Consider the combination of 2 capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$, with $\mathrm{C}_{2}>\mathrm{C}_{1}$, 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, $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$.
Correct Option: 2,
When connected in parallel
$\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$
When in series
$\mathrm{C}_{\mathrm{eq}}^{\prime}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
$\mathrm{C}_{1}+\mathrm{C}_{2}=\frac{15}{4}\left(\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)$
$4\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)^{2}=15 \mathrm{C}_{1} \mathrm{C}_{2}$
$4 \mathrm{C}_{1}^{2}+4 \mathrm{C}_{2}^{2}-7 \mathrm{C}_{1} \mathrm{C}_{2}=0$
dividing by $\mathrm{C}_{1}{ }^{2}$
$4\left(\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\right)^{2}-\frac{7 \mathrm{C}_{2}}{\mathrm{C}_{1}}+4=0$
Let $\frac{C_{2}}{C_{1}}=x$
$4 x^{2}-7 x+4=0$
$b^{2}-4 a c=49-64<0$
No solution exits
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