The two-port network is powered by 10 mV voltage source and has admittance matrix: \$ Y = \left| \begin{matrix} y_{11} & y_{12} \\ y_{21} & y_{22} \end{matrix} \right| = \left| \begin{matrix} 2 \cdot 10^{-3} & -3 \cdot 10^{-5} \\ 0.5 & 2 \cdot 10^{-4} \end{matrix} \right| \$

From admittance matrix I know, that

\$ \left[ \begin{matrix} i_1 \\ i_2 \end{matrix} \right] = \left[ \begin{matrix} 2 \cdot 10^{-3} & -3 \cdot 10^{-5} \\ 0.5 & 2 \cdot 10^{-4} \end{matrix} \right] \times \left[ \begin{matrix} u_1 \\ u_2 \end{matrix} \right] \$

\$ i_1 = 2 \cdot 10^{-3} \cdot u_1 - 3\cdot 10^{-5}\cdot u_2 \\ i_2 = 0.5 \cdot u_1 + 2\cdot 10^{-4}\cdot u_2\$

enter image description here

The task is to draw Thevenin's circuit for this and to compute its parameters. All I can imagine when I hear "Thevenin" is circuit like this: enter image description here

I suppose, the parameters are \$U\$ and \$R_i\$. I would compute \$ R_i \$ as \$ \frac{1}{y_{11}} = \frac{10^3}{2} = 500 \space \Omega\$. Is that right? If not, how to compute that?

And I have no idea how to compute \$U\$. I know it is equal to voltage between \$2\$ and \$2'\$ nodes in this picture: enter image description here

Any hint or explanation?

  • \$\begingroup\$ Can you post the complete circuit? \$\endgroup\$ – Martin Petrei Feb 5 '14 at 15:11
  • \$\begingroup\$ What do you mean? This is all I have. We have no other information. \$\endgroup\$ – user50222 Feb 5 '14 at 15:18
  • \$\begingroup\$ sorry. I thought you had the diagram associated to the matrix. \$\endgroup\$ – Martin Petrei Feb 5 '14 at 15:32

To calculate the Thevenin voltage the port 2 has to be opened, thus the current is zero: \$ i_2=0 \$.
You get: \$ 0.5⋅u_1+2⋅10^{-4}⋅u_2=0 \to u_2= -2500⋅u_1 \to u_2=10mV*2500=25V\$.
To calculate the Thevenin resistance, do the same but with \$ u_2=0 \to i_2=0.5u_1 \to i_2=5mA\$.
Then, since the voltage drop is all on the Thevenin resistance $$ R_{th}=\frac{u_2}{i_2} \to \frac{25}{5m}=5k\Omega$$
Notice it is the \$(y_{22})^{-1}\$.

  • \$\begingroup\$ Do you mean to put \$u_2 = 0\$ in the first equation with \$i_1\$ or in the second, as in your answer? Is \$R_i = \frac{u_1}{i_1}\$ or \$ \frac{u_1}{i_2} \$? I could understand the first case. If the second is right, why? \$\endgroup\$ – user50222 Feb 5 '14 at 15:56
  • \$\begingroup\$ Added what I meant \$\endgroup\$ – Alex Pacini Feb 5 '14 at 16:03

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