find
(a) the current in `3Omega` resistance
(b) the change on the capacitor

Explanation

Loop 1, `A` to `A `path `ACBA`
`V_(A) = (i-i_1)xx4+4+4i_(1) = V_(A)`
`-4i+4i_(1) +4+4i_(1) = V_(A)`
`i-2i_(1) = 1` ..(i)
loop 2, `A` to `A` , path `ABYXA`
`V_A - i_(1) xx 4 - I xx3 -ixx1+8=V_(A)`
`-4i_(1)-3i-i+8=0`
`4i +4i_(1)=8`
`i+i_(1)=2`..(ii)
Solving
`i=5/3A, i_(1)=1/3A`
(b) Branch `XMY`
`V_(X)-V_(Y) = 8-ixx1=8-5/3xx1=19/3 V`
Branch `XLY`:
`V_(X)+3-V_(0)=V_(Y)`
`V_(0)=V_(X)-V_(Y)+3=19/3+3=28/3V`
Charge on capacitor `Q = 6V_(0)=6xx28/3= 56muC`.