Question

Three capacitors each of capacitance $6\mu F$ are arranged What would be the value of maximum and minimum capacitance of combination ?

Answer 1

(i)For maximum capacitance, they are connected in parallel :
$\therefore C_{max}=C_1+C_2+C_3=6\mu F+6\mu F+6\mu F=18\mu F$
(ii) For minimum capacitances, they should be connected in series :
$\frac{1}{C_{min}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{6\mu F}+\frac{1}{6\mu F}+\frac{1}{6\mu F}$
$\therefore \frac{1}{C_{min}}=\frac{1+1+1}{6\mu F}=\frac{3}{6\mu F}\Rightarrow C_{min}=2\mu F$