0
\$\begingroup\$

I was solving this question related to two port parameters. I am confused as to what will be \$I_1\$ when \$V_1=0\$ i.e. the input port is short circuit-ed, since the controlled current source, \$gv_x\$ will be parallel to the short. From one perspective, it seems like it will form a loop in which only \$gv_x\$ will flow, hence making \$I_1 = gv_x\$ and \$Y_{12}=g\$. And from another it seems like magnitude of \$I_1\$ will be difference of current through the \$1\Omega\$ resistor and the current drawn by the current source i.e. \$|I_1|=|I_{1\Omega}-gv_x|\$ as per KCL at the node connecting the controlled source, the 1 ohm resistor and the short. Another perspective I thought of was, since the two input terminals will get shorted, all the current flowing through 1 ohm resistor will flow through the short and the controlled source won't matter, since current prefers path of least resistance and an ideal current source has infinite resistance, so no current would be drawn by the current source.

I find circuit analysis quite confusing, I am not even confident as to whether all these scenarios I described above make complete sense.Any tips on how to get better at it?

enter image description here

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

When in doubt always stick to the basic principles you know.

You, correctly, mentioned KCL, that's the point!

So, why in the world should not apply to the black contour below?

enter image description here

So, now you have your answer

$$ I_1=g\,v_\mathrm{x}\pm I_\mathrm{R1\Omega} $$

Where you should match plus or minus according to the conventional current flow as you choose it in R1.

\$\endgroup\$
2
  • \$\begingroup\$ Quick question regarding current sources, regardless of whether are independent or depend on some other quantity, they will draw the current from some other node right? i.e. when writing KCL equations, is it correct to view them as drawing current from some node? even for independent current sources \$\endgroup\$ Commented May 19 at 14:42
  • \$\begingroup\$ @KoustubhJain what you asking not too clear me. But keep in mind that KCL applies to any closed contour. Even if you draw one around your generator. If I'm not answering you, open a second question on the new topic. \$\endgroup\$
    – carloc
    Commented May 19 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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