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, \[q=C''V=\frac{50}{11}CV\] 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, \[{{q}_{2}}={{q}_{1}}={{q}_{3}}=q'=\frac{6CV}{11}\] 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
