Question

Five identical capacitors each of magnitude $5\mu F$ are arranged with a battery of emf $20V$ as shown in the figure. Initially, the switch is opened. The charge that flows from point $A$ to point $B$, when the switch is closed is given as $-\frac{x}{7}\mu C$. Find $x$

Answer 1

$-\frac{x}{7}\mu C5+5=10\mu F\frac{5\times 5}{5+5}=\frac{25}{10}=2.5\mu F$

$2.5+5=7.5\mu F\frac{7.5\times 10}{7.5+10}=\frac{75\times 10}{17.5}=\frac{30}{7}$

$X=30$