question_answer

Three capacitors each of capacitance \[1\mu F\]are connected in parallel. To this combination, a fourth capacitor of capacitance \[1\mu F\] is connected in series. The resultant capacitance of the system is                                              [MP PMT 1985]

A)                    \[4\mu F\]

B)                                      \[2\mu F\]

C)                    \[\frac{4}{3}\mu F\]  

D)            \[\frac{3}{4}\mu F\]

Correct Answer: D

Solution :

           The circuit can be drawn as follows \[\Rightarrow \,{{C}_{AB}}=\frac{3\times 1}{3+1}=\frac{3}{4}\mu F\]