you would need a really accurate, really small resisitor. The main circuit has a resistance of 0,18 Ohm, if you make a shunt 1/100th f that it will have a voltage drop of 0,22V. But that means a shunt of 0,0018 Ohm. I see 2 ways you could try.
1- take 50 meter of copper wire (i would take my roll of extension cord, though I’m not sure that could take 120A) . Measure the resistance of this ( say that’s 2,0 Ohm) Then make a linear interpolation and cut a short piece from the wire. Use this as resistor (in my example, this would be a piece of 4,5 cm )
Now, this would still give 120A*0,22V=26W ! of heat dissipation in this wire, which is probably too much. And if the wire gets hot, the resistance will change… Hm. Cutting the short wire in 2 isn’t very practical, so lets use 10 equally sized pieces in parallel to build the shunt. This divides the resistance by 10, voltage drop is now 0,022V.
You can use the above multiplying method to build a ‘radiator’ of wires and build op a resistor that has a small resistance and enough surface to be cooled.
Also, if you can get your hands on a spool of constantan wire, that’s very good to make resistors from. This is an alloy designed to have the same resistance as temperature increases.
Method 2: For this method you need an Amp-meter that can measure 120 Amp (mine sure can’t)
Just use a piece of the regular wire in the circuit as shunt. Measure the voltage across this wiie with the circuit described above while you place the Amp-meter in the circuit. You can now see what voltage drop across this wire corresponds to what Amperage. Basically, if you have an amp-meter, you can build one
hope this helps,