A town situated 20 kW away from a power plant generating power at 440 V requires 600 KW of electric power at 200 V. The resistance of two wire lines carrying poweris `0.4 Omega` per km. The town gets power from the line through a 3000-220 V step down transformer at a substation in the town. Find line power losses in the form of heat. How much power must the plant supply assuming that there is negligilbe power loss due to leakage ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the total resistance of the wire
The resistance of the wire is given as \(0.4 \, \Omega\) per km. The distance from the power plant to the town is \(20 \, \text{km}\). Since the power travels to the town and back, we need to consider the total length of the wire, which is \(20 \, \text{km} \times 2 = 40 \, \text{km}\).

\[ \text{Total Resistance} (R) = \text{Length} \times \text{Resistance per km} = 40 \, \text{km} \times 0.4 \, \Omega/\text{km} = 16 \, \Omega \]

### Step 2: Calculate the current flowing through the wire
The town requires \(600 \, \text{kW}\) of electric power at \(200 \, \text{V}\). First, we need to find the current using the formula:

\[ P = V \times I \implies I = \frac{P}{V} \]

Substituting the values:

\[ I = \frac{600 \times 10^3 \, \text{W}}{3000 \, \text{V}} = 200 \, \text{A} \]

### Step 3: Calculate the power loss in the form of heat
The power loss due to the resistance of the wire can be calculated using the formula:

\[ P_{\text{loss}} = I^2 \times R \]

Substituting the values we calculated:

\[ P_{\text{loss}} = (200 \, \text{A})^2 \times 16 \, \Omega = 40000 \times 16 = 640000 \, \text{W} = 64 \, \text{kW} \]

### Step 4: Calculate the total power that must be supplied by the plant
The total power that the plant must supply is the sum of the power required by the town and the power loss in the wires:

\[ P_{\text{total}} = P_{\text{required}} + P_{\text{loss}} \]

Substituting the values:

\[ P_{\text{total}} = 600 \, \text{kW} + 64 \, \text{kW} = 664 \, \text{kW} \]

### Final Answer
The power that must be supplied by the plant is \(664 \, \text{kW}\).

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