Question

In the given circuit switch $K$ is open. The charge on the capacitor is ' $C$ ' in steady state is $q _{1}$. Now key is closed and steady state charge on $C$ is $q _{2} .$ The ratio of charges $q _{1} / q _{2}$ is-

• A. $3 / 2$
• B. $2 / 3$
• C. $1$
• D. $1 / 2$

Answer 1

Answer: (A)

Initially when switch is open

Finally when switch is closed

$I =\frac{ E }{3 R }$;
$\therefore V _{ A }- V _{ B }= I 2 R =( E / 3 R ) \times 2 R =2 E / 3$
$\therefore $ Charge on capacitor $= CV ; q _{2}=\frac{ C 2 E }{3}$
$\therefore \frac{ q _{1}}{ q _{2}}=\frac{ CE }{2 / 3 CE }$ $=3: 2$