Hi there,
It's true that regular solder has much less conductivity than copper, so that a copper trace of a certain width and thickness covered with a layer of regular solder that has the same width and thickness would carry only about 10 percent more current, but only if the electrical conductivity was the only thing to consider here, but there is another consideration. The electrical conductivity is about 10 times less than copper, and unfortunately the thermal conductivity is also about 10 times less than copper. This means when we coat a trace of copper with solder in addition to lowering the resistance we are also decreasing the trace's ability to dissipate heat by covering it with a (somewhat) heat insulator.
The ability of the trace to dissipate heat has a lot to do with it's current carrying capability, so when we reduce it's ability to dissipate heat we necessarily decrease it's ability to conduct current. Since the copper trace has a fiber or glass backing, when already one side of the trace is covered by a pseudo heat insulator, and the edges are not thick enough to have much impact on the heat dissipation, so when we cover the top with another pseudo heat insulator (albeit a somewhat poorer one though) we are again reducing the ability of the trace to dissipate heat and thus lowering it's ability to conduct current.
Just what is the overall impact then, having one parameter lower the resistance and the other decrease the heat dissipation factor?
The conductance of each trace (one being copper and the other solder) would be combined and then the temperature rise of the copper and solder combination would have to be calculated knowing the thermal conductivity is about 10 times less for solder, and that could be compared to copper alone.
With only one 'layer' of solder (a 'layer' being the same thickness as the copper) the current carrying capacity will most likely go DOWN due to the temperature rise, and multiple layers may or may not help the situation until some break point where the sides of the trace begin to have a more significant effect by adding significantly to the total dissipation surface area. It's hard to predict this action exactly though.
So how about a simpler way to deal with all these variables? Luckily, there is a way...
One way for example would be to etch a trace say 6 to 12 inches long on a scrap piece of PC board material that is the same type to be used with the final circuit. Next we would attach terminals to each end and run a DC test current through the trace. We would then measure the current through the trace and measure voltage drop. Knowing the voltage drop and current, we would calculate resistance.
Next we would cover the trace with a layer of solder, then test again and compare results.
A third test might be to try apply a heavy load of solder and then measure again.
Note however that in each case enough time has to pass for the trace to heat up to equilibrium before a measurement is made, and that it would be best to use the actual max current for the test that will be found in the actual application. It will certainly not work to use a small test current like 1ma, but the full 10 or 20 amps if that is what is needed for the finished circuit. A DC meter able to measure down to 1uv would be a good idea too, and it that isnt good enough than the test trace will have to be lengthened.
Silver solder comes in something like 2 percent to 4 percent for electrical use i think, and im not sure that that will decrease the resistance very much. One of the reasons for using that solder is because it flows better than regular solder and probably adheres better too...better conductivity is probably over stated for this kind of product.
Another variable is the surface condition of the added solder. If the solder is left to be very shiny, that will cause less dissipation too.