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 \[\frac{1}{C}=\frac{3}{5}\] or \[C=1\,\,\mu F\]