Five capacitors each of capacity 'C' are connected as shown in figure. If their resultant capacity is $2 \mu…
Five capacitors each of capacity 'C' are connected as shown in figure. If their resultant capacity is $2 \mu \mathrm{F}$, then the capacity of each condenser is
$2.5 \mu \mathrm{F}$
$2\mu \mathrm{F}$
$10\mu \mathrm{F}$
$5\mu \mathrm{F}$
Solution
Given, resultant capacity, $\mathrm{C}_{\mathrm{eq}}=2 \mu \mathrm{F}$
The equivalent capacity of the circuit is given by $\frac{1}{C_{a q}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \ldots(\because$ all capacitors are in series)
$\begin{array}{l}
\frac{1}{C_{e q}}=\frac{5}{C} \\
\frac{1}{2 \mu F}=\frac{5}{C} \\
\Rightarrow C=10 \mu F
\end{array}$
