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