Q197 / 240 · KCET 2026 · Physics
A small town with a demand of $900\text{ kW}$ of electric power at $220\text{ V}$ is situated $20\text{ km}$ away from an electric power generating station. Each wire of the two-wire transmission line has a resistance per unit length of $5 \times 10^{-4} \, \Omega\text{ m}^{-1}$. The town gets power from the line through a $45000\text{ V}$ to $220\text{ V}$ step-down transformer at a substation in the town. The line power loss in the form of heat is:
- A$4$ kW
- Correct answerB$8$ kW
- C$40$ kW
- D$80$ kW
Solution
The total length of the two-wire transmission line is:
$L = 2 \times 20 \text{ km} = 40 \text{ km} = 40000 \text{ m}$
The total resistance of the transmission line is:
$R = L \times \text{resistance per unit length}$
$R = 40000 \text{ m} \times 5 \times 10^{-4} \, \Omega\text{ m}^{-1} = 20 \, \Omega$
The power demanded by the town is $P = 900 \text{ kW} = 9 \times 10^5 \text{ W}$.
The voltage at the primary of the step-down transformer is $V = 45000 \text{ V}$.
Assuming an ideal transformer, the current in the transmission line is:
$I = \dfrac{P}{V} = \dfrac{9 \times 10^5 \text{ W}}{45000 \text{ V}} = 20 \text{ A}$
The line power loss in the form of heat is given by:
$P_{\text{loss}} = I^2 R$
$P_{\text{loss}} = (20 \text{ A})^2 \times 20 \, \Omega = 400 \times 20 \text{ W} = 8000 \text{ W} = 8 \text{ kW}$
Answer: $8$ kW
$L = 2 \times 20 \text{ km} = 40 \text{ km} = 40000 \text{ m}$
The total resistance of the transmission line is:
$R = L \times \text{resistance per unit length}$
$R = 40000 \text{ m} \times 5 \times 10^{-4} \, \Omega\text{ m}^{-1} = 20 \, \Omega$
The power demanded by the town is $P = 900 \text{ kW} = 9 \times 10^5 \text{ W}$.
The voltage at the primary of the step-down transformer is $V = 45000 \text{ V}$.
Assuming an ideal transformer, the current in the transmission line is:
$I = \dfrac{P}{V} = \dfrac{9 \times 10^5 \text{ W}}{45000 \text{ V}} = 20 \text{ A}$
The line power loss in the form of heat is given by:
$P_{\text{loss}} = I^2 R$
$P_{\text{loss}} = (20 \text{ A})^2 \times 20 \, \Omega = 400 \times 20 \text{ W} = 8000 \text{ W} = 8 \text{ kW}$
Answer: $8$ kW
Hence, the correct answer is option: 'B'.