No charge is on the two capacitors before the switch is closed. What is the char

Question

No charge is on the two capacitors before the switch is closed.

What is the charge (as function of time) on each of the two capacitors after switch is closed?

Explanation / Answer

voltage across both capacitors is same there fore Q1/C1 = Q2/C2 Q1= C1Q2/C2 ==> Q1= (4/3) x Q2 Q2 = C2Q1/C1 ==> Q2 = (3/4) x Q1 total charge suplied by battery Q = Q1+Q2 = (1+3/4)Q1=(1+4/3) Q2 R = (3 x 4)/(3+4) = 12/7 i= i1+i2 i= 7/4 i1 i= 7/3 i2 by applying KVL V= (dQ/dt)R+Q1/C1 ------->1 V= (dQ/dt)R+Q2/C2 ------->2 by derivating eq 1 0= (di/dt) R + i1/C1 (7/4) (di1/dt) = -(i1/RC1) di1/i1 = (-7/4) (1/RC1) dt = 0.255 dt i1 =exp (-0.255t) Q1= 3.921(1-exp (-0.255t)) similarly Q2 = 2.207(1-exp (-0.453t))

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