as I said in my previous post but to add to that, I suppose that the total area of the mast, which had to be saturated, the bonding wire connections, which most likely weren't perfect, the total area and length of the bonding wire plus the positive wire in the radar bundle in contact to the mast must have caused a high enough resistance to depress the current down to 6 amps. Remember, I = E/R so the current will be the result of the voltage (say, 13.2 VDC) divided by how ever many ohms of resistance was in the circuit (Mast, wire and connections). If we had 13.2 volts (E), 6 amps known then we can solve for R. R=E/I. 13.2 volts divided 6 amps equals 2.2 ohms of resistance. Ohms law, E=IXR, R=E/I and I=E/R. 2.2 Ohms of resistance is a circuit is a bunch if the circuit didn't include any coils, resistors and such. If you took a single #14 wire and stretched it from the radar at the mast to the tip of the prop shaft and back and put an Ohm meter on it the resistance in that length of wire would be almost nothing. 2.2 ohms is a bunch. Somebody, please check my math. As always, IMHO.