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Electricity and Magnetism Level 2

Let $ax^{5} -bx^{4} +cx^{3} -dx^{2} +ex -f=0$ be an equation which has 5 real roots $α_{1}, α_{2}, α_{3}, α_{4}$ 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).

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

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

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

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

1 solution

Prakkash Manohar
May 30, 2015

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

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

$C'=\frac{C_{1} C_{2} C_{3} C_{4} C_{5}}{C_{2} C_{3} C_{4} C_{5} + C_{1} C_{3} C_{4} C_{5} + C_{1} C_{2} C_{4} C_{5} + C_{1} C_{2} C_{3} C_{5} + C_{1} C_{2} C_{3} C_{4}}$

$C'=\frac{α_{1} α_{2} α_{3} α_{4} α_{5}}{α_{2} α_{3} α_{4} α_{5} + α_{1} α_{3} α_{4} α_{5} + α_{1} α_{2} α_{4} α_{5} + α_{1} α_{2} α_{3} α_{5} + α_{1} α_{2} α_{3} α_{4}}$

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

(USING VIETA'S FORMULA)

Now, $C', C_{5}, C_{4}, C_{3}, C_{2}, C_{1}$ are in PARALLEL.

So, $C_{equivalent} = C'+ C_{5}+ C_{4}+ C_{3}+ C_{2}+ C_{1}$

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

(USING VIETA'S FORMULA)

Note: In the 3rd step, I have replaced $C_{1}$ 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.

50% Maths and 50 % Physics!!

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