JEE Main 2024 · 1 February · evening · Single correct · +4 / −1
A transformer has an efficiency of 80 % and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is :
- (a) 1.33 A
- (b) 13.33 A
- (c) 1.59 A
- (d) 15.1 A
Answer and solution
Answer (b) 13.33 A
Solution
Physics sir · 3 steps
Only 80% of the 4 kW reaches the secondary, so share that output power over 240 V.
- Output power. Efficiency is output power divided by input power. The input is 4 kW, so the output is 0.8 × 4000 = 3200 W.
- Secondary current. The secondary delivers this output at 240 V, so use Pₛ = Vₛ Iₛ. The primary voltage of 10 V is not needed here.
Iₛ = (η Pᵢₙ)/Vₛ = (0.8 × 4000)/240 = 3200/240 = 13.33 A
- So. The current in the secondary coil is 13.33 A, option (b).
The idea to keep. In a real transformer, find the secondary current from output power: Iₛ = η Pᵢₙ / Vₛ. The 100% loss-free assumption only works when no efficiency is given.
Answer from the NTA final answer key. Solution by PrepWiser.