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 the resistor when a charged capacitor is connected across it, we can follow these steps:

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

### Step 2: Calculate the energy stored in the capacitor
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
Substituting the known values into the formula:
\[ U = \frac{1}{2} \times (4 \times 10^{-6}) \times (400)^2 \]

### Step 3: Calculate \(V^2\)
First, calculate \(400^2\):
\[ 400^2 = 160000 \]

### Step 4: Substitute \(V^2\) back into the energy formula
Now substitute \(V^2\) back into the energy equation:
\[ U = \frac{1}{2} \times (4 \times 10^{-6}) \times 160000 \]

### Step 5: Calculate the energy
Now calculate the energy:
\[ U = \frac{1}{2} \times (4 \times 10^{-6}) \times 160000 = 2 \times 10^{-6} \times 160000 = 0.32 \text{ Joules} \]

### Step 6: Conclusion
The heat produced in the resistor when the capacitor discharges is equal to the energy stored in the capacitor, which is:
\[ \text{Heat produced} = 0.32 \text{ Joules} \]

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

---