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)

$C_{p a r a l l e l}=\frac{25}{6}C_{s e r i e s}$
$\Rightarrow C_{1}+C_{2}=\frac{25}{6}\frac{C_{1} C_{2}}{C_{1} + C_{2}}$
$\Rightarrow 6\left(C_{1} + C_{2}\right)^{2}=25C_{1}C_{2}$
$\Rightarrow 6C_{1}^{2}+6C_{2}^{2}+12C_{1}C_{2}=25C_{1}C_{2}$
$\Rightarrow 6\frac{C_{1}^{2}}{C_{2}^{2}}+6+12\frac{C_{1}}{C_{2}}=25\frac{C_{1}}{C_{2}}$
$\Rightarrow 6\frac{C_{1}^{2}}{C_{2}^{2}}-13\frac{C_{1}}{C_{2}}+6=0$
This is quadratic equation
By solving we get $\frac{C_{2}}{C_{1}}=\frac{3}{2}$