In the circuit find the potential difference between point `A and B`. Assume that both the batteries have zero internal resistance

Answer

`11 V`

Explanation

Here, `epsilon_(1) = 20 V, epsilon_(2) = 8 V, R_(1) = 3 Omega , R_(2) = 1Omega`
As the batteries are sending the current in opposite direction so effective emf
`= 20 - 8 = 12 V`
The internal resistance of batteries is zero `R_(1),R_(2)` are in series in the circuit .Therefore Total resistance `R = R_(1)+ R_(2)= 3 + 1= 4 Omega`
Effective current `1 = 12 //4 = 3A`
`V_(AB) = epsilon_(1) - IR_(1) = 20 - 3 xx 3 = 11V`
Also `V_(AB) = epsilon_(2) - IR_(2) = 8 + 3 xx 1 = 11V`