Question

A person has only two capacitors. By using them singly, in parallel or in series, he is able to obtain the capacitance $2\mu F,3\mu F,6\mu F$ and $9\mu F$. What are the capacitance of the capacitors?

• A. $2\mu F$ and $3\mu F$
• B. $3\mu F$ and $6\mu F$
• C. $6\mu F$ and $9\mu F$
• D. $2\mu F$ and $9\mu F$

Answer 1

Answer: (B)

$3\mu F$ and $6\mu F$

For (A), $C_{eq}^s=\frac{2\times 3}{2+3}=\frac{6}{5}\mu F$
$C_{eq}^p=2+3=5\mu F$
For (B), $C_{eq}^s=\frac{3\times 6}{3+6}=2\mu F$
$C_{eq}^p=3+6=9\mu F$
For (C), $C_{eq}^s=\frac{6\times 9}{6+9}=\frac{18}{5}\mu F$
$C_{eq}^p=6+9=15\mu F$
For (D), $C_{eq}^s=\frac{2\times 9}{2+9}=\frac{18}{11}\mu F$
$C_{eq}^p=2+9=11\mu F$
Thus only (B) will give the capacitances which are expectable.