Question

FET & AIIMS 41. In the circuit Gz = 50 MF. IA e circuit below C1 = 20 uF, C2 = 40 uF and - 50 uF. If no capacitor can sustain more than v then maximum potential difference between X and Y is XoHHHoy a, Cz Cz (1) 95 V (2) 75 v (3) 150 V (4) 65 V In the circuit shown below C, = 10 uF, C = Cz ip

NEET/Medical Exams

Physics

Solution

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( E ) Let the charge on each capacitor be 9 then potential against sach capacitar will be [ begin{array}{l} text { for } c_{1}=frac{9}{20} v+c 0 text { for } c_{2}=frac{a}{a 0} text { vin } end{array} ] for ( (3)=frac{9}{30} V ) capacitor chas max. potential doop which accordira to situation can be sov. so ( 9 / 20=50 mathrm{v} ) so ( 9=1000 ) micro conlomb so the potential drop again, ( t quad C ), capacits ( =frac{1000}{40} ) [ =25 mathrm{V} ] while potential drop agairs mill be [ frac{1000}{50}=20 v cdot-6 ] Now all the the capaidre als in sedin so ( a operatorname{ctos} x quad x ) ad ( y=50+25+20 ) poterdial diffirin a crots is aro