A 10 m long nichrome wire having `80 Omega` resistance has current carrying capacity of 5 A. This wire can be cut into equal parts and equal parts can be connected in series or parallel. What is the maximum power which can be obtain as heat by the wire from a 200 V mains supply? (in kW)

Explanation

To solve the problem step by step, we will determine the maximum power that can be obtained from the nichrome wire when connected to a 200 V mains supply.

### Step 1: Calculate the initial current through the wire
Given:
- Voltage (V) = 200 V
- Resistance (R) = 80 Ω

Using Ohm's law:
\[ I = \frac{V}{R} \]

Substituting the values:
\[ I = \frac{200}{80} = 2.5 \, \text{A} \]

### Step 2: Check the current carrying capacity
The current carrying capacity of the wire is given as 5 A, which is greater than the calculated current of 2.5 A. Therefore, the wire can handle this current without overheating.

### Step 3: Calculate the initial power delivered by the wire
Using the formula for power:
\[ P = \frac{V^2}{R} \]

Substituting the values:
\[ P = \frac{200^2}{80} = \frac{40000}{80} = 500 \, \text{W} \]

### Step 4: Cut the wire into equal parts
If we cut the wire into 2 equal parts, the length of each part will be:
\[ L' = \frac{10 \, \text{m}}{2} = 5 \, \text{m} \]

### Step 5: Calculate the resistance of each part
The resistance of each part (R1 and R2) can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
Since the resistance is directly proportional to the length, the resistance of each part will be:
\[ R' = \frac{80 \, \Omega}{2} = 40 \, \Omega \]

### Step 6: Calculate the equivalent resistance when connected in parallel
For two resistors in parallel:
\[ R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2} \]

Substituting the values:
\[ R_{eq} = \frac{40 \cdot 40}{40 + 40} = \frac{1600}{80} = 20 \, \Omega \]

### Step 7: Calculate the maximum power using the equivalent resistance
Using the power formula again:
\[ P_{max} = \frac{V^2}{R_{eq}} \]

Substituting the values:
\[ P_{max} = \frac{200^2}{20} = \frac{40000}{20} = 2000 \, \text{W} \]

### Step 8: Convert power to kilowatts
To convert watts to kilowatts:
\[ P_{max} = \frac{2000}{1000} = 2 \, \text{kW} \]

### Final Answer
The maximum power that can be obtained as heat by the wire from a 200 V mains supply is **2 kW**.

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