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 connected in series. Calculate the ratio of capacitors, $\frac{C_{2}}{C_{1}}$

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$\frac{15}{11}$

$\frac{15}{4}$

$\frac{111}{80}$

None of these

answer is D.

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Detailed Solution

$\begin{array}{l} Given : C_{P_eff} = \frac{15}{4} C_{S_eff} \\ \Rightarrow C_{1} + C_{2} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} \cdot \frac{15}{4} \\ Let \frac{C_{2}}{C_{1}} = x \\ \Rightarrow (1 + x) C_{1} = \frac{15}{4} \frac{x C_{1}^{2}}{(1 + x) C_{1}} \\ \Rightarrow (1 + x^{2} + 2 x) 4 = 15 x \\ \Rightarrow 4 x^{2} - 7 x + 4 = 0 \\ \Rightarrow x = \frac{7 \pm \sqrt{49 - 64}}{2} = imaginary roots \end{array}$