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 \[C=C'+C'\,'\] \[=2+2\] \[C=4\mu F\]You need to login to perform this action.
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