question_answer

The equivalent capacitance in the given circuit is

A) \[\text{3}\mu \text{F}\]                              

B)  \[\text{1}.\text{5 }\mu \text{f}\]

C) \[\text{2}\mu \text{F}\]                              

D)  \[1\mu \text{F}\]

Correct Answer: D

Solution :

                 In the given circuit, two \[1.5\,\,\mu F\] capacitors are connected in parallel, hence equivalent capacitance  is                                 \[C={{C}_{1}}+{{C}_{2}}=1.5\times 2=3\mu F\] Now, the equivalent capacitance of 3 capacitors in series is                 \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}=\frac{3}{3}=\frac{1}{1}\] \[\Rightarrow \]               \[C=1\,\,\mu F\]