How do tolerances combine when equal valued resistors are paralleled? I'm making a project, a NiMH battery charger, in which I'm trying to measure the battery current by measuring the voltage across a low value shunt resistor of about 0.5 ohms. The shunt resistor is just a junkbox 10% wirewound, and I want to estimate its value to quite a bit better than that. (This is a hobby, not a commercial project, so I'm not interested in buying a precision shunt; the point of this project is learning how to do this sort of design.) I have a digital voltmeter that measures using a precision reference of about 0.2% tolerance, and I have a box of several dozen 250 Ohm precision resistors rated at 0.01% tolerance. My thought was to parallel a bunch of the 250 Ohm resistors to make a quite low valued precision resistor and connect that in series with the shunt resistor, then apply a voltage to the series pair to send the same current through both resistors. Then by comparing the voltages across the resistors, the resistance of the shunt can be expressed to much better than its 10% stated tolerance. So if several of the 0.01% resistors are paralleled, how do their tolerances combine? Paralleling two of the 250 Ohm resistors gives a nominal 125 Ohms. The parallel of two maximum toleranced resistors gives 125.0125, and two having the minimum at their stated tolerance gives 124.9875, so with a combined resistance of 125 Ohms +/- 0.0125 Ohms, it seems like the tolerance of the combined pair is unchanged at 0.01%. Is that right? It would be great if it were really that simple, because with that, I could parallel 10 or even 25 of the 250 Ohm resistors and still have a 0.01% combination. Is combining tolerances (of equal valued, equal toleranced resistors) really that easy?