JEE Main 2022 · 28 July · evening · Single correct · +4 / −1
A transformer operating at primary voltage 8 kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be
- (a) 800 Ω and 1.06 Ω
- (b) 10 Ω and 500 Ω
- (c) 800 Ω and 0.32 Ω
- (d) 1.06 Ω and 500 Ω
Answer and solution
Answer (c) 800 Ω and 0.32 Ω
Solution
Physics sir · 4 steps
An ideal transformer passes the full 80 kW to each side, so use R = V²/P on each side separately.
- Same power both sides. An ideal transformer loses no power, and at unity power factor the load is purely resistive. So each winding delivers P = 80 kW at its own voltage.
R = V²/P
- Primary side. Rₚ = (8000)²/80000 = (64 × 10⁶)/(8 × 10⁴) = 800 Ω.
- Secondary side. Rₛ = (160)²/80000 = 25600/80000 = 0.32 Ω. As a check, the turns ratio is 8000/160 = 50, and 50² × 0.32 = 800 Ω.
- So. The loads are 800 Ω in the primary and 0.32 Ω in the secondary, which is option (c).
The idea to keep. In an ideal transformer, power is the same on both sides. A resistance seen through the transformer scales as the square of the turns ratio, so Rₚ = (Nₚ/Nₛ)² Rₛ.
Answer from the NTA final answer key. Solution by PrepWiser.