Minimum number of 2,mu F and 250, V capacitor | vedclass

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Question

(B) Let $m$ capacitors be connected in series in each row,and $n$ such rows be connected in parallel.The equivalent capacitance of one row is $C_{row}

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) Let $m$ capacitors be connected in series in each row,and $n$ such rows be connected in parallel.The equivalent capacitance of one row is $C_{row} = \frac{2\,\mu F}{m}$.The total equivalent capacitance is $C_{eq} = n 	imes C_{row} = n 	imes \frac{2}{m} = 12\,\mu F$.Thus,$n = 6m$.The voltage rating of each row is $V_{row} = m 	imes 250\, V$. To withstand $500\, V$,we must have $m 	imes 250 = 500$,which gives $m = 2$.Substituting $m = 2$ into the equation $n = 6m$,we get $n = 6 	imes 2 = 12$.The total number of capacitors required is $N = n 	imes m = 12 	imes 2 = 24$.

Minimum number of $2\,\mu F$ and $250\, V$ capacitors required to make a combination of $12\,\mu F$ and $500\, V$ is:

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