question_answer

Each capacitor shown in figure is\[2\mu F\]. Then the equivalent capacitance between A and B is

A)  \[2\mu F\]   

B)  \[4\mu F\]  

C)  \[6\mu F\]   

D)  \[8\mu F\]

Correct Answer: A

Solution :

 Equivalent capacitance of upper arms in series \[{{C}_{1}}=\frac{2\times 2}{2+2}=1\mu F\] In lowers arms \[{{C}_{2}}=1\mu F\] Now\[{{C}_{1}}\]and\[{{C}_{2}}\] are in parallel \[\therefore \]Equivalent capacitance\[C={{C}_{1}}+{{C}_{2}}=2\,\mu F\]