Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is.
.

Explanation

let i be the current flowing in the circuit. Apply Kirchhoff's
law in the loop we get
`-3-2i-i+6=0`
`:. 3i = 3`
`:. i = 1 Amp`.
Now let us travel in the circuit from A to B through battery of 6V, we get
`V_(A) - 6 + 1 xx 1 = V_(B)`
`:. V_(A) - V_(B) = 5 volt.