KCET 2026 Physics Solved Paper

Concept:Power loss in transmission lines is calculated using $P_{\text{loss}} = I^2 R$, where $I$ is the line current and $R$ is the total resistance of both wires.Explanation:The town draws $900 \, \text{kW}$ of power, which is supplied at a high voltage of $45000 \, \text{V}$ through the transmission line.Convert the power demand to watts: $P = 900 \, \text{kW} = 900000 \, \text{W}$.The current in the line is given by $I = \frac{P}{V}$.Substitute the values: $I = \frac{900000}{45000} = 20 \, \text{A}$.Each wire has a length of $20 \, \text{km} = 20000 \, \text{m}$.Resistance of one wire: $R = 5 \times 10^{-4} \, \Omega \, \text{m}^{-1} \times 20000 \, \text{m} = 10 \, \Omega$.For a two-wire line, total resistance is $R_{\text{total}} = 10 + 10 = 20 \, \Omega$.Line power loss as heat: $P_{\text{loss}} = I^2 R = (20)^2 \times 20 = 400 \times 20 = 8000 \, \text{W}$.Convert to kilowatts: $P_{\text{loss}} = 8 \, \text{kW}$.Answer:The line power loss is $8 \, \text{kW}$, which corresponds to Option B.