Question

Four capacitors $C_1,C_2,C_3$ and $C_4$ of capacitance $3\mu \text{F},4\mu \text{F},6\mu \text{F}$ and $2\mu \text{F}$ are charged upto a potential difference of $2\text{V},3\text{V},4\text{V}$ and $4\text{V}$ respectively, if terminal $\text{a}$ is connected with $\text{f}$, terminal $\text{d}$ is connected with $\text{g}$, terminal $\text{e}$ is connected to $\text{h}$ and $\text{b}$ is connected $\text{c}$, then find the charge flows in the circuit?

• A. $26\mu \text{C}$
• B. $\frac{156}{16}\mu \text{C}$
• C. $\frac{156}{15}\mu \text{C}$
• D. $18\mu \text{C}$

Answer 1

The correct option is C $\frac{156}{15}\mu \text{C}$

According to the problem and using $Q=CV$ we get,


Let $q$ amount of charge flows in the circuit,


Applying $\text{KVL}$ to the loop $\text{PQRS}$,

$\frac{24-q}{6}+\frac{8-q}{2}+\frac{12-q}{4}+\frac{6-q}{3}=0$

$\Rightarrow 48-2q+48-6q+36-3q+24-4q=0$

$\Rightarrow 156=15q$

$\therefore q=\frac{156}{15}\mu \text{C}$