A battery of emf `10 V` is connected as. Find the potential difference between the point A and B

Answer

`5.0 V`

Explanation

Total resistance of circuit
`R = ((3+1)xx(1+3))/((3+1)+(1+3)) = 2Omega`
Current in circuit, `I = (epsilon)/(R ) = (10)/(2) = 5A`
Since the two parallel branches of network has same resistance `(= 4 Omega)`, so the current in each branch `= 3//2 = 2.5 A`
Now `V_(C) - V_(B) = 2.5 xx 1= 2.5 V`
`V_(C) - V_(B) = 2.5 xx 3= 7.5 V`
`(V_(A) - V_(B)) = (V_(C) - V_(B)) - (V_(C) - V_(A)) = 7.5 - 2.5`
= `5.0 V`