$A$ small town with a demand of $800 \; kW$ of electric power at $220 \; V$ is situated $15 \; km$ away from an electric plant generating power at $440 \; V$. The resistance of the two-wire line carrying power is $0.5 \; \Omega/km$. The town gets power from the line through a $40,000-220 \; V$ step-down transformer at a sub-station in the town.
$(a)$ Estimate the line power loss in the form of heat.
$(b)$ How much power must the plant supply,assuming there is negligible power loss due to leakage?
$(c)$ Characterise the step-up transformer at the plant.
(Neglect,as before,leakage losses). Hence,explain why high-voltage transmission is preferred?

(A) Given: Power $P = 800 \; kW = 8 \times 10^5 \; W$,Voltage at town $V_1 = 40,000 \; V$,Distance $d = 15 \; km$,Resistance per unit length $r = 0.5 \; \Omega/km$.
Total resistance $R = (15 + 15) \times 0.5 = 15 \; \Omega$.
Current in the line $I = P / V_1 = (800 \times 10^3) / 40,000 = 20 \; A$.
$(a)$ Line power loss $P_{loss} = I^2 R = (20)^2 \times 15 = 400 \times 15 = 6,000 \; W = 6 \; kW$.
$(b)$ Total power supplied by the plant $P_{total} = P + P_{loss} = 800 \; kW + 6 \; kW = 806 \; kW$.
$(c)$ Voltage drop in the line $V_{drop} = I \times R = 20 \times 15 = 300 \; V$. Voltage at the output of the step-up transformer $V_{out} = V_1 + V_{drop} = 40,000 + 300 = 40,300 \; V$. Since the plant generates at $440 \; V$,the step-up transformer is $440 \; V - 40,300 \; V$.
High-voltage transmission is preferred because power loss $P_{loss} = I^2 R = (P/V)^2 R$ is inversely proportional to the square of the transmission voltage $V^2$. Increasing $V$ significantly reduces $I^2 R$ losses.