Three condensers of same capacity connected in series has effective capacity 2mF. If they are connected in parallel and charged using a battery of emf 12V, the total energy stored in the combination is

Options

A. 1296 mJ
B. 648 mJ
C. 162 mJ
D. 48 mJ

Answer

To solve the problem step by step, we will follow the outlined process to find the total energy stored in the combination of capacitors.

### Step 1: Determine the capacitance of each capacitor
Given that three capacitors of the same capacity (C) are connected in series and the effective capacitance (C_eq) is 2 mF, we can use the formula for capacitors in series:

\[ \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \]

From this, we can rearrange to find C:

\[ C_{eq} = \frac{C}{3} \implies C = 3 \times C_{eq} \]

Substituting the given value:

\[ C = 3 \times 2 \text{ mF} = 6 \text{ mF} \]

### Step 2: Calculate the effective capacitance when connected in parallel
When the three capacitors are connected in parallel, the effective capacitance (C_parallel) is the sum of the individual capacitances:

\[ C_{parallel} = C + C + C = 3C \]

Substituting the value of C we found:

\[ C_{parallel} = 3 \times 6 \text{ mF} = 18 \text{ mF} \]

### Step 3: Calculate the energy stored in the capacitors
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 we have:

\[ U = \frac{1}{2} \times 18 \text{ mF} \times (12 \text{ V})^2 \]

Converting milliFarads to Farads:

\[ 18 \text{ mF} = 18 \times 10^{-3} \text{ F} \]

Now substituting this into the energy formula:

\[ U = \frac{1}{2} \times 18 \times 10^{-3} \text{ F} \times 144 \text{ V}^2 \]

Calculating the energy:

\[ U = 9 \times 10^{-3} \times 144 = 1296 \times 10^{-3} \text{ J} \]

### Step 4: Convert Joules to milliJoules
Since \(1 \text{ J} = 1000 \text{ mJ}\):

\[ U = 1296 \text{ mJ} \]

### Final Answer
The total energy stored in the combination is **1296 milliJoules**.

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