0
\$\begingroup\$

My main concern is the short circuit on the right. Since I1 won't be zero the dependent current source will have current equal to 5*I1 Amps flowing through. The short circuit, however, implies no voltage drop across the dependent source. Is this possible?

Of course I could start writing down equations to see what's happening but I like trying to understand the circuit by inspection first.

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ "... The short circuit, however, implies no voltage drop across the dependent source. Is this possible? ..." For example an ideal inductor carrying a constant current has no voltage drop (vl = L.di/dt). \$\endgroup\$ – Dirceu Rodrigues Jr Aug 3 '18 at 16:28
1
\$\begingroup\$

There's nothing wrong with having a short circuit across a (theoretical, ideal) current source, independent or dependent. (Similarly, there's no problem having an open circuit on the output of a voltage source, which is a good thing if you have some batteries sitting in a drawer somewhere) It just means all current from the source will go through the source instead of anywhere else in the circuit.

If you want to analyze this mathematically, and know how much current goes out port 2 (for example if you're trying to find the transfer function of the 2-port), you can put a resistor across port 2 instead of a short; then find the limit of the port 2 current as the resistor value goes to 0.

\$\endgroup\$

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.