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(D) Let $n$ be the number of capacitors in each parallel row and $m$ be the number of such rows in series.Each capacitor can withstand $300\ V$. To wi

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$32$

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(D) Let $n$ be the number of capacitors in each parallel row and $m$ be the number of such rows in series.Each capacitor can withstand $300\ V$. To withstand a total potential difference of $1000\ V$,the number of capacitors in series $(m)$ must satisfy $m 	imes 300 \ge 1000$,which gives $m \ge 3.33$. Thus,we need at least $m = 4$ rows in series.The potential difference across each row will be $1000/4 = 250\ V$,which is within the safe limit of $300\ V$.The equivalent capacitance of one row of $n$ capacitors in parallel is $C_{row} = n 	imes 1\ \mu F = n\ \mu F$.Since there are $m = 4$ such rows in series,the total equivalent capacitance is given by $\frac{1}{C_{eq}} = \frac{1}{C_{row}} + \frac{1}{C_{row}} + \frac{1}{C_{row}} + \frac{1}{C_{row}} = \frac{4}{n}$.Given $C_{eq} = 2\ \mu F$,we have $\frac{1}{2} = \frac{4}{n}$,which implies $n = 8$.Total number of capacitors = $m 	imes n = 4 	imes 8 = 32$.

$A$ capacitance of $2\ \mu F$ is required in an electrical circuit across a potential difference of $1.0\ kV$. $A$ large number of $1\ \mu F$ capacitors are available which can withstand a potential difference of not more than $300\ V$. The minimum number of capacitors required to achieve this is

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