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Electronic device is on pulling a bigger load, my meter will not read more than 20 amps?
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You need a meter with a higher range.
The .4 ohm resistor has lowered the current enough to let me keep my amp meter in the circuit.
The difference between 18.6 amps and 14.65 amps for load a = 3.95 amps.
If I add 3.95 amps to 18.6 amp load B = 22.95 amps estimated current with out the .4 ohm resistor. Wonder how accurate this is?
I had to correct the picture I forgot to draw the load.
But your 0.4 ohm resistor has reduced the voltage to the load significantly. While using a shunt resistor is a valid way to measure current, it's value should be low enough that it's insertion loss is negligable. My standard shunt resistor is a four terminal, 50 amp, 1 milliohm, instrument grade shunt similar these.
Even with such a low value, they are accurate due to the kelvin connection the extra two terminals give. As long as your meter can read millivolts, you can get a good measurement. 1 millivolt = 1 amp.
As for just extrapolating the current draw when you device has the B load, that will only be accurate if the voltage to current transfer function of your device is known. If you don't know that, then you need to make a current measurement of all three states with your the correct operating voltage at your devices input terminals.
I expect that, when feed the full 15 volts, your idle, Load A, and Load B currents will all change.
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**broken link removed**
As for the confusion others, including myself had until I re-read your post a few times, it's because your drawings and your text don't match. Your schematic just shows a simple load, but your text describes some unknown device in between the power and one of two loads. It would be more clear if your drawing showed [device?] where you now show the load, and two more wires from it to your A or B end load.