Parallel combination of two capacitors, each of value `10muF` is charged by 200 volt d.c. Total energy of the capictor in joules will be :

Options

A. 0.1
B. 0.2
C. 0.4
D. 0.6

Answer

To find the total energy stored in the parallel combination of two capacitors, each with a capacitance of \(10 \mu F\) and charged by a \(200 V\) DC source, we can follow these steps:

### Step 1: Determine the Equivalent Capacitance
For capacitors in parallel, the equivalent capacitance \(C_{eq}\) is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 \]
Given that both capacitors have a capacitance of \(10 \mu F\):
\[ C_{eq} = 10 \mu F + 10 \mu F = 20 \mu F \]

### Step 2: Convert Capacitance to Farads
Convert microfarads to farads for calculations:
\[ C_{eq} = 20 \mu F = 20 \times 10^{-6} F \]

### Step 3: Use the Energy Formula
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
- \(V\) is the voltage in volts

Substituting the values:
\[ U = \frac{1}{2} \times (20 \times 10^{-6} F) \times (200 V)^2 \]

### Step 4: Calculate \(V^2\)
Calculate \(200^2\):
\[ 200^2 = 40000 \]

### Step 5: Substitute and Calculate Energy
Now substitute \(V^2\) back into the energy formula:
\[ U = \frac{1}{2} \times (20 \times 10^{-6}) \times 40000 \]
\[ U = \frac{1}{2} \times 20 \times 40000 \times 10^{-6} \]
\[ U = 10 \times 40000 \times 10^{-6} \]
\[ U = 400000 \times 10^{-6} \]
\[ U = 0.4 \text{ Joules} \]

### Final Answer
The total energy stored in the capacitors is:
\[ \boxed{0.4 \text{ Joules}} \]