Is this a 1.5 ohm ballast resistor?

[sotxbill]

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?

 

[sotxbill]

I was taught that..................The ballast resistor functions in such a way that as the current flowing through it increases, so does the temperature. As a result of the rise in temperature, the resistance will rise. As a result, increases in resistance will impede current flow across the network.

OR

These high-resistance ballast resistors work by being placed in electrical circuits in series with the load or other electrical devices, thus effectively limiting the current that passes through them. The resistance of ballast resistors can fluctuate with the current. The resistance increases if the current passing through the resistor rises over the threshold level. Hence, when the current declines, the resistance can also do so.


The ballast resistor attempts to maintain a steady current flow across a circuit in this way. Ballast resistors can be easily confused with load resistors in that they are installed the same way. However, ballast resistors act as variable loads in a system, contrary to load resistors whose load is constant with different currents and voltages.

 

[JMOR]

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?

 

[JMOR]

What the meter shunt does in dropping current/altering circuit does not matter in what George is doing in his original Dec 17 post, as that is a "four-wire Ohmmeter", where the current is fed to resistor being measured by one pair of leads and a separate set of leads (carrying no current beyond the microamp needed to register the meter circuitry) measures the voltage across the resistor under test. He then has accurate current and accurate voltage by which R=E/I. No matter if current would have been higher without the meter shunt.