rs14smith
Member
Actually it is pretty simple.
The shunt is no more that a very high power low resistance resistor. When current runs through the low resistance there is a small subsequent voltage drop. That voltage drop is proportional to the current passing through to the load. The tie points on the block for the meter are precision points across the resistance. I can post a few images of shunts I have laying around if it would help but what you see is what you get.
Ron
Yeah I figured it worked similar to that, just did not appear that way. So would I have to worry about my nonstandard 36v if I just go with an analog meter?
Like if I found a 50A > analog meter (includes shunt), would it be as simple as hooking it up in series with my motor, and I'm done? Or will I still probably have to use a regulator of some sort?