I can't though comprehend how solder on a seemingly random spot, with no care taken to ensure the solder joint is uniform will decrease the resistance, especially when ideally the cross-sectional area of the original shunt would ideally be uniform. If the original shunt has a uniform cross-section, wouldn't the resistance be the same along the length of the shunt?
The blob of solder makes the cross sectional area at the point of the solder no longer uniformly the same as the cross-sectional area everywhere else
The formula for resistance of a uniformly circular wire is
![](http://latex.codecogs.com/gif.latex?R&space;=&space;\rho\times{{\ell}\over{A}}&space;=&space;\rho\times{{\ell}\over{\pi}r^2})
where
![](http://latex.codecogs.com/gif.latex?\rho)
is the resistivity of the medium,
![](http://latex.codecogs.com/gif.latex?\ell)
is the length of the wire in meters and
![](http://latex.codecogs.com/gif.latex?A)
is the cross sectional area in square meters. Thus
![](http://latex.codecogs.com/gif.latex?r)
is the radius of the wire.
This works because it's uniform, and you can multiply it by the length. But the solder blob adds a cross sectional area that is changing along the length of the wire, now you have to integrate it. And the resistivity
![](http://latex.codecogs.com/gif.latex?\rho)
is different at that point too, so you would have to do the two regions separately to get the total resistance, and these two regions are in parallel. The math for an unknown blob of solder would be hard, at best. But the resistance will still decrease as you increase the total cross sectional area at that point where the solder blob is. Basically, it's adding a small parallel resistance at that point, it has its own resistivity
![](http://latex.codecogs.com/gif.latex?\rho)
different from that of the base shunt wire, and it has its own tempco, and its own total resistance along its length, which is hard to calculate because it's oddly shaped.
Adding a blob of solder may not seem precise, but they are tuning it at the factory, so they are watching the measurements as they tune it.
As an example that is easier to visualize, consider the image below, I've soldered 4 chip resistors together, the size doesn't matter, assume they are all value 100 ohms. You have no trouble believing that the structure in the image is 250 ohms: 100+50+100. without the top resistor, it is 100+100+100 = 300 ohms. So adding the top resistor lowers the resistance because there is a parallel resistance there. You can visualize increasing the length of the lower section with more and more 100 ohm resistors in series, but the section with two in parallel will always be less resistance, i.e. 50 ohms, at that spot.
![](http://i.imgur.com/vzw8gPe.png)
Dave, do an experiment for us?
Yes, Dave can prove it with an experiment