Find current through each cell.
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Explanation

Here cells cannot be replaced by single cell because terminals of cells are not joined to same point. This problem can be solved by Kirdhhoff's laws.
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First, assume current in different braches keeping in mind junction law.
Lo op ①, B to B, path BCDXYB`
`V_B - i_1 xx 2 - (i_1 + i_2) xx 2 - i xx 2 + 1 = V_B`
`4 i_1 + 2 i_2 + 2 i = 1`...(i)
`Lo op ②, B to b, path BACB
`V_B -(i - i_1) xx 1- 2 - i_2 xx 1 + i_1 xx 2 = V_B`
`3 i_1 - i_2 - i = 2`..(ii)
`Lo op ③, A to A, path ADCA`
`V_A - (i - i_1 - i_2) xx 1 + 2 (i_1 + i_2) + i_2 xx 1 + 2 = V_A`
`3 i_1 + 4 i_2 - i = -2` ...(iii)
(i)+(ii)+(iii) `rArr 10 i_1 + 5 i_2 = 1` ....(a)
(iii)-(ii) `rArr 5 i_2 = -4 rArr i_2 = -(4)/(5) A = -0.8 A`
Putting `i_2 = -(4)/(5) in (a)`
`10 i_1 + 5(-(4)/(5)) = 0 rArr i_1 = 0.4 A`
`4 i_1 + 2i_2 - 2i = 1`
`4 xx 0.4 + 2xx (-0.8) + 2 i = 1`
`i = (1)/(2) A = 0.5 A`
Current through `1 V : 0.5 A, 2 V : 0.8 A`.