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.