Exam Frequency Analysis
Past paper frequency (2018 to 2024)
This topic accounts for approximately 8% of your exam marks.
stable
Low
Stable8%
Transformer equations, generator principles and Lenz's Law appear as multi-mark questions.
Voltage-and-turns ratio
For an ideal transformer, the voltage on each side of the iron core is proportional to the number of turns on that side:
Vp / Vs = Np / Ns
- where:
- Vp = voltage across the primary coil (V)
- Vs = voltage across the secondary coil (V)
- Np = number of turns on the primary coil
- Ns = number of turns on the secondary coil
- The equation can equally be written upside down (Vs / Vp = Ns / Np). Pick the form that puts the unknown on the top of a fraction so you have less rearranging to do
- The key fact: the ratio of voltages equals the ratio of turns
Example — a mains adapter has a primary coil of 1200 turns and a secondary coil of 60 turns. The primary coil is connected to a 240 V mains supply. Calculate the output voltage at the secondary.
- Vs / Vp = Ns / Np
- Vs = Vp × (Ns / Np) = 240 × (60 / 1200) = 240 × 0.05 = 12 V
- This is a step-down transformer (fewer turns on the secondary, lower output voltage). Typical of a wall-wart power supply
Conservation of energy
- An ideal transformer is 100 % efficient, with no energy lost as heat in the coils, the core, or eddy currents
- Real transformers reach about 99 % efficiency; the small loss appears as gentle warming of the iron core
- For an ideal transformer, the input electrical power equals the output electrical power:
Vp × Ip = Vs × Is
- where:
- Ip = current in the primary coil (A)
- Is = current in the secondary coil (A)
- This is just the statement P = V × I applied to both sides of the transformer
- A direct consequence: if the voltage is stepped up by some factor n, the current is stepped down by the same factor n, and vice versa
Example — a high-voltage transmission transformer steps the line voltage down from 25 000 V to 230 V for a town's mains supply. The town draws 80 A from the secondary. Treating the transformer as ideal (100 % efficient), find the current that flows into the primary from the high-voltage line.
- Vp × Ip = Vs × Is
- Ip = (Vs × Is) / Vp = (230 × 80) / 25 000
- Ip = 18 400 / 25 000 = 0.736 A (about 0.74 A)
- The current is much smaller on the high-voltage side, exactly as expected for the same power output
