WHEN.... you measure current.. your meter is changing the circuit SO MUCH that you are no longer getting the real reading that would be there with the meter out. IF.... the resistor is 1 ohm, and the coil is 1.5 ohms, and the meter dropping resistor is what?? one ohm maybe??? IF so, you have increased the resistance in the circuit by 1/3 and dropped your current by one third of what it is with the meter out of the circuit. SO... what is the resistance of your meter... in the 10 amp position... that you are now putting into a 2.5 ohm circuit (assuming the resistor is 1 ohm, and the coil is 1.5 ohms).. USing the old "ten times" rule, the meter resistance would have to be .25 ohms or lower to give you a half chance of getting close and correct reading. Anyone have a wheatstone bridge?