A thermal power plant produces electric power of 600 kW at 4000 V , which is to be transported to a

Question

A thermal power plant produces electric power of 600 kW at 4000 V , which is to be transported to a place 20 km away from the power plant for consumers' usage. A step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. The power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.\In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is $1: 10$. If the power to the consumers has to be supplied at 200 V , the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is\

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(a) : For a step-up transformer, Here, $\frac{N_{p}}{N_{s}}=\frac{1}{10}, V_{p}=4000 \mathrm{~V}$\ $	herefore \quad V_{s}=V_{p} \frac{N_{s}}{N_{p}}=(4000 \mathrm{~V})=40,000 \mathrm{~V}$\ For a step-down transformer,\ Here, $\frac{N_{p}}{N_{s}}=$ ?, $V_{p}=40,000 \mathrm{~V}, \mathrm{Vs}=200 \mathrm{~V}$\ $	herefore \frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}=\frac{40,000 \mathrm{~V}}{200 \mathrm{~V}}=\frac{200}{1}$\