To start with this is not a numerical but merely a basic concept I fail to understand. I will quote what the CIE book says "Consider a battery of emf 3 V and of internal resistacne 1 ohm. The maximum current that can be drawn from this battery is I=e/r = 3/1=3 A The terminal p.d of the battery dpeends on the resistance of the external resistor. For an external resistance of 1 ohm the terminal pd is 1.5v (1/2 of emf). The terminal p.d approaches the value of emf when the external resistance R is very much greater than internal resistance. E,g a resistor of resistance 1k oh connected to battery will give a terminal p.d of 2.997V. this is almost equal to emf of battery. The more current a battery supplies, the more its terminal pd will decrease." The last line (more current = less p.d ) doesnt go well with the formula V=IR which says more voltage will result in more current. If more voltage was to yield less current wouldnt Voltage and Current be inversely proportional? Another confusion is "A resistance of 1k will give a terminal p.d of 2.997V", that doesn't make much sense either. It is a known fact that resistors decrease the amount of current flowing and a higher resistance will result in a lower current and since voltage and current are directly proportional wouldn't the voltage be low too?