentrate the field.

The changing current in the primary solenoid produces a changing magnetic field, and by the right-hand grip rule this field is strongest inside the coil but also extends (fringes) into the surrounding space.
If the secondary solenoid sits within that fringing field, some of the field lines pass through its turns, linking flux to it.
Because that linked flux is changing, Faraday's Law still applies: an emf is induced in the secondary, $\varepsilon_s = -N_s \dfrac{\Delta \Phi}{\Delta t}$.
Unlike a transformer, there is no iron core to channel the field, so only a small fraction of the primary's flux ever reaches the secondary. Most of the field simply spreads out into space and misses it entirely.

Feature	Transformer (shared iron core)	Two separate air-core solenoids
Flux path	Channelled and concentrated through both coils by the core	Simply extends through open space to reach the second coil
Coupling strength	Very strong, nearly all primary flux reaches the secondary	Weak, most field lines miss the secondary entirely
Core material	Soft iron (high relative permeability)	Air (relative permeability of about 1)
Governing law	Faraday's Law, $\varepsilon_s = -N_s\Delta\Phi/\Delta t$	The same Faraday's Law, just with much smaller linked flux

Key insight

A transformer is not a different phenomenon from two separate solenoids inducing current in each other, it is the same mutual induction, engineered to be efficient. The iron core exists purely to concentrate the flux so that almost none of it is wasted. Remove the core and separate the coils, and you are left with the weaker, general case, still real induction, just with far less of the primary's flux actually linking the secondary.

HSC Tip

If asked why two nearby coils with no shared core show a much smaller induced effect than a transformer, do not say "no induction occurs." Explain that induction still occurs by Faraday's Law, but without a core to channel the field, only a small fraction of the changing flux links the second coil, so the induced emf is much weaker.

Two separate air-core solenoids near each other still show mutual induction, a changing current in one induces an emf in the other via the changing flux linking it ($\varepsilon_s = -N_s\Delta\Phi/\Delta t$). This is weaker than a transformer because there is no shared iron core to channel/concentrate the flux, the field simply extends through space, and only a small fraction of it reaches the second coil.

Pause, copy the highlighted separate-solenoids comparison into your book before moving on.

Two separate air-core solenoids placed near each other, with no shared core, cannot show any mutual induction at all.

In a transformer, the shared iron core channels and concentrates the flux so that almost all of it reaches the secondary coil.

Two discrete solenoids with no shared core rely on the magnetic field simply extending through space to reach the second coil, giving much weaker coupling than a transformer.

AC, switched DC and ideal limits

A transformer requires changing flux. Steady DC produces no secondary emf after the switching transient, but switching or pulsed DC can change flux and induce voltage. The turns ratio is the ideal/near-ideal voltage relation; the inverse current ratio follows only when input and output powers are treated as equal. For a real transformer, use efficiency $\eta=P_{\text{out}}/P_{\text{in}}$ and account for copper, eddy-current, hysteresis and leakage losses.

Apply ideal calculations and real-loss reasoning, then consolidate the AC/DC condition.

Activity 1, Transformer Calculator

ApplyBand 3

Use the interactive above to explore and calculate

Set $N_p = 200$, $N_s = 400$, $V_p = 120$ V. Record $V_s$. Is this step-up or step-down?
Swap the turns: $N_p = 400$, $N_s = 200$. What is $V_s$ now? Explain why it changed.
Set $N_p = N_s = 300$. What is $V_s$? What is this type of transformer called?
A power station generates electricity at 25 kV. It is stepped up to 330 kV for transmission. If the primary has 1000 turns, how many turns does the secondary have?

Activity check, for a transformer with $N_p = 500$ and $N_s = 1500$, if the primary voltage is 100 V, the secondary voltage (in V) is _____.

Activity 2, Efficiency Analysis

UnderstandBand 5

Link energy losses to real-world design decisions

A student connects a 12 V battery to the primary of a transformer and expects 120 V from the secondary.

Explain why this will not work and what the student would actually measure at the secondary output. Then, for each energy loss below, state one design strategy that reduces it: (a) eddy currents, (b) copper losses, (c) hysteresis.

Connect the ideas, transformer summary

Voltage ratio: $V_p/V_s = N_p/N_s$, step-up if $N_s > N_p$, step-down if $N_s < N_p$

Power conservation: $V_p I_p = V_s I_s$, voltage up means current down

AC required: Changing flux needed; DC gives zero rate of change → zero emf

Real losses: copper losses ($I^2R$), eddy currents, flux leakage, hysteresis, all reduce efficiency below 100%

In a step-up transformer, which statement is correct?

Practicemixed bank + 2 SAQs

Check ratios, power, efficiency and changing-flux reasoning with mixed practice.

Quick recall, transformers and electromagnetic induction

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 7 marks

+5 XP

ApplyBand 4(3 marks) 1. A step-down transformer has 1200 primary turns and 60 secondary turns. The primary is connected to 230 V AC and draws 0.50 A. Calculate (a) the secondary voltage, and (b) the secondary current.

1 mark: correct $V_s$ with turns ratio applied · 1 mark: correct $I_s$ using power conservation · 1 mark: correct units throughout

AnalyseBand 5(4 marks) 2. A transformer has $V_p=25$ kV, $V_s=400$ kV and 500 primary turns. Its input power is 20 MW and its efficiency is 96%. (a) Calculate the secondary turns. (b) Calculate the ideal secondary current if input and output powers were equal. (c) Calculate the real output power and explain why it is smaller than the input power.

1 mark: correct $N_s$ · 1 mark: ideal $I_s=50$ A · 1 mark: real output power $19.2$ MW · 1 mark: identifies energy transferred to the surroundings through real loss mechanisms

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (3 marks): (a) $V_s = V_p(N_s/N_p) = 230(60/1200) = 11.5$ V (1 mark). (b) For the ideal model, $I_s=(V_pI_p)/V_s=(230\times0.50)/11.5=10$ A (1 mark). Units: V and A throughout (1 mark).

Q2 (4 marks): (a) $N_s=N_p(V_s/V_p)=500(400/25)=8000$ turns (1 mark). (b) In the ideal equal-power model, $I_s=P_{in}/V_s=20\times10^6/(400\times10^3)=50$ A (1 mark). (c) $P_{out}=\eta P_{in}=0.96(20\text{ MW})=19.2$ MW (1 mark). The missing 0.8 MW is transferred to the surroundings through copper heating, eddy currents, hysteresis and flux leakage rather than delivered to the load (1 mark).

Reviewretrieval + reflection

Revisit ideal-versus-real reasoning and complete the retrieval check.

Check what actually stuck

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quiz

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How did your thinking change?

At the start you predicted what happens when the primary is connected to steady DC rather than AC.

The complete answer is time-dependent: switching DC can produce a brief secondary emf while flux changes, but once the current and core flux are steady, $d\Phi/dt=0$ and the secondary emf returns to zero. AC supplies continuous flux change, so transformer action is sustained.

Extend your thinking: Compare an ideal transformer with a 96%-efficient real device. Which loss pathways account for the power difference, and which engineering changes reduce each one?

I have completed this lesson

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