Question

( 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

# 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

Solution