Question

If 3 capacitors of values 1, 2 and $3\mu F$ are available. The maximum and minimum values of capacitance one can obtain by different combinations of the three capacitors together are respectively:

• A. $6\mu F,\frac{6}{11}\mu F$
• B. $6\mu F\frac{11}{6}\mu F$
• C. $3\mu F,1\mu F$
• D. $4\mu F,2\mu F$

Answer 1

The correct option is A $6\mu F,\frac{6}{11}\mu F$

minimum capacity is obtained when they are in series
$\frac{1}{C_{eff}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$
$\therefore C_{eff}=\frac{6}{11}\mu F$
maximum capacity is obtained when they are in parallel
$C_{eff}=1+2+3$
$C_{eff}=6\mu F$