question_answer

Three condensers of capacitance \[2\mu F\] each are connected in series. The resultant capacitance is:

A) \[6\mu F\]                         

B) \[3/2\mu F\]

C)  \[2/3\mu F\]                    

D)  \[5\mu F\]

Correct Answer: C

Solution :

The resultant capacitance of three capacitors in series is                 \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\]