An ideal battery cell would maintain it's voltage regardless of the current draw. This is not the case, however, in the real world since the terminal voltage drops as current draw increases. A "real world" cell can be modeled as an "ideal" cell with a cell voltage of Voc (Open Circuit Voltage) and a series resistance Ri (Internal Resistance). Since this resistance is part and parcel of the cell itself (materials, chemical characteristics, mechanical characteristics) it is refereed to as "internal". Obviously, some cell types have lower internal resistance's than others, Ni-Cd has a lower Ri than a carbon zinc cell for example. Even cells of the same type have differing Ri values due to manufacturing and chemical differences.

The power lost to a cell's internal resistance is expressed as I^2Ri, that is, the current draw squared times the internal resistance value. As you can see, a cell with a lower Ri is able to deliver more power to the load (load resistance is usually designated Rl, by the way) as less power is lost to internal resistance. And as the power loss is propositional to the current *squared*, you can see that a high Ri cell has trouble delivering the punch to the motor under high load; in turns, off the start. That's not so say a cell with a low Ri is always good, cell runtimes can be reduced because more energy is consumed during high current draw situations than is consumed in a cell with higher Ri.

Ri is calculated by measuring the cell voltage without any current draw (measuring the open circuit voltage, Voc) and by measuring the voltage under load (Vl at I amps). Then the Ri can be computed as:

Ri = (Voc-Vl)/I

(it's actually a little more complicated than this as one must take into account the drop in the cell's electrochemical potential when under load as well, ie. discharge)

Tony