0
\$\begingroup\$

I'm still new to EE, and taking a introduction class right now. I have a simple question regarding nodal analysis.

Suppose that we have three things connecting in parallel,

  • One 12 voltage source with a resistor in series
  • One resistor by itself
  • One dependent voltage source where its voltage is equal to the voltage across the first resistor

These three are connected in parallel, the negative terminals are connected each other for the voltage sources.

Suppose I want to use node voltage to analyze this circuit.

What I have so far is, choose the node connecting the negative terminals as the reference node, and the node connecting the positive terminals as Node 1.

Then $$I_1+I_2+I_3 = 0$$

But how can I calculate the I_3 when there is only one voltage source there without any resistor?

\$\endgroup\$
  • \$\begingroup\$ Suppose that we have four things connecting in parallel, One 12 voltage source with a resistor in series, One resistor by itself, One dependent voltage source where its voltage is equal to the voltage across the first resistor. FOUR? \$\endgroup\$ – Dilip Sarwate Feb 20 '12 at 4:35
  • \$\begingroup\$ miscount sorry.; \$\endgroup\$ – geraldgreen Feb 20 '12 at 4:36
  • 2
    \$\begingroup\$ Could you draw up this circuit and post it in your question? It would help to confirm your description. \$\endgroup\$ – Jim Feb 20 '12 at 10:28
  • \$\begingroup\$ You should add a Homework tag if this is homework (it certainly sounds like it!) Hint: \$I_3 = -I_1 - I_2\$. You know \$I_2 = V/R_2\$, you know that \$V\$ equals the voltage of the dependent source and so is equal to the voltage across \$R_1\$. Now write an expression for \$V\$ by going from the reference node to node 1 through the branch with the independent voltage source and \$R_1\$, adding up voltages as you go. \$\endgroup\$ – Dilip Sarwate Feb 20 '12 at 13:07
  • \$\begingroup\$ @DilipSarwate - We use \$ as a TeX delimiter here; the $ sign was used too often when discussing prices and causing whole paragraphs to get \$rendered like this.\$ I've fixed your comment, sorry for the confusion. \$\endgroup\$ – Kevin Vermeer Feb 21 '12 at 3:49
3
\$\begingroup\$

The verbal description of the circuit sounds like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Unfortunately, this circuit is unsolvable. VCVS1, configured with gain 1, is asked reproduce the voltage present across R1, while simultaneously being forced to 12V plus the voltage across R1. VCVS1 needs non-unity gain in order for this to be solvable.

\$\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.