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 \[{{q}_{4}}={{C}_{v}}\times V=4CV\] The series combination of C1, C2 and C3 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 \[C''=C'+{{C}_{4}}\] \[=\frac{6C}{11}+4C=\frac{50\,C}{11}\] Net charge \[q=C''V\] \[=\frac{50}{11}CV\] Total charge flowing through \[{{C}_{1}},\,{{C}_{2}},\,{{C}_{3}}\] will be \[q'=q-{{q}_{4}}\] \[=\frac{50}{11}CV-4C\,V=\frac{6CV}{11}\] Since, \[{{C}_{1}},\,{{C}_{2}}\] and \[{{C}_{3}}\] are in series combination hence, charge flowing through these will be same. Hence, \[{{q}_{2}}={{q}_{1}}={{q}_{3}}=q'=\frac{6\,C\,V}{11}\] Thus, \[\frac{{{q}_{2}}}{{{q}_{4}}}=\frac{6CV/11}{4CV}=\frac{3}{22}\]
You need to login to perform this action.
You will be redirected in
3 sec
