2
\$\begingroup\$

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?

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

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}\$.

\$\endgroup\$
  • \$\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

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.