In current electricity , potential difference V is used rather than potential at a point in Electric field topic. In a circuit with a resistor of 8 ohms and a battery of e.m.f E of 9V and internal resistance 1 ohms. You can easily work out that the current flowing should be 9V divide by 9 ohms , that will give you 1 ampere. If you measure the potential difference across the battery you could only obtain 8V. The puzzle is that the battery has 9V . Why do you have 1 V missing? You measure the potential across the external resistor , p.d = i R , again you only have 8V. Check the terminals of the battery and across the resistor R , the voltmeter reading is consistently 8V. Check inside the battery, the left side terminal potential is higher than the right hand side terminal by 8V .Lets put some values called potential at a point to the left side , say 9V and 0V as you go across the battery.Since current flow leftwards across the resistor r, the right is higher by a potential difference of ir = 1V, and since the left is 0V the right is 1V . In conclusion the difference between the left (9V) and the right (1V) is 8V.
