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 $\frac{15}{4}$ times the equivalent capacitance of the same capacitors connected in series. Calculate the ratio of capacitors, $\frac{C_2}{C_1}$.

• A. $\frac{15}{11}$
• B. $\frac{29}{15}$
• C. $\frac{15}{4}$
• D. Imaginary

Answer 1

The correct option is D Imaginary

Given:
$C_{\text{eq}}\left(\text{parallel combination}\right)=\frac{15}{4}{C}_{\text{eq}}\left(\text{series combination}\right)$

$\Rightarrow (C_1+C_2)=\frac{15}{4}\left(\frac{C_1{C}_2}{C_1+C_2}\right)$

$\Rightarrow 15C_1{C}_2=4(C_1+C_2{)}^2$

$\Rightarrow 15C_1{C}_2=4C_1^2+4C_2^2+8C_1{C}_2$

$\Rightarrow 4C_1^2+4C_2^2-7C_1{C}_2=0$

$\Rightarrow 4+4{\left(\frac{C_2}{C_1}\right)}^2-7\left(\frac{C_2}{C_1}\right)=0$

$\Rightarrow 4+4x^2-7x=0$

$[$Assuming, $x=\frac{C_2}{C_1}]$

$\Rightarrow 4x^2-7x+4=0$

Discriminant,
$D=(-7{)}^2-4\times 4\times 4$
$\Rightarrow D=-15=$ Negative

Hence,
$x=\frac{C_2}{C_1}=$ Imaginary