A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5 A and its efficiency is 90 %. The winding of transformer is made of copper. The output current of transformer is _________ A.

Solution

Physics sir · 3 steps

Efficiency is about power, so find the output power first and then divide by the output voltage.

1. Input power. The primary takes Vₚ Iₚ = 2300 V × 5 A = 11500 W.
2. Apply efficiency. Only 90% of this reaches the secondary, so Pₒᵤₜ = 0.9 × 11500 = 10350 W.

   Iₛ = (η Vₚ Iₚ)/Vₛ = (0.9 × 2300 × 5)/230

3. So. Iₛ = 10350/230 = 45 A. The output current is 45.

The common slip: 50. Treating the transformer as ideal and using Iₛ = Iₚ Vₚ / Vₛ gives 50 A, which ignores the 10% power loss.

The idea to keep. For a real transformer, the current ratio is not just the voltage ratio. Write η Vₚ Iₚ = Vₛ Iₛ and the losses are handled for you.

Answer from the NTA final answer key. Solution by PrepWiser.