Question

Four identical capacitors are connected as shown in diagram. When a battery of $6\text{V}$ is connected between A and B, the charge stored is found to be $1.5\mu \text{C}$. The value of $C_1$ is: -

• A. $2.5\mu \text{F}$
• B. $15\mu \text{F}$
• C. $1.5\mu \text{F}$
• D. $0.1\mu \text{F}$

Answer 1

The correct option is D $0.1\mu \text{F}$

The given circuit is,

It can be redrawn as,


Now equivalent capacitance between A and B,

$C_{\text{eq}}=\frac{C_1}{2}+C_1+C_1=\frac{5C_1}{2}$

Net charge of the circuit is,

$Q=C_{\text{eq}}V$

i.e. $1.5=\frac{5C_1}{2}\times 6$

$C_1=0.1\mu \text{F}$