question_answer

Five capacitors are connected as shown in the figure. The equivalent capacitance between A and B is :                    

A)  4\[\mu \]F                                        

B)  3\[\mu \]F

C)  2\[\mu \]F                                        

D)  1\[\mu \]F

Correct Answer: C

Solution :

                Equivalent capacitance of \[2\mu F\] and \[2\mu F\] connected   in  series  is  given  by \[=\frac{2\times 2}{2+2}=1\mu F\]                 Now, equivalent capacitance ot \[1\mu F\] and \[1\mu F\] connected in parallel is \[=1+1=2\mu F\] equivalent circuit Now \[2\mu F\] and \[2\mu F\] are in connected in series                                    \[=\frac{2\times 2}{2+2}=1\mu F\] Equivalent capacitance between A and B is given by,                 \[C=1+1=2\mu F\]