question_answer

Three condensers of capacitance 2uF 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 \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\] Given, \[{{C}_{1}}={{C}_{2}}={{C}_{3}}=2\mu F\] \[\therefore \] \[\frac{1}{C}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\] \[\Rightarrow \] \[C=\frac{2}{3}\mu F\]