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, $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$
Note: NTA has dropped this question in the final official answer key.
Correct Option: , 4
(4)
$C_{1}+C_{2}=\frac{15}{4}\left(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\right)$
$4\left(C_{1}+C_{2}\right)^{2}=15 C_{1} C_{2}$
$4 C_{1}^{2}+4 C_{2}^{2}-7 C_{1} C_{2}=0$
$4+4\left(\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\right)^{2}-7 \frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=0$
$4\left(\frac{C_{2}}{C_{1}}\right)^{2}-7 \frac{C_{2}}{C_{1}}+4=0$
$\frac{C_{2}}{\mathrm{C}_{1}}$ has not real value
$\frac{C_{2}}{C_{1}}=$ imaginary
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