Minimum number of $2\,\mu F$ and $250\, V$ capacitors used to make a combination of $12\,\mu F$ and $500\,V$ are
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Let m capacitors are joined in series and n such groups are joined in parallel.
$\mathrm{So}, C=2 / m$ and $C_{\text {equi }}=n \times \frac{2}{m}=12$
or, $n=6 m$ Potentiai of arrangement, $m V=500$
or, $\quad m=\frac{500}{250}=2 . \quad \therefore \quad n=2 \times 6=12$
So, total number of capacitors required $=n m= 2\times 12=24$
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