The heat generated through `4 Omega and 9 Omega` resistances separately , when a capacitor pf `100 my F` capacity charged to `200 V` is discharged one by one , will be

Options

A. `2 J and 8 J, respectively
B. `8 J and 2 J, respectively
C. `2 J and 4 J, respectively
D. `2 J and 2 J, respectively

Explanation

To solve the problem of calculating the heat generated through 4 ohm and 9 ohm resistances when a capacitor of 100 µF capacity charged to 200 V is discharged, we can follow these steps:

### Step 1: Understand the Energy Stored in the Capacitor
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where:
- \( C \) is the capacitance in farads (F),
- \( V \) is the voltage in volts (V).

### Step 2: Substitute the Given Values
In this case:
- \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- \( V = 200 \, V \)

Substituting these values into the formula:
\[ U = \frac{1}{2} \times (100 \times 10^{-6}) \times (200)^2 \]

### Step 3: Calculate the Energy
Now, we calculate \( U \):
\[ U = \frac{1}{2} \times 100 \times 10^{-6} \times 40000 \]
\[ U = \frac{1}{2} \times 100 \times 10^{-6} \times 4 \times 10^4 \]
\[ U = \frac{1}{2} \times 100 \times 4 \times 10^{-2} \]
\[ U = 200 \times 10^{-2} = 2 \, J \]

### Step 4: Heat Generated in the Resistors
The energy stored in the capacitor is completely converted into heat when it discharges through the resistors. Therefore, the heat generated through both resistances (4 ohm and 9 ohm) will be equal to the energy stored in the capacitor.

### Conclusion
Thus, the heat generated through both the 4 ohm and 9 ohm resistors when the capacitor is discharged will be:
- Heat through 4 ohm resistor: \( 2 \, J \)
- Heat through 9 ohm resistor: \( 2 \, J \)

**Final Answer:** The heat generated through both resistances is \( 2 \, J \) and \( 2 \, J \) respectively.

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