question_answer

A network of four capacitors of capacity equal to \[{{C}_{1}}=C,\,\,{{C}_{2}}=2C,\,{{C}_{3}}=3C\] and \[{{C}_{4}}=4C\] are conducted in a battery as shown in the figure. The ratio of the charges on \[{{C}_{2}}\] and \[{{C}_{4}}\] is                                                                  [CBSE PMT 2005]

A)            \[\frac{22}{3}\]

B)            \[\frac{3}{22}\]

C)            \[\frac{7}{4}\]

D)            \[\frac{4}{7}\]

Correct Answer: B

Solution :

           The given circuit can be redrawn as follows                     \[{{C}_{eq}}=\frac{{{C}_{1}}{{C}_{2}}{{C}_{3}}}{{{C}_{1}}{{C}_{2}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{1}}}=\frac{6C}{11}\] \[\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{{{C}_{A}}}{{{C}_{B}}}=\frac{6C/11}{4C}=\frac{3}{22}\]