A capacitor having capacity of `2 muF` is charged to `200 V` and then the plates of the capacitor are connected to a resistance wire. The heat produced in joule will be

Options

A. `2xx10^(-2)`
B. `4xx10^(-2)`
C. `4xx10^(4)`
D. `4xx10^(10)`

Answer

To solve the problem of calculating the heat produced when a capacitor is discharged through a resistance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance (C) = 2 µF = \(2 \times 10^{-6}\) F
- Voltage (V) = 200 V

2. **Use the formula for energy stored in a capacitor:**
The energy (U) stored in a capacitor can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \]

3. **Substitute the values into the formula:**
\[ U = \frac{1}{2} \times (2 \times 10^{-6}) \times (200)^2 \]

4. **Calculate \(V^2\):**
\[ V^2 = 200^2 = 40000 \]

5. **Plug \(V^2\) back into the energy formula:**
\[ U = \frac{1}{2} \times (2 \times 10^{-6}) \times 40000 \]

6. **Simplify the expression:**
\[ U = (1 \times 10^{-6}) \times 40000 = 4 \times 10^{-2} \text{ joules} \]

7. **Conclusion:**
The heat produced when the capacitor is discharged through the resistance is \(4 \times 10^{-2}\) joules.

### Final Answer:
The heat produced in joules is \(0.04\) joules or \(4 \times 10^{-2}\) joules.

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