a x 5 − b x 4 + c x 3 − d x 2 + e x − f = 0 be an equation which has 5 real roots α 1 , α 2 , α 3 , α 4 and α 5 , which are the values of capacitance of the 5 capacitors shown in the figure (not necessarily in order). Find the capacitance of a single capacitor which can replace the combination and store the same charge (as the combination).
Let
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We just have to find the equivalent capacitance. If we replace the topmost row by C', C ′ = C 1 1 + C 2 1 + C 3 1 + C 4 1 + C 5 1 1
(Because, C 1 , C 2 , C 3 , C 4 , C 5 are in SERIES.)
C ′ = C 2 C 3 C 4 C 5 + C 1 C 3 C 4 C 5 + C 1 C 2 C 4 C 5 + C 1 C 2 C 3 C 5 + C 1 C 2 C 3 C 4 C 1 C 2 C 3 C 4 C 5
C ′ = α 2 α 3 α 4 α 5 + α 1 α 3 α 4 α 5 + α 1 α 2 α 4 α 5 + α 1 α 2 α 3 α 5 + α 1 α 2 α 3 α 4 α 1 α 2 α 3 α 4 α 5
C ′ = e / a f / a = e f
(USING VIETA'S FORMULA)
Now, C ′ , C 5 , C 4 , C 3 , C 2 , C 1 are in PARALLEL.
So, C e q u i v a l e n t = C ′ + C 5 + C 4 + C 3 + C 2 + C 1
C e q u i v a l e n t = C ′ + [ C 5 + C 4 + C 3 + C 2 + C 1 ] = e f + a b
(USING VIETA'S FORMULA)
Note: In the 3rd step, I have replaced C 1 by α 1 , C 2 by α 2 and likewise, but, it is not necessary that C 1 = α 1 , C 2 = α 2 and so on. Even then, the expression will evaluate to C ′ = product of roots / sum of product of roots taken 4 at a time.