If n capacitors are connected in parallel to a V volt source then total energy of the system is:

Options

A. CV
B. `(1)/(2)CV^(2)`
C. `(1)/(2)nCV^(2)`
D. `(1)/(2n)CV^(2)`

Answer

To find the total energy of a system of n capacitors connected in parallel to a V volt source, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Parallel Connection**:
When capacitors are connected in parallel, the total or equivalent capacitance (C_eq) of the system can be calculated by simply adding the capacitances of all the individual capacitors. If each capacitor has a capacitance C, then for n capacitors connected in parallel, the equivalent capacitance is:
\[ C_{eq} = nC \]

2. **Using 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 and V is the voltage across the capacitor.

3. **Substituting the Equivalent Capacitance**:
Now, substituting the equivalent capacitance (C_eq) into the energy formula, we have:
\[ U = \frac{1}{2} C_{eq} V^2 \]
Substituting \( C_{eq} = nC \):
\[ U = \frac{1}{2} (nC) V^2 \]

4. **Final Expression for Total Energy**:
Therefore, the total energy stored in the system of n capacitors connected in parallel to a V volt source is:
\[ U = \frac{1}{2} n C V^2 \]

### Final Answer:
The total energy of the system is \( \frac{1}{2} n C V^2 \).

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