A) \[\frac{22}{3}\]
B) \[\frac{3}{22}\]
C) \[\frac{7}{4}\]
D) \[\frac{4}{7}\]
Correct Answer: B
Solution :
The charge in\[{{C}_{4}},{{q}_{4}}\times V=4CV\]the capacitors \[{{C}_{1}},{{C}_{2}}\]and \[{{C}_{3}}\]are connected in series. Let the resultant of these capacitors is \[C'.\] \[\frac{1}{C'}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}\] \[=\frac{6+3+2}{6C}=\frac{11}{6C}\] \[\Rightarrow \] \[C'=\frac{6C}{11}\] Now, C? and \[{{C}_{4}}\]form parallel combination giving \[{{C}^{n}}=C'+{{C}_{4}}\] \[=\frac{6C}{11}+4C=\frac{50C}{11}\] Net charge \[q=C''\,V\] \[=\frac{50}{11}CV\] Total charge flowing trough \[{{C}_{1}},{{C}_{2}}\]and \[{{C}_{3}}\]will be \[q'=q-{{q}_{4}}\] \[=\frac{50}{11}CV-4CV=\frac{6CV}{11}\] Since, \[\,{{C}_{1}},{{C}_{2}}\]and \[{{C}_{3}}\]are in series combination, hence, charge flowing through these will be the same. Hence, \[{{q}_{2}}={{q}_{1}}={{q}_{3}}=q'=\frac{6CV}{11}\] Thus, \[\frac{{{q}_{2}}}{{{q}_{4}}}=\frac{6CV/11}{4CV}=\frac{3}{22}\]
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