Question

In the circuit shown in the figure, initially switch S is open. When the switch is closed, the charge passing through the switch is _________ $\mu C$ in the direction A to B.

Answer 1

When switch is open $2\mu F$ and $3\mu F$ are in series, so their equivalent is given as

$\frac{1}{C_{eq\cdot }}=\frac{1}{2\mu F}+\frac{1}{3\mu F}$

$=\frac{3+2}{6}$

$\Rightarrow C_{eq\cdot }=\frac{6}{5}\mu F$

Cells are in series with same polarity, so, total voltage across combination is $(60V+60V)$

$\therefore$ charge on each capacitor

$=(60V+60V)\left(\frac{6}{5}\mu F\right)(q=cv)$

$=144\mu C$

Total charge in loop = 0



On key is closed, potential differnce across $2\mu F$ and $3\mu F$ is 60 V and 60 V respectively

$\therefore$ charge of $2\mu F$ = $\left(2\mu F\right)\left(60V\right)$

$=120\mu C$

Charge on $3\mu F=\left(3\mu F\right)\left(60V\right)$

$=180\mu C$



$\left(Q_{total}\right)$ in loop = $+180\mu C-120\mu C=60\mu C$

Thus, $+60\mu C$ flows from A to B, as this is the only path thrugh which charge can enter the plates inside the loop.