A `4muF` capacitor is charged to 400 V. If its plates are joined through a resistance of `2kOmega`, then heat produced in the resistance is

Options

A. 0.16 J
B. 0.32 J
C. 0.64 J
D. 1.28 J

Explanation

To solve the problem, we need to calculate the heat produced in the resistance when a charged capacitor discharges through it. We can use the formula for the energy stored in a capacitor, which is given by:

\[ H = \frac{1}{2} C V^2 \]

where:
- \( H \) is the heat produced (or energy stored),
- \( C \) is the capacitance in farads,
- \( V \) is the voltage in volts.

### Step-by-Step Solution:

1. **Identify the values given in the problem:**
- Capacitance, \( C = 4 \mu F = 4 \times 10^{-6} F \)
- Voltage, \( V = 400 V \)

2. **Convert the capacitance to SI units:**
- \( C = 4 \mu F = 4 \times 10^{-6} F \)

3. **Substitute the values into the formula for heat produced:**
\[ H = \frac{1}{2} C V^2 \]
\[ H = \frac{1}{2} \times (4 \times 10^{-6}) \times (400)^2 \]

4. **Calculate \( V^2 \):**
\[ V^2 = 400^2 = 160000 \]

5. **Substitute \( V^2 \) back into the equation:**
\[ H = \frac{1}{2} \times (4 \times 10^{-6}) \times 160000 \]

6. **Calculate the product:**
\[ H = \frac{1}{2} \times (4 \times 160000 \times 10^{-6}) \]
\[ H = \frac{1}{2} \times (640000 \times 10^{-6}) \]
\[ H = 320000 \times 10^{-6} \]
\[ H = 0.32 \text{ Joules} \]

### Final Answer:
The heat produced in the resistance is \( 0.32 \) Joules.