Phy problem...electricity2

Let current delivered by batteries 2 and 3 be I2 and I3 respectively, and assume that their directions are upward.
Hence, current through battery 1 is (I1+I2) and is in the downward direction.

Consider the right loop,
I2(R2) = I1.(2.R1)
i.e. 2(I2) = 2(I1)
I2 = I1

Consider the left loop,
I2.(R2) + (I1+I2).(2.R1) = 4 - 2
2.(I2) + 2.(I2).(2) = 2
6.(I2) = 2
I2 = 2/6 A = 1/3 A
Thus, current delivered from batteries 1 and 2 are each equals to 1/3 A (up)
current entered into battery 1 = (1/3 + 1/3) A = 2/3 A (down)

(d) Va - Vb = (-(1/3).(2) + 4) v = 3.33 v

(e) Initial charge on capacitor Q1 = 60 x 3 uC = 180 uC

After S is closed, current through the two resistors (by Ohm's Law),
= (3-1)/(0.2+0.8) A = 2 A
Hence, voltage across capacitor
= (3 - 2 x 0.8) v = 1.4 v
Charge on capacitor = 60 x 1.4 uC = 84 uC

Reduction in charge = (180 - 84) uC = 96 uC