The equivalent capacitance of two capacitors (C2>C1) connected in parallel is 15/4 times the equivalent resistance when they are connected in series. Find C2/C1
Answers
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
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