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 analyze the situation and calculate the maximum power that can be obtained from the nichrome wire when connected to a 200 V mains supply.

### Step 1: Understand the given data
- Length of the nichrome wire (L) = 10 m
- Resistance of the wire (R) = 80 Ω
- Current carrying capacity (I_max) = 5 A
- Voltage supply (V) = 200 V

### Step 2: Calculate the initial power using the total resistance
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 3: Determine the maximum current that can be carried
The maximum current that can be carried by the wire is given as 5 A.

### Step 4: Determine the configuration for maximum power
To maximize the power, we can cut the wire into equal parts. If we cut the wire into two equal parts, each part will have:
- Length of each part = \( \frac{10}{2} = 5 \) m
- Resistance of each part = \( \frac{80}{2} = 40 \) Ω

### Step 5: Calculate the equivalent resistance when connected in parallel
When two resistors of equal resistance (R1 = R2 = 40 Ω) are connected in parallel, the equivalent resistance (R_eq) is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
\[ \frac{1}{R_{eq}} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20} \]
Thus,
\[ R_{eq} = 20 \, \Omega \]

### Step 6: Calculate the maximum power using the equivalent resistance
Now we can calculate the maximum power using the equivalent resistance:
\[ P_{max} = \frac{V^2}{R_{eq}} \]
Substituting the values:
\[ P_{max} = \frac{200^2}{20} = \frac{40000}{20} = 2000 \text{ W} \]

### Step 7: 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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