question_answer

In the given Figure the capacitors C1, C3, C4 and C5 have a capacitance of 4\[\mu \]F each,. If the capacitor C2 has a capacitance of \[4\mu F\]if, then effective capacitance between A and B is:  

A) 8\[\mu \]F                                         

B) 6\[\mu \]F

C) 4\[\mu \]F                                         

D) 2\[\mu \]F

Correct Answer: C

Solution :

Key Idea: The setup is a balanced Wheat stones, bridge. The ratio of capacitances in the opposite arms is same, hence it is a balanced Wheatstone bridge. Therefore,   middle capacitor  becomes ineffective. For resultant capacitance in series \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{4}}}\] \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{4}}}\] \[\Rightarrow \]   \[C'=\frac{{{C}_{1}}{{C}_{4}}}{{{C}_{1}}+{{C}_{4}}}=\frac{4\times 4}{4+4}=2\mu F\] Similarly, resultant of \[{{C}_{3}}\] and \[{{C}_{5}}\] is \[C'\,'=\frac{{{C}_{3}}{{C}_{5}}}{{{C}_{3}}+{{C}_{5}}}=2\mu F\] Now C' and C' ' are in parallel. \[\therefore \]  Resultant capacitance \[C=C'+C'\,'\] \[=2+2\] \[C=4\mu F\]