question_answer

The equivalent capacitance between the points A and B in the following circuit is:

A) \[1\mu F\]                                         

B)         \[2\mu F\]

C)         \[4\mu F\]                                        

D)         \[8\mu F\]

Correct Answer: A

Solution :

The two capacitors each of value \[1.5\,\mu F\]are in parallel, so their equivalent capacitance. \[{{C}_{1}}=1.5+1.5=3\,\mu F\] Now, three capacitors each of value \[3\,\mu F\]are in series. Hence, their equivalent capacitance is given by \[\frac{1}{C}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\] or                            \[\frac{1}{C}=\frac{3}{3}\] or                            \[C=1\,\mu F\]