Question

Figure shows a network of five capacitors connected to a supply voltage 'V'. The equivalent capacitance and the energy stored in the network is respectively

• A. \(4C\) , \(2CV^2\)
• B. \(6C\) , \(3CV^2\)
• C. \(9C\) , \(4CV^2\)
• D. \(11C\) , \(9CV^2\)

Hint:

Trace the nodes. The bottom wires join the left end of the 2C capacitor to the battery's left terminal and its right end to the right terminal.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The figure shows five capacitors: two of 3C, one of 2C on the right branch, one of 1C, and one of 2C in the lower middle, connected across a supply of voltage \(V\) at the bottom. First we need to see which junctions are the same point electrically.

Step 2: Key Formula or Approach:
Parallel: \(C = C_1 + C_2\). Series: \(\dfrac1C = \dfrac{1}{C_1} + \dfrac{1}{C_2}\). Energy: \(U = \dfrac12C_{eq}V^2\).

Step 3: Detailed Explanation:
The bottom wire joins the left battery terminal to the bottom end of the vertical 3C capacitor, and the right battery terminal to the bottom end of both the 1C and the right 2C capacitors. So call the left terminal \(L\), the right terminal \(R\), and the common top wire \(T\).
Between \(T\) and \(L\): the top-left 3C and the vertical 3C are in parallel, giving \(3C + 3C = 6C\).
Between \(T\) and \(R\): the 1C and the right 2C are in parallel, giving \(1C + 2C = 3C\).
These two groups are in series from \(L\) to \(R\) through \(T\):
\[ \frac{6C\times3C}{6C+3C} = 2C \]
The lower 2C capacitor lies directly between \(L\) and \(R\), so it is in parallel with this combination:
\[ C_{eq} = 2C + 2C = 4C \]
Energy:
\[ U = \frac12(4C)V^2 = 2CV^2 \]

Final Answer:
The equivalent capacitance is \(4C\) and the energy is \(2CV^2\), option (A). \[ \boxed{4C,\ 2CV^2 \text{ (A)}} \]