Question

The connections shown below is established with the switch $\text{S}$ which is open at present. How much charge will flow through the swiitch if it is closed?

• A. $15\mu \text{C}$
• B. $20\mu \text{C}$
• C. $10\mu \text{C}$
• D. $30\mu \text{C}$

Answer 1

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

$\text{When S is open}$

$C_1$ and $C_2$ are in series and let their equivalent capacitance is $C^{\prime }$. Similarly, $C_3$ and $C_4$ are in series and their equivalent capacitance $C^{\prime \prime }$ and $C^{\prime }$ are in parallel.

Let charge flows through $C_1$ and $C_2$ be $q_1$ and charge through $C_3$ and $C_4$ be $q_2$.

$\therefore q_1=C^{\prime }V=\frac{3\times 6}{3+6}\left(10\right)=20\mu \text{C}$

$\therefore q_2=C^{\prime }v=\frac{3\times 6}{3+6}\left(10\right)=20\mu \text{C}$

Thus, $q_1=q_2=q$

We can distribute cahrge $q_1$ and $q_2$ in the given circuit as,


$\text{When switch S is closed}:$


From the above fig, we can observe that


By symmetry the potential difference at $\text{x}$ will be half of emf given,
$\therefore x=5\text{V}$

Using $\text{VL}$ at juction $\text{x}$, shown in fig (ii)

$q_1+q_2+q_3=0....\left(1\right)$

The new charge across capacitors$C_1$ and $C_2$ are,

$q_1^{\prime }=3(x-10)=3(5-10)=-15\mu \text{C}$

$q_3^{\prime }=6(x-0)=6(5-0)=30\mu \text{C}$

$q_1$ and $q_2$ is the difference of final and initial charge on capacitor$C_1$ and $C_2$ respectiely.

$q_1=q_1^{\prime }-(-q)=-15-(-20)=5\mu \text{C}$

$q_3=q_3^{\prime }-q=30-20=10\mu \text{C}$

Substituting the value of $q_1$ and $q_2$in eqn (1), we get

$5+q_2+10=0$

$\therefore \left|q_2\right|=15\mu \text{C}$

Hence, option (a) is correct answer.