All batteries have internal resistance, which is a function of the battery’s state of charge, its chemical composition, the number of cells in the battery, the connections between the cells, the construction and size of the electrodes in the cells, and so on. You can think of this internal resistance as a lumped resistance in series with an ideal voltage source, whose voltage is equal to the open circuit terminal voltage of the battery (meaning, fully charged, but with no current flowing through it). If we call this open circuit voltage Voc, and the internal resistance of the battery as Rint, then the maximum current that will flow from the battery if you connect (short-circuit) the positive terminal to the negative terminal is Isc = Voc/Rint (in an ideal battery with no internal resistance (Rint ~ 0), Isc would theoretically be infinite).

Consider the actual circuit of the above battery feeding a load resistance RL.

By Ohm’s Law, Is = Voc/Rt = Voc/(Rint + RL)

The voltage across the load, VL = Is ٠ RL = Voc ٠ RL / (Rint + RL)

The voltage “drop”, is the voltage lost across the internal resistance of the battery.:

Vdrop = Voc - VL = Is ٠ Rint = Voc - [Voc ٠ RL/(Rint + RL)]

or Vdrop = Voc [ 1 - RL/(Rint + RL)]

As you can see, the higher the internal resistance of the battery (Rint), the greater the voltage drop will occur across Rint for a given load current Is, and the less voltage you’ll have available across the load (VL). Similarly, the smaller the load resistance RL, the larger effect the internal voltage drop across Rint will have on the voltage available across the load (VL).

I was tempted to show you how to solve the above relationship for Rint, the internal resistance of a battery by (a) measuring its open circuit voltage Voc, and then (b) its terminal voltage VL when you connect a load resistance RL across the battery, but I suspect that if you want that information, you can readily do the algebra yourself.