The operation of a transformer (A-level only) — AQA A-Level Physics

This topic covers how a transformer uses electromagnetic induction to change the size of an alternating voltage. You need to understand the construction of a transformer, including the primary and secondary coils and the laminated soft iron core, and how a changing magnetic flux in the core induces an e.m.f. in the secondary coil. The key equations are the turns ratio equation (V_s/V_p = N_s/N_p) and the power conservation equation for an ideal transformer (V_p I_p = V_s I_s).

Transformers are essential in the national grid for transmitting electrical power at high voltages to reduce energy losses, and in many everyday devices to step down voltage to safe levels. This topic links directly to electromagnetic induction, Faraday's law, and Lenz's law, and it is a common context for exam questions on efficiency and power losses. Understanding transformers also helps you appreciate why a.c. is used for power distribution rather than d.c.

• →A transformer consists of two coils (primary and secondary) wound on a laminated soft iron core. An alternating current in the primary coil produces a changing magnetic flux in the core, which links the secondary coil and induces an alternating e.m.f. across it.
• →For an ideal transformer, the ratio of secondary voltage to primary voltage equals the ratio of secondary turns to primary turns: V_s/V_p = N_s/N_p. This means a step-up transformer has more turns on the secondary coil, and a step-down transformer has fewer.
• →In an ideal transformer, power is conserved: V_p I_p = V_s I_s. Therefore, if voltage is stepped up, current is stepped down, and vice versa.
• →Real transformers are not 100% efficient. Energy losses occur due to resistance heating in the coils, eddy currents induced in the core, and hysteresis losses as the core magnetises and demagnetises. These are reduced by using thick copper wires, a laminated core, and a soft magnetic material.
• →Transformers only work with alternating current (or a changing direct current) because a changing magnetic flux is required to induce an e.m.f. A steady direct current produces a constant flux and no induced e.m.f.

• Resistance in the primary and secondary coils causes heating, often called I²R or copper losses.
• Eddy currents are induced in the core by changing magnetic flux and dissipate energy as heat.
• Hysteresis losses occur because energy is required to magnetise and demagnetise the core each cycle.
• Flux leakage occurs when some magnetic field lines from the primary do not link the secondary coil.
• Efficiency = (output power ÷ input power) × 100%, and real efficiency is always less than 100%.
• Laminated cores reduce eddy currents; soft magnetic materials reduce hysteresis losses.
• Power transmitted is related to voltage and current by P = VI, so I = P ÷ V.
• Power loss in a transmission line of resistance R is P_loss = I²R.
• For a fixed transmitted power, increasing voltage decreases current and therefore decreases power loss.
• Transformers are used to step up voltage for transmission and step down voltage for distribution and domestic use.
• Calculations may require combining P = VI with P_loss = I²R to find current, resistance or lost power.
• The efficiency of transmission can be found from useful output power divided by input power, often as a percentage.

• 💡When asked for causes, name the mechanism and state the energy transfer, for example eddy currents transfer energy by heating the core.
• 💡Use the efficiency equation with consistent power units and give the answer as a percentage when appropriate.
• 💡Link each cause to a practical design feature, such as laminated core for eddy currents or soft iron for hysteresis.
• 💡Write down the transmitted power and the line resistance before substituting into equations.
• 💡Show the current calculation separately so that an error in squaring does not lose all method marks.
• 💡State clearly that high voltage reduces current, and link this to reduced heating in the cables.
• 💡Always state that the transformer relies on electromagnetic induction and that a changing magnetic flux is essential. Mention that the core is laminated to reduce eddy currents and made of soft iron to reduce hysteresis losses.
• 💡When calculating with the transformer equation, write down the equation clearly and substitute values carefully. Check whether the transformer is step-up or step-down by comparing turns.
• 💡For efficiency questions, use the formula: efficiency = (useful power output / total power input) x 100%. Remember that power input = V_p I_p and power output = V_s I_s for an ideal transformer, but for a real transformer, output power is less than input power.

• Thinking that a transformer creates energy or that output power can exceed input power; correct by applying conservation of energy so output power is always less than input power in a real transformer.
• Confusing eddy currents with the current in the coils; correct by stating that eddy currents are induced within the core itself and cause heating.
• Ignoring flux leakage and attributing all inefficiency to resistance; correct by listing flux leakage as a separate cause where magnetic field lines fail to link both coils.
• Using P_loss = V²R instead of P_loss = I²R; correct by using the current in the line and the line resistance.
• Assuming that increasing voltage increases power loss; correct by explaining that for fixed transmitted power, higher voltage means lower current and lower I²R loss.
• Forgetting to square the current in P_loss = I²R; correct by calculating the current first and then squaring it before multiplying by resistance.
• Students often think a step-up transformer increases power. In fact, for an ideal transformer, power is conserved; the voltage increases but the current decreases proportionally.
• Students sometimes believe transformers work with direct current. They only work with alternating current because a changing magnetic flux is needed to induce an e.m.f. in the secondary coil.
• Students may confuse the turns ratio with the current ratio. Remember: V_s/V_p = N_s/N_p, but I_s/I_p = N_p/N_s.

1. 1Start by reviewing electromagnetic induction and Faraday's law. Make sure you understand how a changing magnetic field induces an e.m.f. in a coil.
2. 2Learn the construction of a transformer and the function of each part (primary coil, secondary coil, laminated soft iron core). Draw a labelled diagram and annotate it.
3. 3Memorise and practise using the transformer equations: V_s/V_p = N_s/N_p and V_p I_p = V_s I_s. Solve numerical problems, including efficiency calculations.
4. 4Study the causes of energy losses in a transformer and how they are reduced. Make a table of loss mechanism and method of reduction.
5. 5Attempt past paper questions on transformers, focusing on explanations and calculations. Mark your answers using the mark scheme and note recurring mistakes.

• 📋Calculation questions: Use the turns ratio equation to find an unknown voltage or number of turns, or use power conservation to find current. Always show your working and give units.
• 📋Explanation questions: Describe how a transformer works, why it needs a.c., and how energy losses are reduced. Use key terms like 'changing magnetic flux', 'induced e.m.f.', 'laminated core', and 'eddy currents'.
• 📋Efficiency and power loss questions: Calculate efficiency or power lost, and explain how real transformers are made more efficient. Remember to use power in = power out + losses.
• 📋Graph or data analysis: Interpret a graph of voltage against time for primary and secondary coils, or analyse data on efficiency. Relate the shape of the graph to the alternating nature of the input.