A `4mu F` capacitor is charged to 400 volts and then its plates are joined through a resistor of resistance `1K Omega`. The heat produced in the resistor is

Options

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

Answer

To solve the problem of finding the heat produced in a resistor when a charged capacitor discharges through it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance (C) = 4 µF = \(4 \times 10^{-6}\) F
- Voltage (V) = 400 V
- Resistance (R) = 1 kΩ = \(1000\) Ω

2. **Calculate the energy stored in the capacitor:**
The energy (U) stored in a capacitor can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \]
Plugging in the values:
\[ U = \frac{1}{2} \times (4 \times 10^{-6} \, \text{F}) \times (400 \, \text{V})^2 \]

3. **Perform the calculations:**
First, calculate \(V^2\):
\[ V^2 = 400^2 = 160000 \, \text{V}^2 \]
Now substitute this back into the energy formula:
\[ U = \frac{1}{2} \times (4 \times 10^{-6}) \times 160000 \]
\[ U = \frac{1}{2} \times 0.64 \, \text{J} = 0.32 \, \text{J} \]

4. **Understanding the heat produced:**
When the capacitor discharges through the resistor, all the stored energy is converted into heat. Therefore, the heat produced (Q) in the resistor is equal to the energy stored in the capacitor:
\[ Q = U = 0.32 \, \text{J} \]

### Final Answer:
The heat produced in the resistor is **0.32 Joules**.

---