Question

A power station delivers 750 kW of power at 12,000 V through wires with a total resistance of 3.0 Ohm. How much less power is wasted if the electricity is delivered at 50,000 V rather than 12,000 V?

Answer 1

The amount of current passing through the wire is

\(I=\frac{P}{V}=\frac{750 \times 10^{3} \mathrm{~W}}{12000 \mathrm{~V}}=62.5 \mathrm{~A}\)

The power loss is

\(P_{\mathrm{bs}}=I^{2} R=(62.5 \mathrm{~A})^{2}(3.0 \Omega)=11718.75 \mathrm{~W}\)

The current transmission for \(50,000 \mathrm{~V}\) is

\(I^{\prime}=\frac{P}{V^{\prime}}=\frac{750 \times 10^{3} \mathrm{~W}}{50,000 \mathrm{~V}}=15 \mathrm{~A}\)

The power loss is

\(P_{\mathrm{bs}}^{\prime}=I^{2} R=(15 \mathrm{~A})^{2}(3.0 \Omega)=675 \mathrm{~W}\)

The wasted power is

\(=1718.75 \mathrm{~W}-675 \mathrm{~W}=1.104 \times 10^{4} \mathrm{~W}=11.04 \mathrm{~kW}\)