Three capacitors ( 3 mu F, 6 mu F ) and ( 6 mu F ) are connected in series to a source of ( 120 mathrm{V} ). The potential difference, in volt, across the ( 3 mu mathrm{F} ) capacitor will be
(a) ( 24 mathrm{V} )
(b) ( 30 mathrm{V} )
(c). ( 40 mathrm{V} )
(d) ( 60 mathrm{V} )

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( 3 mu F, 6 mu f, 6 mu f ) series [ frac{1}{c_{e q}}=frac{1}{3}+frac{1}{6}+frac{1}{6}=frac{2}{3} ] ( C_{q}=frac{3}{2}: ) [ q=C_{e q} V=frac{3}{2} times 120=180 mu C ] Vacross 3 Mμcapaitor ( =frac{q}{3} ) [ =frac{180}{3}=60 mathrm{V} ] Option d is correct answer.

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