Transformers — Theory and Operation
In 1895, the Tesla/Westinghouse Niagara Falls power station ran three 2.25 MW generators at 2,200 V, stepped the voltage up to 22,000 V using transformers, then transmitted the power 35 km to Buffalo, New York. At 2,200 V the resistive losses would have been ~62% of generated power; at 22,000 V they dropped to ~2%. This was the decisive moment that ended the War of Currents and proved that AC electricity — with transformers — was the only practical way to transmit power over long distances.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
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
- DC produces constant flux → no induced emf → no transformation
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
Core Content
Induction between coupled coils
Plug a 12 V DC power supply into a transformer's primary coil and connect a voltmeter to the secondary: the voltmeter reads zero — nothing happens. Now plug in a 12 V AC supply and the voltmeter immediately jumps to a non-zero reading, even though the two coils are not electrically connected. The AC continuously changes the magnetic flux in the iron core; that changing flux induces an emf in the secondary by Faraday's Law. DC produces constant flux — no change, no induction, no output. This is why transformers only work with AC.
- 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:
$\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$ — 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 |
A transformer has 400 primary turns and 800 secondary turns. The primary voltage is 240 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:
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:
- DC produces a constant current and therefore a constant magnetic field.
- Constant field means constant flux through the secondary.
- Constant flux means zero rate of change → zero induced emf.
The secondary coil would act like a plain piece of wire with no voltage across it. This is why DC cannot be transformed directly and must first be converted to AC (or switched) for voltage conversion.
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 $\Phi$ → induced emf in secondary. DC: constant $B$ → constant $\Phi$ → $\Delta\Phi/\Delta t = 0$ → zero emf. Both coils share the same changing flux: each turn adds the same emf, so $V \propto N$.
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 transformer tool with 100 primary turns and 500 secondary turns. A 240 V input gives an output of:
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 240 V AC and draws 0.50 A. Calculate (a) secondary voltage, (b) secondary current, and (c) power delivered.
- 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.
- 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.
- 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:
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.
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.
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 _____.
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.
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?
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.
ApplyBand 4(3 marks) 1. A step-down transformer has 1200 primary turns and 60 secondary turns. The primary is connected to 240 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 power station generates electricity at 25 kV. A transformer steps this up to 400 kV for long-distance transmission. The primary coil has 500 turns. (a) Calculate the number of secondary turns. (b) If the transmission current is 50 A, calculate the power being transmitted. (c) Explain why power is transmitted at high voltage.
1 mark: correct $N_s$ from turns ratio · 1 mark: correct power calculation · 1 mark: high voltage means low current · 1 mark: low current means less $I^2R$ loss in transmission lines
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) $V_s = V_p \times (N_s/N_p) = 240 \times (60/1200) = 12$ V (1 mark). (b) Using power conservation: $V_p I_p = V_s I_s$, so $I_s = (240 \times 0.50)/12 = 10$ A (1 mark). Units: V and A throughout (1 mark).
Q2 (4 marks): (a) $N_s = N_p \times (V_s/V_p) = 500 \times (400\,000/25\,000) = 8000$ turns (1 mark). (b) $P = V_s \times I_s = 400\,000 \times 50 = 20 \times 10^6$ W $= 20$ MW (1 mark). (c) High voltage means lower current for the same power ($P = VI$). Lower current means much less power lost as heat in the resistance of the transmission lines ($P_{loss} = I^2 R$), since loss scales with the square of current. Transmitting at 400 kV instead of 25 kV reduces the current by a factor of 16, cutting line losses by a factor of 256 (1 mark for each of the two marking points).
Five timed questions on transformers and electromagnetic induction. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaAt the start you were asked about the 1895 Niagara Falls power station (Tesla/Westinghouse): if the station had transmitted at 2,200 V instead of stepping up to 22,000 V, how many times greater would the losses have been?
The answer: at the same power, lowering voltage by 10× raises current by 10×. Power loss $P = I^2R$, so losses increase by $10^2 = 100$ times. The station's step-up transformers reduced transmission losses from ~62% to ~2% — a factor of ~31 improvement. This is why the War of Currents ended: no amount of DC engineering could replicate this, and every national grid since 1895 uses transformers.
Extend your thinking: A real grid-scale transformer is about 98–99% efficient. Given that it handles hundreds of megawatts, even 1% of energy lost as heat is enormous. Which two strategies would you prioritise to push efficiency as high as possible, and why?