# Inaccurate two-port network problem?

#### ElectricMagician

Joined Jul 26, 2012
57
The solution apparently involves assuming that the two networks ARE in series, and that there is a current going through the resistor! Almost everyone got the same answer: ~19.5.

This is against any knowledge I have of circuit theory. I just want to confirm that this is inaccurate before bringing it up with the professor.

Any remarks highly appreciated.

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#### t_n_k

Joined Mar 6, 2009
5,455
It's not clear to me what the original problem statement is and what in particular is troubling you about the problem.

#### ElectricMagician

Joined Jul 26, 2012
57
My apologies, the full problem is attached now.

My problem is that the black currents are supposed to be the same (since the networks are to be considered a Series-Parallel combination), and this is impossible with the red current going through the resistor.

Further more, we cannot assume no current goes through the resistor, and if we do it means that Vin = Vout (which is not only different than the "correct" answer, but also shorts the circuit.

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#### t_n_k

Joined Mar 6, 2009
5,455
Why can't the black input currents (i.e. through the series connected network inputs) be the same?

#### ElectricMagician

Joined Jul 26, 2012
57
Because black = blue and blue = black+red
and red <> 0

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#### t_n_k

Joined Mar 6, 2009
5,455
You are mistaken. Black does not equal Blue. That is not a necessary condition for the circuit to have a solution. You simply need to ensure Kirchoff's current law is consistent at the node.
Blue=Black+Red.
If you are not convinced then redraw the schematic with h-parameter equivalent circuits replacing the two matrix representations.

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#### t_n_k

Joined Mar 6, 2009
5,455
Hi ElectricMagician,

You are too easily discouraged!
Why not make your point more forcefully?
Clearly, there is a condition where blue equals black and (thus) Kirchoff's Law is not contravened. This is surely true if the red current is zero.
The two port h-parameter network ideally has galvanic isolation between input and output ports. Unless the input source and the output load are both ground referenced (or they share a common reference) then current cannot flow in the 1k feedback resistor.
So, I would venture that two solution are possible - with ground /common referencing applied or not applied.

#### ElectricMagician

Joined Jul 26, 2012
57
Well I was told by the marker that the red current is not zero so I can't argue much with that.

It has always been my understand that a voltage source has the same current going in/out its terminals.

Otherwise, separate connections to potential and reference would be used.

#### The Electrician

Joined Oct 9, 2007
2,887
The solution apparently involves assuming that the two networks ARE in series, and that there is a current going through the resistor! Almost everyone got the same answer: ~19.5.

This is against any knowledge I have of circuit theory. I just want to confirm that this is inaccurate before bringing it up with the professor.

Any remarks highly appreciated.

http://en.wikipedia.org/wiki/Two-port_network

Especially understand the part about the "port condition".

About two thirds of the way down, under the sub-heading "Combinations of two-port networks" is some relevant discussion.

The port condition can always be satisfied by the use of ideal transformers. If you place an ideal 1 to 1 transformer at the input of each two-port, things will work out.

Then you can add the two matrices to get an overall matrix, add the effect of the feedback and load resistors and get an answer.

If I do this, I get a voltage transfer ratio of -19.5, and an input impedance of 47.84 ohms.