question_answer

In the given circuit when key S is closed then charge on the capacitors A and B are respectively

A)  3q, 6q           

B)  6q, 3q

C)  4.5q, 4.5q       

D)  5q, 4q

Correct Answer: B

Solution :

 Let charges is\[{{q}_{1}}\]and\[{{q}_{2}}\]after closing key S then \[V=\frac{{{q}_{1}}}{6C}=\frac{{{q}_{2}}}{3C}\] \[\Rightarrow \] \[\frac{{{q}_{1}}}{2}={{q}_{2}}\] \[\Rightarrow \] \[{{q}_{1}}=2{{q}_{2}}\]             ...(i) Charge is always conserved \[\therefore \] \[{{q}_{1}}+{{q}_{2}}=3q+6q\] or      \[{{q}_{1}}+{{q}_{2}}=9q\]                  ...(ii) From Eqs. (i) and (ii) \[2{{q}_{2}}+{{q}_{2}}=9q\] \[\Rightarrow \] \[3{{q}_{2}}=9q\] From Eq. (i) \[{{q}_{1}}=2\times 3q\] \[\Rightarrow \] \[{{q}_{1}}=6q\] \[(\because {{q}_{1}}=6q,{{q}_{2}}=3q)\]