1
\$\begingroup\$

I am asked to find the power of the real current source I1:

enter image description here

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:

schematic

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

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.