Recent content by The Electrician

Proceed like this. First note that the 6 ohm resistor is connected across voltage sources so it has no effect on the values of v1, v2 and v3; remove it--that means i6=0. Treating v1, v2 and v3 as a supernode means that you don't write 3 KCL equations for each of v1, v2 and v3. Instead you...

You are forgetting what BvU said in post #15: "No worries about the directions of the currents: the correct sign will come out in the analysis." The arrows on the diagram may not be showing the direction of physical current flow. What if i2 has a negative value?

Look at the diagram in post #15. i2, i4 and i5 are shown.

v2 receives i2, i4, i5.

Fist you tell me if you have even heard the term "supernode", much less studied it. https://en.wikipedia.org/wiki/Supernode_(circuit)

Have learned about the concept of a "supernode"? As I said in post #19: This network needs only one KCL equation and two constraint equations. The 3 nodes can be treated as one supernode.

Note that the 6 ohm resistor does nothing. It can have any value, or it can be missing. Also note that the problem statement requires a nodal solution.

In other words, i5 is not indpendent of i. Likewise, v2 is not independent of v1. These are constraint equations. Since a network with 3 nodes like this would need at most 3 equations to solve it, given more than 3 equations they are not all independent. This network needs only one KCL...

In order to get a unique solution, the equations must be independent. The nine you propose are not independent. Besides which the TS seems to be gone away.

These might be helpful: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/serres.html http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html#c1

Those blue things are trimpots. They do the calibration. Like this:

Your source transformation method didn't find I1, I2 and I3.

Babadag is showing how to setup the loop equations in matrix form using a text description. Using standard linear algebra math symbols and using matrix inverse rather than Cramer's rule:

Reference this: https://en.wikipedia.org/wiki/Two-port_network What I'm getting at is that you are using the output port of the h-parameter two-port you created as the input you're feeding the opamp output into, and taking signal fed into the minus input of the opamp from the input of your...

Why have you chosen the orientation of your h parameter two port so the signal flow is from left to right? Don't you want the feedback to flow from the opamp output back to the minus input, which would be right to left?