Sorry, can not open your image :(
i put spaces between the "h t t p :" because it wouldnt let me post it otherwise. if you delete the spaces it should load (i hope!).
I call the current coming out the top of the left battery i1
I call the current coming out the top of the right battery i2
Then the current headed down from a to b is (i1+i2) (Kirkhoff)
In the left loop
12 = 4 i1 + 6(i1+i2)
In the right loop
12 = 3 i2 + 6(i1+i2)
12 = 10 i1 + 6 i2
12 = 6 i1 + 9 i2
solve those two. I will multiply the first by three and the second by 2
36 = 30 i1 + 18 i2
24 = 12 i1 + 18 i2
12 = 18 i1
i1 = 2/3
12 = 10(2/3) + 6 i2
12 = 20/3 + 6 i2
36 = 20 + 18 i2
16 = 18 i2
i2 = 8/9
i1+i2 = 2/3 + 8/9 = 14/9
times the resistance which is 6 = 28/3 volts a above b
2/3 = 6/9 = i1
8/9 = i2
so i2>i1 so the one on the right will run out first
(8/9) t = 20 amp hr
t = 9*20/8 = 9*5/2 = 22.5 hr
at present i1 = 6/9 and i2 = 8/9
clearly if I multiply the voltages by 9 I will get 6 amps and 8 amps
then if I divide both voltages by two I will get 3 amps and 4 amps
so multiply the voltage by 9/2
(9/2)12 = 54 volts
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