Question

Consider the combination of $2$ capacitors $C_1$ and $C_2,$ with $C_2>C_1,$ when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ time the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $\frac{C_2}{C_1}$

• A. $\frac{15}{11}$
• B. $\frac{111}{80}$
• C. $\frac{29}{15}$
• D. None of these

Answer 1

Answer: (D)

When connected in parallel
$C_{eq}=C_1+C_2$
When in series
$C_{eq}^{\prime }=\frac{C_1{C}_2}{C_1+C_2}$
$C_1+C_2=\frac{15}{4}\left(\frac{C_1{C}_2}{C_1+C_2}\right)$
$4{\left(C_1+C_2\right)}^2=15C_1{C}_2$
$4C_1^2+4C_2^2-7C_1{C}_2=0$
dividing by $C_1{}^2$
$4{\left(\frac{C_2}{C_1}\right)}^2-\frac{7C_2}{C_1}+4=0$
Let $\frac{C_2}{C_1}=x$
$4x^2-7x+4=0$
$b^2-4ac=49-64<0$
No solution exits