Question:
A small town with a demand of 900 kW of electric power at 220 V is situated 20 km away from an electric power generating station. The two-wires line has resistance per unit length of \( 5 \times 10^{-4} \ \Omega \text{m}^{-1} \). The town gets power from the line through 45000 V to 220 V stepdown transformer at a substation in the town. The line power loss in the form of heat is
A small town with a demand of 900 kW of electric power at 220 V is situated 20 km away from an electric power generating station. The two-wires line has resistance per unit length of \( 5 \times 10^{-4} \ \Omega \text{m}^{-1} \). The town gets power from the line through 45000 V to 220 V stepdown transformer at a substation in the town. The line power loss in the form of heat is
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To minimize power loss (\( I^2 R \)), electricity is transmitted at very high voltages. High voltage means low current for the same power demand (\( P = VI \)), which significantly reduces the energy lost in the resistance of the long transmission cables.
Updated On: Apr 23, 2026
- 4 kW
- 8 kW
- 40 kW
- 80 kW
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The goal is to calculate the power wasted as heat in the transmission lines (line loss) before the power reaches the town's step-down transformer.
Step 2: Key Formula or Approach:
1. Total line resistance \( R = \text{resistance per meter} \times \text{total length of wire} \).
2. Transmission current \( I = \frac{P_{\text{demand}}}{V_{\text{transmission}}} \).
3. Line power loss \( P_{\text{loss}} = I^2 R \).
Step 3: Detailed Explanation:
Part 1: Calculate the total line resistance.
The station is 20 km away. A "two-wires line" means current goes through 20 km and returns through another 20 km.
Total length \( D = 2 \times 20 \text{ km} = 40 \text{ km} = 40,000 \text{ m} \).
Resistance per meter \( r = 5 \times 10^{-4} \ \Omega/\text{m} \).
Total resistance \( R = r \times D = (5 \times 10^{-4} \ \Omega/\text{m}) \times (40,000 \text{ m}) = 20 \ \Omega \).
Part 2: Calculate the transmission current.
The power demand is \( P = 900 \text{ kW} = 900,000 \text{ W} \).
The voltage at which power is transmitted over the lines (entering the substation) is \( V = 45,000 \text{ V} \).
Current in the line \( I = \frac{P}{V} = \frac{900,000}{45,000} = \frac{900}{45} = 20 \text{ A} \).
Part 3: Calculate the line power loss.
[ P_{\text{loss}} = I^2 R = (20)^2 \times 20 ]
[ P_{\text{loss}} = 400 \times 20 = 8,000 \text{ W} = 8 \text{ kW} ]
Step 4: Final Answer:
The line power loss in the form of heat is 8 kW, which corresponds to option (2).
The goal is to calculate the power wasted as heat in the transmission lines (line loss) before the power reaches the town's step-down transformer.
Step 2: Key Formula or Approach:
1. Total line resistance \( R = \text{resistance per meter} \times \text{total length of wire} \).
2. Transmission current \( I = \frac{P_{\text{demand}}}{V_{\text{transmission}}} \).
3. Line power loss \( P_{\text{loss}} = I^2 R \).
Step 3: Detailed Explanation:
Part 1: Calculate the total line resistance.
The station is 20 km away. A "two-wires line" means current goes through 20 km and returns through another 20 km.
Total length \( D = 2 \times 20 \text{ km} = 40 \text{ km} = 40,000 \text{ m} \).
Resistance per meter \( r = 5 \times 10^{-4} \ \Omega/\text{m} \).
Total resistance \( R = r \times D = (5 \times 10^{-4} \ \Omega/\text{m}) \times (40,000 \text{ m}) = 20 \ \Omega \).
Part 2: Calculate the transmission current.
The power demand is \( P = 900 \text{ kW} = 900,000 \text{ W} \).
The voltage at which power is transmitted over the lines (entering the substation) is \( V = 45,000 \text{ V} \).
Current in the line \( I = \frac{P}{V} = \frac{900,000}{45,000} = \frac{900}{45} = 20 \text{ A} \).
Part 3: Calculate the line power loss.
[ P_{\text{loss}} = I^2 R = (20)^2 \times 20 ]
[ P_{\text{loss}} = 400 \times 20 = 8,000 \text{ W} = 8 \text{ kW} ]
Step 4: Final Answer:
The line power loss in the form of heat is 8 kW, which corresponds to option (2).
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