question_answer

Three capacitors of capacitances 3\[\mu \]F, 9\[\mu \]F and 18\[\mu \]F are connected once in series and another time in parallel. The ratio of equivalent capacitances in the two cases will be

A)  1 : 15                                    

B)  15 : 1

C)  1 : 1                                      

D)  1 : 3

Correct Answer: A

Solution :

                 The equivalent capacitance in parallel is                 \[{{C}_{p}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]                 = 3 + 9 + 18 = 30 \[\mu F\]                 Equivalent capacitance in series is                 \[\frac{1}{{{C}_{s}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\]                 \[=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}=\frac{1}{2}\] \[\Rightarrow \]               \[{{C}_{s}}=2\,\mu F\] \[\therefore \]  \[\frac{{{C}_{s}}}{{{C}_{p}}}=\frac{2}{30}=\frac{1}{15}\]