# Find the power of the real current source?

I am asked to find the power of the real current source I1: For that I would need to know the voltage of I1, and I want to find the Thevenin equivalent for the rest of the circuit. Now in my textbook it says that I can apply this theorem to replace some part of the circuit between "two points" A and B, however here I'm not sure where to put A and B, my assumption was that since I1-R4 are in parallel the Voltage/Potential between them is the same so I can do something like this: simulate this circuit – Schematic created using CircuitLab

And now just find the Voltage between A and B using nodal analysis (also Rth), are my assumptions correct?

After which $$P_{Ig}=U_{Ig}Ig-\frac{U_{Ig}^2}{R_4}$$

## 1 Answer

Just remove the current source. Do not remove R4 (If you remove R4 it will distort the result).

Then determine the Thevenin Equvalent of the remaining circuit: get equivalent circuit of voltage source $V_{Th}$ with a series resistance $R_{Th}$.

The voltage across the current source then is $V_{CS} = V_{Th}$ ± 1mA * $R_{th}$

(The ambiguity of the ± sign comes from the fact that it is not clear whichg side (A or B) is the GND side of the Thevenin Eq.)

Now you have the voltage across $V_{CS}$ and the current through $I_{CS}$ (=1mA) the current source to calculate the power.