# Consider the combination of 2 capacitors

Question:

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

1. $\frac{15}{11}$

2. $\frac{111}{80}$

3. $\frac{29}{15}$

4. $\frac{15}{4}$

Correct Option: 2,

Solution:

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