You really need to have this answered by an electorphysiologist, probably not me, but you’re talking about electrical, not mechanical resistance. For a given applied voltage (here millivoltage I would guess) and resistance, you will have a certain amount of current flowing. In this case—and here I am really just guessing—that the larger the “hole” represented by the pipet, the greater will be the conductance of the path through it. That is, I expect the resistance of the pipet to be lower if the apature is larger, but if you pull it too far and make the apature too small, you get a higher resistance (lower conductance) than you want. I’m not sure, but I think the measurement goes someting like a null potentiometer; you apply a voltage across the membrane that opposes the normal voltage gradient until the current becomes zero. The potential at that point has to be equal to the potential of the cell membrane, but in the opposite direction. You have to know the resistance of your pipet in order to know what the actual potential is, and that resistance varies with the way you pull your electrode/pipet. I don’t know what I’m talking about here, so you should take my explanation with a pretty large grain so salt, but at least it makes physical sense to me.

oh you’re right blcr, the resistivity of the pipe is inversely proportional to cross-sectional area(diameter) and a large diameter indicates lower resistivity so a resistivity less than 3 represents too large of a pipe. it is also proportional to the length of the pipe so a longer pipe has higher resistance but I’m assuming you have that normalized somehow.

As I understand it, it’s an art to pull your own electrodes–and get them to work, anyway. I think there are machines now that can make them somewhat automatically.