source
I know that source transformations work as shown in the top part of the attached image but what happens when the resistor is placed in a different location?
Does this give the same result? If so, how do I know which where to place the resistor if I performed the transformation the other way around i.e. from Norton eqv to Thevenin eqv?
Sorry for the poor quality of the image.
Traditionally you would want to use an instrumentation amp of some kind.
Here's a circuit that uses a ZXCT1009 instead. Accuracy is up to the 4.7 / 47 ohm resistor matching and the ZXCT's transconductance accuracy, about 1%.
Here's the way I teach it:
Adding or removing a resistor in parallel with an ideal voltage source changes only the current through (and thus the power associated with) the voltage source - no other circuit variable is changed.
Adding or removing a resistor in series with an ideal current source changes only the voltage across (and thus the power associated with) the current source - no other circuit variable is changed.
Thus, to solve for the other circuit variables, one may remove any resistor in parallel with a voltage source and short any resistor in series with a current source without affecting the circuit solution (except as mentioned above).
As an aside, do note that Thevenin and Norton equivalent circuits are equivalent only from the perspective of what an attached circuit 'sees'.
That is to say, if one had two black boxes with two exposed terminals each, one a Thevenin equivalent and the other the dual Norton equivalent, the terminal voltage and current would be identical for a given attached circuit.
However, one could easily tell the two apart with a thermometer. Do you see why this is so?
Best Answer
If inside a black box was a battery and series resistor, it would be impossible to tell which lead of the battery contained the series resistor.
In short, the equivalent circuit is the same.