Question

Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are

• A. $2\mu F$ and $18\mu F$
• B. $5\mu F$ and $5\mu F$
• C. $7\mu F$ and $3\mu F$
• D. $8\mu F$ and $2\mu F$

Answer 1

The correct option is A $2\mu F$ and $18\mu F$

the minimum capacitance can be obtained by connecting all capacitors in series. Hence this can be calculated as follows:-

$\frac{1}{C}=$ $\frac{1}{6}+$$\frac{1}{6}+$$\frac{1}{6}=\frac{1}{2}$

$C=2\mu F$

Also the maximum capacitance can be obtained by connecting all capacitors in parallel. Hence this can be calculated as follows:-

$C=6+6+6=18\mu F$

So, the correct answer is option (A).