Question : A capacitor of 20µF is charged to 30 V. It is then connected to an uncharged 30µF capacitor. Calculate the charges on each capacitor in equilibrium.
CBSE 2024
Solution :
C1=20µF
C2=30µF
C2=30µF
V1=30 V
V2=?
Initially
Q1=C1V1
Q1=20µF×30
Q1=600µC
Q2=C2V2
Q2=C2×0
Q2=0
As per the law of conservation of charge
Q1+Q2=Q1'+Q2'
600µC=Q1'+Q2' — (1)
Q2=C2×0
Q2=0
As per the law of conservation of charge
Q1+Q2=Q1'+Q2'
600µC=Q1'+Q2' — (1)
We know,
Q=CV
Q∝C [At equilibrium V=Common Potential= Const.]
Q=CV
Q∝C [At equilibrium V=Common Potential= Const.]
Q1'/Q2'=C1/C2
Q1'/Q2'=20/30
Q1'/Q2'=2/3 — (2)
Using Equation (1) and (2)
Q1'/Q2'=20/30
Q1'/Q2'=2/3 — (2)
Using Equation (1) and (2)
Q1'=[2/5]×600µC
Q1'=1200/5 µC
Q1'=240 µC
Q1'=1200/5 µC
Q1'=240 µC
Q2'=[3/5]×600µC
Q2'=1800/5 µC
Q2'=360µC
Q2'=1800/5 µC
Q2'=360µC
Hence, at equilibrium, the charghes on each capacitor would be 240µC and 360µC.
