Addy buddy..
If it was a foot of wire.. ( 1/1000)(1000)... ( 1)
4.2v power source at one end..
and the resultant voltage read at the other end of the wire..As 3.7v... ( losing 0.5v, through 1 foot, a resistance of 200mOh, @ )
That wire is .... very very small... 1.25w ( 0,5(v) x (2.55(A))... would be dissipated in the circuit and the wire would be between 32g and 33guage wire.. and.. would not support the 2.5A.. but only about 0.5A... before.. getting really hot really fast.
A 18 gage wire does NOT drop 0.5v in 1 foot. VoltagePotential.
A 33g wire does. 33g wire is 206mOh / foot. (0.2Ohm per foot. ) Applya voltage (X) to a 33g wire, and.. in a foot, length, you will measure (X-0.5)v
I very highly doubt he has 32g multimeter wires.
I betcha he has 18gage multimeter wires. 18g should not drop half a volt (0,5v) potential/foot. Maybe there was a (hidden?) break in the wire that was a bottleneck/high resistance point.. some broken strands, not making good contact... .. a failing wire.. . ( that would create a hotspot.. Too)... Then when ( He, She) replaced wires, everything was fine again.. IDk
Im not saying " Try another wire" was not a good idea. It was, and is. The wire may have been dropping (v) but @ 2.5A it would have certainly burnt.. if it was... 200mOh.. ( resistance req for our mystery 0.5v drop)
200mOh resistance WILL burn (dissipate power calculations) a 33g wire at 2.55A/3.7v..... Certainly. The wire (33g) cannot dissipate 1.25w...( waste power) ( dissipate power).... when it is feeding 10w through it.. NOOOPe...It wont last long trust me... It can only handle like... 0,5A before getting hot... ( 1.85w MAX for a foot of 33g)( then magic smoke, above that.. ).
Thats all I'm seeing. Given the basic nature of the equations. Given what we know offhand.