Retrieval Practice

Transformers — AQA A-Level Physics

Q1. Describe the basic structure of a transformer.
  • A transformer consists of a primary coil and a secondary coil, both wound on a shared soft iron core.
  • The coils are electrically isolated from each other.
Q2. Explain why a transformer only works with AC and not DC.
  • AC produces a continuously changing current, which creates a continuously changing magnetic flux in the core.
  • This changing flux induces an EMF in the secondary coil (Faraday's law).
  • DC produces a constant current, so the flux is constant and no EMF is induced.
Q3. State the transformer equation and define each term.
Vs/Vp = Ns/Np, where Vs = secondary voltage, Vp = primary voltage, Ns = number of turns on the secondary coil, Np = number of turns on the primary coil.
Q4. What is the difference between a step-up and a step-down transformer?
  • A step-up transformer has more turns on the secondary than the primary (Ns > Np), so the output voltage is greater than the input voltage.
  • A step-down transformer has fewer turns on the secondary (Ns < Np), so the output voltage is less than the input.
Q5. State the power conservation equation for an ideal transformer.
  • VpIp = VsIs.
  • Input power equals output power.
  • If voltage is stepped up, current is stepped down by the same factor.
Q6. Write the equation for transformer efficiency.
  • Efficiency = (VsIs)/(VpIp) x 100%.
  • This is output power divided by input power, expressed as a percentage.
Q7. State four causes of energy loss in a real transformer.
  • 1.
  • Resistive heating (I²R losses) in the copper coils. 2.
  • Eddy currents induced in the iron core. 3.
  • Hysteresis losses from repeated magnetisation and demagnetisation of the core. 4.
  • Flux leakage where not all flux links both coils.
Q8. How are eddy current losses reduced in a transformer?
  • By laminating the core.
  • The core is made from thin sheets of iron separated by layers of insulation.
  • This breaks up the eddy current paths, reducing their magnitude and the associated I²R heating.
Q9. Explain why the National Grid transmits electricity at high voltage.
  • For a given power P = IV, increasing voltage reduces current.
  • Power lost in cables is P_loss = I²R, so reducing current greatly reduces losses.
  • Doubling the voltage halves the current and quarters the power loss.
Q10. Show that power loss in transmission cables is inversely proportional to the square of the transmission voltage.
  • P = IV, so I = P/V.
  • Substituting into P_loss = I²R gives P_loss = (P/V)²R = P²R/V².
  • Since P and R are constant, P_loss is proportional to 1/V².
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