Consider the combination of 2 capacitors C_{1 | vedclass

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(D) For capacitors in parallel,the equivalent capacitance is $C_{p} = C_{1} + C_{2}$.For capacitors in series,the equivalent capacitance is $C_{s} = \

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(D) For capacitors in parallel,the equivalent capacitance is $C_{p} = C_{1} + C_{2}$.For capacitors in series,the equivalent capacitance is $C_{s} = \frac{C_{1}C_{2}}{C_{1} + C_{2}}$.According to the problem,$C_{p} = \frac{15}{4} C_{s}$.Substituting the expressions,we get $C_{1} + C_{2} = \frac{15}{4} \left( \frac{C_{1}C_{2}}{C_{1} + C_{2}} \right)$.This simplifies to $4(C_{1} + C_{2})^2 = 15C_{1}C_{2}$.Expanding the square,$4(C_{1}^2 + C_{2}^2 + 2C_{1}C_{2}) = 15C_{1}C_{2}$,which gives $4C_{1}^2 + 4C_{2}^2 + 8C_{1}C_{2} = 15C_{1}C_{2}$.Rearranging the terms,$4C_{1}^2 - 7C_{1}C_{2} + 4C_{2}^2 = 0$.Dividing by $C_{1}^2$ and letting $x = \frac{C_{2}}{C_{1}}$,we get $4x^2 - 7x + 4 = 0$.The discriminant $D = b^2 - 4ac = (-7)^2 - 4(4)(4) = 49 - 64 = -15$.Since the discriminant is negative,there are no real solutions for the ratio $\frac{C_{2}}{C_{1}}$. Therefore,the correct option is $D$.

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 capacitors connected in series. Calculate the ratio of the capacitors,$\frac{C_{2}}{C_{1}}$.

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