question_answer

The equivalent capacitance between points A and B of circuit shown in the figure is  9uF         luF  4.5 |af         6^F

A)  \[9\mu F\]        

B)  \[1\mu F\]

C)  \[4.5\mu F\]                     

D)  \[6\mu F\]

Correct Answer: A

Solution :

                 The given circuit is equivalent to three capacitors connected in parallel, hence                 \[C={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]                 C = 3 + 3 + 3 = 9 \[\mu F\]