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Three capacitors of capacitance \mathrm{3\, \mu f,9\, \mu f\: and\: 18\, \mu f} are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases \mathrm{\left(\frac{C_{s}}{C_{p}}\right)} will be -

Option: 1

1: 15


Option: 2

15: 1


Option: 3

1: 1


Option: 4

1: 3


Answers (1)

best_answer

Given,

\mathrm{c_{1}=3 \, \mu \mathrm{f}}
\mathrm{c_{2}=9 \, \mu \mathrm{f}}
\mathrm{c_{3}=18 \, \mu \mathrm{f}}

Case I  All are in series -
              \mathrm{\frac{1}{c_{s}}=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}}
               \mathrm{\frac{1}{c_{s}}=\frac{6+2+1}{18}=\frac{9}{18}}
              \mathrm{\frac{1}{c_{s}}=\frac{1}{2} \Rightarrow c_{s}=2 \mu \mathrm{F}}

Case II   All are in parallel
              \mathrm{C_{p}=3+9+18=30\, \mu \mathrm{F}}

Now,
          \mathrm{\left(\frac{C_{s}}{C_{p}}\right)=\frac{2\, \mu f}{30\, \mu f}=\frac{1}{25} }  Ans

       

Posted by

SANGALDEEP SINGH

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