Question

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, $\frac{ C _{2}}{ C _{1}}$

• A. $\frac{15}{11}$
• B. $\frac{111}{80}$
• C. $\frac{29}{15}$
• D. None of these

Answer 1

Answer: (D)

When connected in parallel
$C_ {eq }=C_{1}+C_{2}$
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 exits