Three capacitors of 1μF , 2μF and 3μF are connected first in series. The same capacitors are again connected in parallel. Compare their equivalent capacitance. Which one of these combinations gives a larger value of capacitance.

Given,

Capacitance of there capacitors are,

$C_1 = 1 \mu F$;          $C_2 = 2 \mu F$;       $C_3 = 3 \mu F$

When they are connected in series, their equivalent capacitance $C_s$ is,

$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6}{11}\mu F$

⇒ $C_s = \frac{11}{6}\mu \; F$

When they are parallel, their equivalent capacitance $C_p$ is

$C_p = C_1 + C_2 + C_3 = 1 + 2 + 3 = 6 \; \mu F$

Now, to compare the capacitance of series and parallel combination,

$\frac{C_p}{C_s} = \frac{6} {\frac{6}{11}}$ $= 11$

$C_p = 11\;C_s$

The capacitance of the parallel combination is 11 times greater than that of the series combination.

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