An ideal transformer steps 240 V rms up to 12.0 kV rms to transmit 48.0 kW through two wires whose combined resistance is 8.00 Ω. Determine the turns ratio, transmission current, line power loss and percentage of generated power lost. Compare with transmitting the same useful power at 240 V, and state why an ordinary transformer cannot step up a steady DC supply.

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N_s/N_p = 50.0; high-voltage current = 4.00 A; line loss = 128 W, or about 0.266% of generated power. At 240 V the current would be 200 A and loss 320 kW, so that operating point is impossible for 48 kW delivery. For an ideal transformer V_s/V_p = N_s/N_p, so the turns ratio is 12000/240 = 50.0. At the high-voltage side, ideal transmitted power gives I = P/V = 48000/12000 = 4.00 A. Wire heating is I²R = (4.00)²(8.00) = 128 W. Relative to roughly 48128 W generated, the loss is 128/48128 × 100% = 0.266%. At 240 V, supplying 48.0 kW would require 200 A before accounting for losses, producing I²R = 320 kW. The assumption of delivering 48 kW at that voltage is self-inconsistent because the line loss exceeds the proposed input by a wide margin. Transformer emf is produced by changing magnetic flux. A steady DC current eventually creates constant flux, so after the switching transient there is no induced secondary voltage and the primary can overheat.

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