question_answer

A network of four capacitors of capacity equal to \[{{C}_{1}}={{C}_{1}}=2C,\,{{C}_{3}}=3C\]and\[{{C}_{4}}=4C\]are connected to in a battery as shown in the figure. The ratio of the charges on \[{{C}_{2}}\]and \[{{C}_{4}}\]is:

A) \[\frac{22}{3}\]                                  

B)  \[\frac{3}{22}\]

C) \[\frac{7}{4}\]            

D)      \[\frac{4}{7}\] 

Correct Answer: B

Solution :

Given circuit can be reduce to For series combination of \[{{C}_{1}},{{C}_{2}},{{C}_{3}}\]resultant capacity \[{{C}_{s}}=\frac{6C}{6C+3C+2C}\]or \[{{C}_{s}}=\frac{6C}{11}\] Now \[\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{C}_{s}}}{{{C}_{4}}}=\frac{6C/11}{4C}=\frac{3}{22}\] Hence, the correction option is [b].