question_answer

A network of six identical capacitors, each of value C, is made as shown in the figure. The equivalent capacitance between the points A and B is:

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                 \[\frac{1}{C\,\,}=\frac{1}{C}+\frac{3}{4C}+\frac{1}{C}\]                 \[\frac{1}{C\,\,}=\frac{11}{4C}\] \[\Rightarrow \]               \[C\,\,\,=\frac{4C}{11}\]