Question

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

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

Answer 1

Answer: (D)

$5$ in parallel, $2$ in series connected in series

For A: $C_p=2+2=4\mu F$ and $C_s=(1/2+1/2+1/2+1/2+1/2{)}^{-1}=2/5\mu F$ so $C_{eq}=\frac{C_p{C}_s}{C_p+C_s}=\frac{4\times 2/5}{4+2/5}=\frac{8}{5}\times \frac{5}{22}=\frac{4}{11}\mu F$.

For B: $C_p=2+2+2=6\mu F$ and $C_s=(1/2+1/2+1/2+1/2{)}^{-1}=1/2\mu F$ so $C_{eq}=\frac{C_p{C}_s}{C_p+C_s}=\frac{6\times 1/2}{6+1/2}=3\times \frac{2}{13}=\frac{6}{13}\mu F$.

For C: $C_p=2+2+2+2=8\mu F$ and $C_s=(1/2+1/2+1/2{)}^{-1}=2/3\mu F$ so $C_{eq}=\frac{C_p{C}_s}{C_p+C_s}=\frac{8\times 2/3}{8+2/3}=\frac{16}{3}\times \frac{3}{26}=\frac{8}{13}\mu F$.

For D: $C_p=2+2+2+2+2=10\mu F$ and $C_s=(1/2+1/2{)}^{-1}=1\mu F$ so

$C_{eq}=\frac{C_p{C}_s}{C_p+C_s}=\frac{10\times 1}{10+1}=\frac{10}{11}\mu F$.