Question

Consider the combination of 2 capacitors C1 and C2 with C2>C1, when connected in parallel, the equivalent capacitance is 15/4 times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors,C1/C2​

Answer 1

Given :  combination of 2 capacitors C1 and C2 with C2>C1, when connected in parallel, the equivalent capacitance is 15/4 times the equivalent capacitance of the same connected in series.

To Find : the ratio of capacitors,C1/C2​

Solution:

C1 and C2 connected in parallel

= C1 + C2

C1 and C2 connected in Series

=  1/(1/C1 + 1/C2)

= (C1 . C2)/(C1 + C2)

connected in parallel, the equivalent capacitance is 15/4 times the equivalent capacitance of the same connected in series

=> C1 + C2  = (15/4) C1 . C2)/(C1 + C2)

=> 4(C1 + C2)² =  15 C1 . C2

Let say C1/C2​ = K      and K < 1   as C2>C1

=> C1 = kC2  

=> 4(kC2 + C2)² =  15 kC2 . C2

=>4(k + 1)²C2² = 15kC2²

=> 4(k² + 2k + 1) = 15k

=> 4k² + 8k + 4 = 15k

=> 4k² - 7k + 4 = 0

=> k =  (- 7 ± √49 - 64)/2(4)

=> k = (- 7 ± √ -15)/8

k is imaginary

looks issue in data

15/4 times  must be > 4 times   as min 4 times is when both capacitors are same

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Answer 2

When capacitors in parallel:- C1+C2

When capacitors in series:- C1C2/C1+C2

Given,

C1+C2=15/4(C1C2/C1+C2)

=>4(C1+C2)^2=15C1C2

After solving we get,

4C1^2+4C2^2-7C1C2=0

Dividing the equation by C1 we get,

4+4(C1/C2)^2-7(C2/C1)=0

Let C2/C1=x

Then we get a Quadratic equation like:-

4x^2-7x+4=0

which has a determinant less than zero.

i.e Δ<0

Therefore, The ratio of C2/C1 has no value since it has imaginary roots