Question

The amount of charge flowing through branches $\text{A}$, $\text{B}$ and $\text{C}$ when switch $S_n$ is closed as shown in figure are repectively
(Assume that the charge delivered by capacitor is negative)

.

• A. $50\mu \text{C},100\mu \text{C},100\mu \text{C}$
• B. $-50\mu \text{C},100\mu \text{C},+50\mu \text{C}$
• C. $100\mu \text{C},-100\mu \text{C},50\mu \text{C}$
• D. $-100\mu \text{C},50\mu \text{C},100\mu \text{C}$

Answer 1

The correct option is B $-50\mu \text{C},100\mu \text{C},+50\mu \text{C}$

When the switch $s_n$ is open, the equivalent circuit is shown in the figure below:

$\frac{1}{C_{eq}}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}\Rightarrow c_{eqn}=5\mu \text{F}$

Since, $C_1$ and $C_2$ are in series, therefore same amount of charge will flow across them

$Q_{initial}=C_{eqn}V=5\left(30\right)=150\mu \text{C}$

Now, when the switch $s_n$ is closed, the equivalent circuit is as shown below:



Now potential difference $C_1$ and $C_2$ becomes $10\text{V}$ and $20\text{V}$ respectively.

Thus, $Q_{final}$ across $C_1$ and $C_2$ becomes $100\mu \text{C}$ and $200\mu \text{C}$ respectively.

$\Delta q$ across branch $\text{A}$:

$Q_{final}-q_{initial}=(100-150)\mu C=-50\mu C$

$\because$ $-ve$ sign indicates that $C_1$ releases $50\mu$C

$\Delta q$ across branch $\text{C}$:

$Q_{final}-q_{initial}=(200-150)\mu C=+50\mu C$

$\therefore$ $+ve$ sign indicates that $C_2$ recieves $50\mu \text{C}$, this required amount of charge will be pushed by battery of $20\text{V}$.

Thus, the final charge distribution becomes,

By applying $KCL$ at the junction, the charge flowing through the branch $\text{B}$ is given by

$q=50+50=100\mu \text{C}$

Hence, option $\left(b\right)$ is the correct answer.