Please solve question 22 please Share with your friends Share 0 Vipra Mishra answered this Dear Student Charge on first capcitor when connected to batteryQ1 = C1V1 =2×50×10-6 = 100μCCharge on second capcitor when connected to batteryQ2 = C2V2 =3×100×10-6 = 300μCNow when battery is disconnected and capacitors are connected in parallel(As paltes of same charges are connected together)Total charge Q =400μCC =C1 +C2 = 5μFV =QC =400×10-65×10-6 =80V Regards 0 View Full Answer
