current measurement
I'm not sure how to interpret the current value in this basic schematic. If you simulate it, CircuitLab will plot a value for the current flowing through the Output node. Why is it a negative value? What convention does it use?
simulate this circuit – Schematic created using CircuitLab
You have found the total current when you short out any two voltage sources, but you also need the current going the other branches of the circuit.
your i_1 is i11, your i_2 is i22 in the diagrams above.
After you have solved the circuit for each source individually you can add together the currents of each branch to get the total current of the final circuit with all sources.
So the current through V1 with all sources would be i11+i21+i31. You can then use the total current to find any voltage drops you need.
Pervious:
I it is not possible, with out solving the circuit in some fashion to know the directions of current you are taking about in your questions. It would be a good idea to include a diagram like on fo the below or add an up/down, left/right annotation to your current where appropriate.
You can draw your current arrows in any direction you want, if the current is actually flowing in the opposite direction of your arrow then then it will just end up being a negative current in your solution.
example diagrams:
[This started as a comment, but I've run out of room.]
First of all, Welcome to EE.SE.
Second, I'd like to comment against the voltage dividers at the inputs. Suppose, the resistors in the dividers are 1%, then voltage difference caused by the mismatch between the dividers will be comparable (not much smaller) to the voltage drop across the sense resistor. You can test this hypothesis by connecting the high sides of the voltage dividers together (i.e. shoring out the sense resistor) and observing the output of the InAmp.
Another thing to try, is to test your InAmp circuit without the dividers. (V3 would have to be set to a lower voltage.)
Typically, high-side current sensing in a situation like this is done with help of a current mirror IC. See this application note, and another one.
Best Answer
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) As drawn. (b) A more usual power flow. (c) Alternative configuration.
The confusion may be caused by CircuitLab's default current source configuration which is (arguably) upside down for this simple circuit. If we follow the current flow (red arrow) in Figure 1a we can see that it is going up through R1 and therefore the top terminal must be at a lower potential than the bottom.
Figure 1b shows a more typical schematic with clockwise current flow and positive rail at the top.
Figure 1c shows a more typical negative current source configuration.
From CircuitLab Basics:
While this doesn't directly answer your question it does give a clue that each circuit element node will either have current in or out of it depending on the nodes relative voltages. In your example the top of the resistor is at lower potential than the bottom so current must be flowing out of it and hence the negative sign.