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Let a x 5 b x 4 + c x 3 d x 2 + e x f = 0 ax^{5} -bx^{4} +cx^{3} -dx^{2} +ex -f=0 be an equation which has 5 real roots α 1 , α 2 , α 3 , α 4 α_{1}, α_{2}, α_{3}, α_{4} and α 5 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).

a f + e b \frac{a}{f} + \frac{e}{b} b c + a d \frac{b}{c} + \frac{a}{d} d b + c a \frac{d}{b} + \frac{c}{a} f e + b a \frac{f}{e} + \frac{b}{a}

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1 solution

Prakkash Manohar
May 30, 2015

We just have to find the equivalent capacitance. If we replace the topmost row by C', C = 1 1 C 1 + 1 C 2 + 1 C 3 + 1 C 4 + 1 C 5 C'=\frac{1}{\frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + \frac{1}{C_{4}} + \frac{1}{C_{5}}}

(Because, C 1 , C 2 , C 3 , C 4 , C 5 C_{1}, C_{2}, C_{3}, C_{4}, C_{5} are in SERIES.)

C = C 1 C 2 C 3 C 4 C 5 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'=\frac{C_{1} C_{2} C_{3} C_{4} C_{5}}{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 = α 1 α 2 α 3 α 4 α 5 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 C'=\frac{α_{1} α_{2} α_{3} α_{4} α_{5}}{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}}

C = f / a e / a = f e C'=\frac{f/a}{e/a} = \frac{f}{e}

(USING VIETA'S FORMULA)

Now, C , C 5 , C 4 , C 3 , C 2 , C 1 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_{equivalent} = 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 ] = f e + b a C_{equivalent} = C'+ [ C_{5}+ C_{4}+ C_{3}+ C_{2}+ C_{1} ] = \frac{f}{e} + \frac{b}{a}

(USING VIETA'S FORMULA)

Note: In the 3rd step, I have replaced C 1 C_{1} by α 1 α_{1} , C 2 C_{2} by α 2 α_{2} and likewise, but, it is not necessary that C 1 = α 1 , C 2 = α 2 C_{1} = α_{1}, C_{2} = α_{2} and so on. Even then, the expression will evaluate to C C' = product of roots / sum of product of roots taken 4 at a time.

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