Three capacitors of $2\,\mu F,\,3\,\mu F$ and $6\,\mu F$ are joined in series and the combination is charged by means of a $24\, volt$ battery. The potential difference between the plates of the $6\,\mu F$ capacitor is.......$volt$
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(a) $\frac{1}{{{C_{eq}}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \Rightarrow {C_{eq}} = 1\,\mu \,F$
Total charge $Q = C_{eq.}\,\,V = 1 × 24 = 24\, µC$
So $p.d.$ across $6\, µF$ capacitor = $\frac{{24}}{6} = 4\,volt$
Total charge $Q = C_{eq.}\,\,V = 1 × 24 = 24\, µC$
So $p.d.$ across $6\, µF$ capacitor = $\frac{{24}}{6} = 4\,volt$
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