To form a composite 16,mu F, 1000,V capacitor | vedclass

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Question

(B) Let the given capacitor have capacitance $C = 8\,\mu F$ and voltage rating $V = 250\,V$. The required composite capacitor has capacitance $C' = 16

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Let the given capacitor have capacitance $C = 8\,\mu F$ and voltage rating $V = 250\,V$. The required composite capacitor has capacitance $C' = 16\,\mu F$ and voltage rating $V' = 1000\,V$.Suppose $m$ rows of capacitors are connected in parallel,and each row contains $n$ capacitors connected in series.The potential difference across each capacitor in a series row is $V = \frac{V'}{n}$.Substituting the values: $250 = \frac{1000}{n} \implies n = 4$.The equivalent capacitance of the network is $C' = \frac{mC}{n}$.Substituting the values: $16 = \frac{m 	imes 8}{4} \implies 16 = 2m \implies m = 8$.The total number of capacitors required is $N = n 	imes m = 4 	imes 8 = 32$.Alternatively,using the formula: $N = \frac{C'}{C} 	imes \left( \frac{V'}{V} ight)^2 = \frac{16}{8} 	imes \left( \frac{1000}{250} ight)^2 = 2 	imes (4)^2 = 2 	imes 16 = 32$.

To form a composite $16\,\mu F, 1000\,V$ capacitor from a supply of identical capacitors marked $8\,\mu F, 250\,V$,we require a minimum number of capacitors:

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