Calculate the current through the resistance connected across MN and the current supplied by each of the batteries in the circuit diagram shown in

Explanation

Loop AFMN : `20 = 5I_(1) + 12(II_(1) - I_(2))`

or ` 20 = 17I_(1) - 12I_(2)` (i)
Loop AFBEA: `20 +20 = 5I_(1) + 5I_(2)` or `8 = I_(1) + I_(2)` (ii)
Solving, we get `I_(1) = 4 A, I_(2) = 4A`
So current through `MN is I_(1) - I_(2) = 4 - 4 = 0`
And current through each of the battereis is 4A.