Question

The effective capacitance of two capacitors of capacitances $C_1$ and $C_2\left(C_2>C_1\right)$ connected in parallel is $\frac{25}{6}$ times the effective capacitance when they are connected in series. The ratio $\frac{C_2}{C_1}$ is

• A. $\frac{3}{2}$
• B. $\frac{4}{3}$
• C. $\frac{5}{3}$
• D. $\frac{25}{6}$

Answer 1

Answer: (A)

$C_{parallel}=\frac{25}{6}{C}_{series}$
$\Rightarrow C_1+C_2=\frac{25}{6}\frac{C_1{C}_2}{C_1+C_2}$
$\Rightarrow 6{\left(C_1+C_2\right)}^2=25C_1{C}_2$
$\Rightarrow 6C_1^2+6C_2^2+12C_1{C}_2=25C_1{C}_2$
$\Rightarrow 6\frac{C_1^2}{C_2^2}+6+12\frac{C_1}{C_2}=25\frac{C_1}{C_2}$
$\Rightarrow 6\frac{C_1^2}{C_2^2}-13\frac{C_1}{C_2}+6=0$
This is quadratic equation
By solving we get $\frac{C_2}{C_1}=\frac{3}{2}$