Physics • Year 12 • Module 6 • Lesson 15
Transformers — Theory and Operation
Lock in the key vocabulary, the two transformer equations, and the four energy-loss mechanisms before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: transformer, primary coil, secondary coil, step-up transformer, step-down transformer, laminated iron core, eddy currents, hysteresis loss, flux linkage, ideal transformer. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A device that uses electromagnetic induction between two coils on a shared core to change AC voltage. | |
| 1.2 | The input coil connected to the AC power source. | |
| 1.3 | The output coil where the induced voltage is delivered to the load. | |
| 1.4 | A transformer in which the secondary voltage is greater than the primary voltage (Ns > Np). | |
| 1.5 | A transformer in which the secondary voltage is less than the primary voltage (Ns < Np). | |
| 1.6 | A core built from thin sheets of iron insulated from each other, which reduces energy losses from swirling currents. | |
| 1.7 | Swirling induced currents within the iron core that dissipate energy as heat. | |
| 1.8 | Energy dissipated as heat during repeated magnetisation and demagnetisation of the iron core each AC cycle. | |
| 1.9 | The degree to which the magnetic flux from the primary coil passes through the secondary coil. | |
| 1.10 | A model transformer assumed to transfer all power from primary to secondary without any energy loss (Pp = Ps). |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 A transformer with twice as many turns on the secondary as the primary will produce twice the input voltage at the secondary. T / F
2.2 Transformers work equally well with both AC and DC power supplies. T / F
2.3 In an ideal step-up transformer, secondary current is greater than primary current. T / F
2.4 Laminating the iron core reduces hysteresis losses by increasing electrical resistance to eddy currents. T / F
2.5 For an ideal transformer, the ratio Vp/Vs equals the ratio Np/Ns. T / F
2.6 Using thicker copper wire in the coils reduces eddy current losses. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
alternating · changing · constant · eddy currents · Faraday · flux · iron core · turns ratio
A transformer uses ___________ current (AC) in the primary coil to produce a continuously ___________ magnetic field in the shared ___________. According to ___________'s Law, this changing magnetic ___________ induces an electromotive force in the secondary coil. If DC were used instead, the magnetic field would be ___________, so no emf would be induced. The ratio of primary to secondary voltage equals the ___________. One energy loss mechanism in the core is caused by ___________ — swirling induced currents that dissipate energy as heat.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 Why must the current in the primary coil be alternating (AC) for a transformer to function?
4.2 State the function of the iron core in a transformer.
4.3 State the two transformer equations and define each symbol.
4.4 Explain how laminating the iron core reduces energy losses. Name the specific loss it targets.
5. Complete the transformer summary table
Fill in every empty cell. 10 marks (1 each)
| Type | Turns condition | Voltage effect | Current effect | Real-world example |
|---|---|---|---|---|
| Step-up | ||||
| Step-down |
Q1 — Term–definition match
1.1 transformer • 1.2 primary coil • 1.3 secondary coil • 1.4 step-up transformer • 1.5 step-down transformer • 1.6 laminated iron core • 1.7 eddy currents • 1.8 hysteresis loss • 1.9 flux linkage • 1.10 ideal transformer.
Q2 — True / false with correction
2.1 True. Vs/Vp = Ns/Np = 2, so Vs = 2Vp.
2.2 False. Transformers only work with AC. DC produces a constant magnetic field in the core, so the flux through the secondary is constant, meaning the rate of change of flux is zero and no emf is induced.
2.3 False. In a step-up transformer Vs > Vp. For an ideal transformer power is conserved (VpIp = VsIs), so if voltage increases, current decreases: Is < Ip.
2.4 False. Laminating the core targets eddy current losses (not hysteresis losses). Lamination breaks up large current loops, increasing resistance to eddy currents. Hysteresis losses are reduced by using soft magnetic materials.
2.5 True.
2.6 False. Thicker copper wire reduces resistive (copper/I2R) losses in the coil windings, not eddy current losses. Eddy current losses are reduced by laminating the core.
Q3 — Cloze paragraph
In order: alternating / changing / iron core / Faraday / flux / constant / turns ratio / eddy currents.
Q4.1 — Why AC is required
AC in the primary produces a continuously changing magnetic field in the iron core. By Faraday's Law, a changing magnetic flux through the secondary coil induces an emf. DC produces a constant field (constant flux), giving zero rate of change and therefore zero induced emf in the secondary.
Q4.2 — Function of the iron core
The iron core channels and concentrates the magnetic flux from the primary coil through the secondary coil, maximising flux linkage so that nearly all the changing flux produced by the primary passes through the secondary. It also provides a low-reluctance path for the magnetic field.
Q4.3 — Transformer equations
Equation 1 (voltage–turns ratio): Vp/Vs = Np/Ns, where Vp = primary voltage (V), Vs = secondary voltage (V), Np = number of primary turns, Ns = number of secondary turns.
Equation 2 (power conservation / current ratio): VpIp = VsIs, or equivalently Ip/Is = Ns/Np, where Ip = primary current (A) and Is = secondary current (A). This holds for an ideal transformer.
Q4.4 — Laminated core and the loss it targets
Laminating the core builds it from many thin sheets of iron, each electrically insulated from its neighbours. This breaks up large eddy current loops into smaller ones in each thin sheet, greatly increasing resistance to eddy currents and so reducing the energy they dissipate as heat. The specific loss targeted is eddy current loss (iron/core loss).
Q5 — Transformer summary table
Step-up: Ns > Np • Vs > Vp (voltage increases) • Is < Ip (current decreases) • Example: power station stepping up to transmission voltage (e.g. 25 kV → 330 kV).
Step-down: Ns < Np • Vs < Vp (voltage decreases) • Is > Ip (current increases) • Example: substation or phone charger stepping 240 V mains down to 5–12 V.