A) \[\frac{4C}{11}\]
B) \[3C/4\]
C) \[3C/2\]
D) \[3C\]
E) \[4C/3\]
Correct Answer: A
Solution :
In the given circuit capacitors (1) (2) and (3) are connected in series, hence equivalent capacitance is \[\frac{1}{C}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}\] \[\Rightarrow \] \[C=\frac{C}{3}\]. This is connected in parallel with (4). \[\therefore \]\[C\,=C+C=\frac{C}{3}+C=\frac{4C}{3}.\] The three capacitors (5),\[\frac{4C}{3},\](6) are now connected in series. \[\therefore \] Equivalent capacitance is \[\frac{1}{C\,\,}=\frac{1}{C}+\frac{3}{4C}+\frac{1}{C}\] \[\frac{1}{C\,\,}=\frac{11}{4C}\] \[\Rightarrow \] \[C\,\,\,=\frac{4C}{11}\]You need to login to perform this action.
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