Three capacitors of capacitance, \( 3\,\mu\mathrm{F} \), \( 9\,\mu\mathrm{F} \) and \( 18\,\mu\mathrm{F} \) respectively are connected once in series and then in parallel. The ratio of equivalent capacitance in series to parallel combination will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1 : 15 \)
The equivalent capacitance is the single capacitance that stores the same charge at the same voltage as the whole combination. In series the reciprocals of the capacitances add; in parallel the capacitances add.
\( \displaystyle \frac{1}{C_{s}} = \frac{1}{3} + \frac{1}{9} + \frac{1}{18} = \frac{6 + 2 + 1}{18} = \frac{9}{18} = \frac{1}{2} \)
\( C_{s} = 2\,\mu\mathrm{F} \)
\( C_{p} = 3 + 9 + 18 = 30\,\mu\mathrm{F} \)
\( \displaystyle \frac{C_{s}}{C_{p}} = \frac{2}{30} = \frac{1}{15} \)
So the ratio of series to parallel capacitance is \( 1 : 15 \).