Question

Three capacitors $A,B$ and $C$ are connected in a circuit as shown in figure. What is the charge in $\mu C$ on the capacitor $B$?

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $1$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

Here $B$ and $C$ are in parallel and this group $(B+C)$ and $A$ are in series.

Thus equivalent capacitance $,C_{eq}=\frac{(C_C+C_B)C_A}{(C_C+C_B)+C_A}=\frac{(3+4)2}{(3+4)+2}=\frac{14}{9}\mu F$

$Q_{eq}=C_{eq}(7-6)=\frac{14}{9}\mu C$

As group $(B+C)$ and $A$ are in series so charge on both $(B+C)$ and $A$ is equal to $Q_{eq}$. As $B$ and $C$ are in parallel so $Q_{eq}$ is divided into $B$ and $C$.

Thus charge on $B$ is $Q_B=\frac{C_B}{C_B+C_C}{Q}_{eq}=\frac{3}{3+4}\times \frac{14}{9}=\frac{3}{7}\times \frac{14}{9}=\frac{2}{3}\mu C$