Question

Five capacitors, each of capacity ' \(C\) ' are connected as shown in the figure. The resultant capacity between A and B is \(14 \mu \mathrm{~F}\). The capacity of each capacitor is

• A. \(2\mu \text{ }\mathrm{F}\)
• B. \(3.5\mu \text{ }\mathrm{F}\)
• C. \(4\mu \text{ }\mathrm{F}\)
• D. \(2.8\mu \text{ }\mathrm{F}\)

Hint:

Three capacitors in parallel add, the series pair gives C over 2, and the two branches are in parallel.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Parallel capacitors add, \(C_p=C_1+C_2+C_3\). Series capacitors combine as \(\dfrac1{C_s}=\dfrac1{C_1}+\dfrac1{C_2}\).

Step 2: Top branch
The three capacitors are in parallel: \(C_{top}=3C\).

Step 3: Bottom branch
The two capacitors in series give \(C_{bottom}=\dfrac{C}{2}\).

Step 4: Total
The two branches are in parallel between A and B: \(C_{AB}=3C+\dfrac{C}{2}=\dfrac{7C}{2}=14\ \mu\text{F}\).
\(C=4\ \mu\text{F}\).

Final Answer:
Each capacitor has capacitance 4 microfarad. \[ \boxed{\text{(C) }4\,\mu\text{F}} \]