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: Simplify using equal potentials:
Label the battery terminals $P$ (left) and $Q$ (right). The two bottom wires mean the lower ends of the vertical 3C and the left side of the 2C are at potential $P$, and the lower ends of the 1C and the right 2C and the right side of the lower 2C are at potential $Q$.

Step 2: Reduce stage by stage:
Both 3C capacitors run from the top wire to $P$: $3C + 3C = 6C$.
The 1C and the right 2C both run from the top wire to $Q$: $1C + 2C = 3C$.
The top wire is the middle node of a series pair: $\dfrac{6\cdot3}{6+3}C = 2C$.
The lower 2C goes straight from $P$ to $Q$, so it adds: $2C + 2C = 4C$.

Step 3: Energy:
$U = \tfrac12\cdot4C\cdot V^2 = 2CV^2$.

Final Answer:
$4C$ and $2CV^2$, option (A). \[ \boxed{4C,\ 2CV^2 \text{ (A)}} \]