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 $\frac{15}{4}$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $\frac{C_2}{C1}$

• A. Imaginary
• B. $\frac{15}{11}$
• C. $\frac{29}{15}$
• D. $\frac{15}{4}$

Answer 1

The correct option is A

Imaginary

Step 1. Given data

Equivalent Capacitance of the capacitors connected in parallel $=\frac{15}{4}\times$Equivalent capacitance of the capacitors connected in series

Step 2. Finding the ration of both the capacitance, $\frac{C_2}{C_1}$

Equivalent Capacitance in series combination$=\frac{C_1{C}_2}{C_1+C_2}$

Equivalent Capacitance in parallel combination$=C_1+C_2$

According to question,

$C_1+C_2=\frac{15}{4}\times \frac{C_1\times C_2}{C_1+C_2}$

$4{\left(C_1+C_2\right)}^2=15C_1{C}_2$

$4C_1^2+4C_2^2-7C_1{C}_2=0$

$4{\left(\frac{C_2}{C_1}\right)}^2-7\left(\frac{C_2}{C_1}\right)+4=0$

$4x^2-7x+4=0$ [Let $x=\frac{C_2}{C_1}$]

On solving the above equation, we get

$x=imaginary$

$\frac{C_2}{C_1}=imaginary$

Hence, the correct option is $\left(A\right)$.