An ideal transformer with purely resistive load operates at 12 kV on the primary side. It supplies electrical energy to a number of nearby houses at 120 V. The average rate of energy consumption in the houses served by the transformer is 60 kW. The value of resistive load (Rs) required in the secondary circuit will be ___________ m Ω.

Solution

Physics sir · 3 steps

The houses only see the secondary side, so use 120 V and 60 kW in R = V²/P.

1. Pick the right side. The load sits in the secondary circuit, so it works at 120 V, not at 12 kV. The primary voltage is not needed here.
2. Power in a resistor. For a purely resistive load, the average power is P = V²/R, where V is the rms voltage across it. Rearrange to get R.

   Rₛ = Vₛ²/P = (120)²/(60 × 10³) = 14400/60000 Ω

3. So. Rₛ = 0.24 Ω = 240 mΩ, so the answer is 240.

The common slip: 0.24. Stopping at ohms gives 0.24, not 240. The blank asks for milliohms, so multiply by 1000.

The idea to keep. For transformer questions, first decide which side the asked quantity lives on. A load resistance only cares about the voltage and power on its own side.

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