A number of capacitors,each of capacitance 1, | vedclass

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(C) Let $n$ be the number of capacitors in each series branch and $m$ be the number of such parallel branches.For each series branch,the total potenti

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$72$ component capacitors

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(C) Let $n$ be the number of capacitors in each series branch and $m$ be the number of such parallel branches.For each series branch,the total potential difference is $V_{total} = n 	imes V_{max} = n 	imes 500 = 3000\,V$.Thus,$n = 3000 / 500 = 6$.The equivalent capacitance of one series branch is $C_s = (1\,\mu F) / n = 1/6\,\mu F$.For $m$ such branches in parallel,the total capacitance is $C_{eq} = m 	imes C_s = m 	imes (1/6) = 2\,\mu F$.Solving for $m$,we get $m = 2 	imes 6 = 12$.The total number of capacitors required is $N = n 	imes m = 6 	imes 12 = 72$.

$A$ number of capacitors,each of capacitance $1\,\mu F$ and each of which gets punctured if a potential difference exceeding $500\,V$ is applied,are provided. An arrangement suitable for giving a capacitance of $2\,\mu F$ across which $3000\,V$ may be applied requires at least:

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