Question

Five capacitors $C_1,C_2,C_3,C_4$ and $C_5$ of capacitance $6\mu \text{F}$, $3\mu \text{F}$, $4\mu \text{F}$, $3\mu \text{F}$ and $6\mu \text{F}$ are charged upto potential difference $5\text{V},4\text{V},4\text{V},4\text{V}$ and $6\text{V}$ respectively. If terminal $\text{a}$ is connected with $\text{f}$, terminal $\text{e}$ is connected with $\text{h}$, terminal $\text{g}$ is connected with $\text{d}$, $\text{b}$ is connected with $\text{c}$ and terminal $\text{bc}$ is connected with $\text{eh}$, then find the charge flows through the branch $\text{bc}$ and $\text{eh}$?

• A. $30\mu \text{C}$
• B. $36\mu \text{C}$
• C. $12\mu \text{C}$
• D. $15\mu \text{C}$

Answer 1

The correct option is C $12\mu \text{C}$

The charges of each capacitors are calculated using the formula, $Q=CV$ shown in the figure below,


Let $q$ amount of charge flows from capacitor $C_2$ to juction $\text{(c, b)}$ and it gets divided as $q_1$ and $q_2$ respectively as shown in figure.


As we know, in parallel branches, charge get divided w.r.t effective capcitance in the path.

Therefore,
$q_1=\left(\frac{C_3}{\frac{C_4{C}_5}{C_4+C_5}+C_3}\right)q$

$\Rightarrow q_1=\left(\frac{4}{\frac{3\times 6}{3+6}+4}\right)q$

$\Rightarrow q_1=\frac{4}{2+4}q=\frac{2}{3}q$

$\Rightarrow q_1=\frac{2}{2+4}=\frac{1}{3}q$

Applying $\text{KVL}$ to the loop $\text{bcafehbc}$, we get

$\frac{30-q}{6}+\frac{12-q}{3}+\frac{12-q_1}{4}=0$

$\Rightarrow 60-2q+48-4q+36-3q_1=0$

As, $q_1=\frac{2}{3}q$

$\Rightarrow 144=8q$

$\therefore q=18\mu \text{C}$

Charge through branch $\text{be, eh}$ is,

$q_1=\frac{2}{3}q=\frac{2}{3}\left(18\right)=12\mu \text{C}$

Hence, option (c) is correct answer.