A) \[\,\frac{22}{3}\]
B) \[\,\frac{3}{22}\]
C) \[\,\frac{7}{4}\]
D) \[\,\frac{4}{7}\]
Correct Answer: B
Solution :
Key Idea: Charge on a capacitor is the product of capacitance and potential difference across it. The charge flowing through \[{{C}_{4}}\] is \[{{q}_{4}}={{C}_{4}}\times V=4CV\] The series combination of \[{{C}_{1}},\,{{C}_{2}}\] and \[{{C}_{3}}\] gives \[\frac{1}{C}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}\] \[=\frac{6+3+2}{6C}=\frac{11}{6C}\] \[\Rightarrow \] \[C=\frac{6C}{11}\] Now, C and \[{{C}_{4}}\] form parallel combination giving \[C=C+{{C}_{4}}\] \[=\frac{6C}{11}+4C=\frac{50C}{11}\] New charge \[q=C\] \[V=\frac{50}{11}CV\] Total charge flowing through \[{{C}_{1}},{{C}_{2}},{{C}_{3}}\] will be \[q=q-{{q}_{4}}\] \[=\frac{50}{11}CV-4CV\] \[=\frac{6\,CV}{11}\] Since, \[{{C}_{1}},{{C}_{2}}\]and \[{{C}_{3}}\] are in, series combination, hence charge flowing through these will be 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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