You are seeing the voltage drop across the resistor. 4.1V-3.75=350 millivolts.
The 3.75 is the drop across the internal resistance of the battery.
I think it was working fine.
If you are getting interested in electronics, this is a good lesson in Ohm's Law and Kirchoff's Voltage law.
You know it might not seem it but I am pretty familiar with those. The problem that I have is that in my undergrad the courses specifically dealing with these types of problems were not practical at all and thus were pretty ineffective. I have what I consider a `working knowledge` with hands-on circuits and do my day-to-day work as a wireless sensors guy using premade Si chips instead of discrete components.
These types of little projects are exactly what I need to sort of fill in some of the holes I have so that I have the same good sense in looking at something and figured it out with these as I do with other things.
The people on this forum are a big help!
One thing though. I was considering that it may be the voltage drop however when I turned up the source voltage slightly I did not notice a change. If it was the voltage drop only, wouldn't you expect this to affect the voltage I measured? I suppose that it is possible that I am recalling that incorrectly, but I believe that is accurate to what I saw.
Either way, I will go back tomorrow and check it out again.
Edit: I just remade it again and what I said above is correct. I will illustrate it here:
4.1V (prototyping board) -> top of resistor -> bottom of resistor -> (RED METER LEAD) -> Pad 4
And
Pad 1 -> (BLACK METER LEAD) -> ground (prototyping board)
With that setup the meter goes:
Condition 1) Measure as shown -> 3.75V
Condition 2) Increase 4.1V to 4.5V -> 3.75V
Condition 3) Turn OFF 4.1V source -> 3.75V
Condition 4) Pull batter, measure from bottom of resistor to ground -> 4.1V
Does this make sense?