0
\$\begingroup\$

circuit circuit-workings I am carrying out node voltage analysis on a circuit that has 2 voltage sources.

  • One source is independent
  • The other is dependent on a current through a resistor in the circuit

After i carry out my analysis i have equations for the unknown node voltages however one equation contains the dependent source as per KCL.

Question

In essence i need to know what i am missing in my analysis,

  • should the current be found first through superposition?

all help is greatly appreciated.

\$\endgroup\$
  • \$\begingroup\$ Sharing the circuit you're analyzing and your work so far would be helpful. \$\endgroup\$ – skrrgwasme Apr 25 '16 at 16:17
  • \$\begingroup\$ I will post that now. \$\endgroup\$ – Eoghan_Mulcahy Apr 25 '16 at 16:17
  • \$\begingroup\$ Have you tried making a supernode around the dependent source the same way you would around an independent source? \$\endgroup\$ – The Photon Apr 25 '16 at 16:21
  • \$\begingroup\$ You method of converting to Norton equivalents should also work. But you need to define what is "I3". You ought to be able to write I3 in terms of node voltages. \$\endgroup\$ – The Photon Apr 25 '16 at 16:23
  • \$\begingroup\$ Hi @ThePhoton that is my problem at the moment i am not sure how to define it. \$\endgroup\$ – Eoghan_Mulcahy Apr 25 '16 at 16:24
1
\$\begingroup\$

Without dependent (voltage) sources one could normally define a matrix equation to solve for the values in the network:
[Z][I] = [V]

In this case though we must add in dependent voltage sources
[Z][I] = [V] + [k][I]

Apologies, I renamed I3 above as I2 below, per the usual custom of ordering currents left to right as shown in sketch. enter image description here

From this diagram, we extract the matrix equation suggested above. enter image description here

This rearranges as
[Z][I] - [k][I] = [V]

Rearranging again [I] = ([Z]-[k])^-1 · [V]

Solution of the matrix math is standard matrix math left to the reader. It produces an equation for each of the 3 currents defined in the sketch as a function of the single independent voltage.

This gives the currents as a function of the one independent voltage.

At this point
Va = (I1 - I2)·R2

And Vb = (I2 - I3)·R4

\$\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.