question_answer

Each capacitor shown in figure is 2uF. 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 :

The figure is a bellowed Wheatstone bridge. So, diagonal capacitor will be in effective. So, the equivalent circuit will be as shown in the figure. Equivalent capacitor of upper arms in series \[{{C}_{1}}=\frac{2\times 2}{2+2}=1\mu F\] Similarly for lower arm \[{{C}_{2}}=\mu F\] \[{{C}_{AB}}={{C}_{1}}+{{C}_{2}}\] \[=1+1=2\mu F\]