Question

Four identical capacitors are connected as shown in the figure. When a battery of $9\text{V}$ is connected across $\text{A}$ and $\text{B},$ the charge stored is found to be $3\mu \text{C}$. The value of capacitance $C_1$ is

• A. $\frac{1}{15}\mu \text{F}$
• B. $1\mu \text{C}$
• C. $15\mu \text{C}$
• D. $\frac{2}{15}\mu \text{F}$

Answer 1

The correct option is D $\frac{2}{15}\mu \text{F}$

The middle branch has two capacitors each having capacitance $C_1$ are in series.

Equivalent capacitance of this middle branch:

$C_m=\frac{C_1\times C_1}{C_1+C_1}=\frac{C_1}{2}$

Now, all the three branches are in parallel.

Hence, equivalent capacitance across$\text{AB}$:

$C_{AB}=C_1+C_m+C_1$

$\Rightarrow C_{AB}=C_1+\frac{C_1}{2}+C_1=\frac{5C_1}{2}$

Given that when a battery of $9\text{V}$ is connected across $\text{A}$ and $\text{B},$ the charge stored is found to be $3\mu \text{C}$.

Now, $Q=C_{AB}V$

$\Rightarrow 3=\frac{5C_1}{2}\times 9$

$\Rightarrow C_1=\frac{2}{15}\mu \text{F}$

Hence, option (d) is the correct answer.

Key concept: If potential difference is the same across capacitors, they are in parallel combination. If $Q$ is the same, the capacitors are in a series combination.