Consider the combination of \(2\) capacitors \(C _{1}\) and \(C _{2},\) with \(C _{2}> C _{1},\) when connected in parallel, the equivalent capacitance is \(\frac{15}{4}\) time the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, \(\frac{ C _{2}}{ C _{1}}\)

When in series

\(C _{ eq }^{\prime}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}\)

\(C _{1}+ C _{2}=\frac{15}{4}\left(\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}\right)\)

\(4\left( C _{1}+ C _{2}\right)^{2}=15 C _{1} C _{2}\)

\(4 C _{1}{ }^{2}+4 C _{2}^{2}-7 C _{1} C _{2}=0\)

dividing by \(C _{1}{ }^{2}\)

\(4\left(\frac{ C _{2}}{ C _{1}}\right)^{2}-\frac{7 C _{2}}{ C _{1}}+4=0\)

Let \(\frac{ C _{2}}{ C _{1}}= x\)

\(4 x^{2}-7 x+4=0\)

\(b ^{2}-4 ac =49-64<0\)

No solution exits981-s782