Suppose a battery of negligible internal resistance which creates a potential difference V across its terminals is connected to a resistor R through wires of negligible resistance then it creates voltage V across the resistor too.
Its emf is the energy given by itself to the unit charge to move it from negative terminal to positive terminal (current being conventional). Then there are two places where energy is being lost by the charges 1) due to resistance of resistor and 2) due to the work done by electrostatic force due to the electric field between Terminals of battery which opposes the motion of charge as it moves from negative to positive terminal and hence this work is negative.
So,
$$emf × q = (V × q) + (qEd)$$ $$\implies emf= V+Ed$$ Where,
q is the charge on which battery does work in a given time and in the same time energy is being given out by q passing through resistor.
d is the straight distance between two terminals. I have supposed that the charge q moves from negative to positive terminal along the shortest distance between them.
E is the electric field intensity between terminals directed from positive to negative terminal. I have supposed it is uniform at any point between two terminals.
The problem is that my book says different as compared to what I wrote in the equation. It says:
$$emf × q = V × q$$ or$$emf=V$$