Question

Seven capacitors, each of capacitance $2\mu F$ are to be connected to obtain a capacitance of $10/11\mu F$. Which of the following combinations is possible?

• A. $5$ in parallel $2$ in series
• B. $4$ in parallel $3$ in series
• C. $3$ in parallel $4$ in series
• D. $2$ in parallel $5$ in series

Answer 1

Answer: (A)

$5$ in parallel $2$ in series

When 5 in parallel 2 in series, $C_{eq}=\frac{(5\times 2)\left(2/2\right)}{(5\times 2)+\left(2/2\right)}=\frac{10}{10+11}=\frac{10}{11}\mu F$
When 4 in parallel 3 in series , $C_{eq}=\frac{(4\times 2)\left(2/3\right)}{(4\times 2)+\left(2/3\right)}=\frac{16/3}{26/3}=\frac{8}{13}\mu F$
When 3 in parallel 4 in series , $C_{eq}=\frac{(3\times 2)\left(2/4\right)}{(3\times 2)+\left(2/4\right)}=\frac{3}{13/2}=\frac{6}{13}\mu F$
When 2 in parallel 5 in series , $C_{eq}=\frac{(2\times 2)\left(2/5\right)}{(2\times 2)+\left(2/5\right)}=\frac{8/5}{22/5}=\frac{4}{11}\mu F$