question_answer

The equivalent capacitance in the following circuit is :

A)  \[2\mu F\]                                        

B) \[1\mu F\]

C) \[1.5\mu F\]                                      

D) \[3\mu F\]

Correct Answer: B

Solution :

                 The given circuit consists of two \[1.5\mu F\] capacitors connected in parallel and this arrangement is connected to two\[3\mu F\] capacitors in series. Resultant capacitance C in parallel is \[C={{C}_{1}}+{{C}_{2}}\] \[=1.5\mu F+1.5\mu F=3\mu F\]                 \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{C}\]                 \[=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1\] \[\Rightarrow \]               \[C=1\,\mu F\]