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I have the following circuit as shown in the picture. My question is how can I perform Node analysis in this circuit in the node \$V_{0}\$?

I want to find \$i_{1}\$ but without KCL or KVL. Just the node analysis in the node \$V_{0}\$.

I tried : \$\frac{V_{0}}{6} +\frac{V_{0}}{4}-1 = 0 \$ ignoring the ccvs. If I take it into account the dependent ccvs how the node analysis will be?

Any help?

(Side note: I know that KCL in node \$V_{0}\$ and KVL in the left sub circuit solves it but I want to see the node analysis how will be changed if I take into account the dependent source. The current source 1A is the circuit excitation in the terminal A,B)

enter image description here

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    \$\begingroup\$ Well, if you ground the bottom node then Vo/6 is i1. So the dependent source is Vo/3. Does that make sense, so far? \$\endgroup\$
    – periblepsis
    Commented Sep 19, 2023 at 11:10
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    \$\begingroup\$ So the voltage over the 4 Ohm resistor must be Vo minus the dependent source, which is Vo/3. So that means Vo - Vo/3. But that is just 2/3*Vo. \$\endgroup\$
    – periblepsis
    Commented Sep 19, 2023 at 11:13
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    \$\begingroup\$ So the current sinking into the 4 Ohm must be [(2/3)*Vo]/4 or 1/6 Vo. And the current sinking into the 6 Ohm must be Vo/6 or i1, as already mentioned. The sum of these two is Vo/3, not so? And these two sinking currents must be equal to 1 A. If Vo/3 = 1, then what is Vo? \$\endgroup\$
    – periblepsis
    Commented Sep 19, 2023 at 11:16
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    \$\begingroup\$ Just put it all together. You've got all the pieces now. The full expression would be written: Vo/6 {the current in the 6 Ohm} + (Vo - 2*i1)/4 {the current in the 4 Ohm} = 1 A {the current sourced into the node}. Then substitute i1=Vo/6 to get: Vo/6 + (Vo - 2*Vo/6)/4 = 1 A. \$\endgroup\$
    – periblepsis
    Commented Sep 19, 2023 at 11:22
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    \$\begingroup\$ You could just as well swap positions of the dependent source and the 4 Ohm resistor, since they are both in series with each other. The equation would be written out a little bit differently. But the result would be the same. See if you can do it that way. \$\endgroup\$
    – periblepsis
    Commented Sep 19, 2023 at 11:29

2 Answers 2

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I want to find i1 but without KCL or KVL.

I'd convert the dependent voltage source into a dependent current source like this: -

enter image description here

Then, it's clear that the current through the 4 Ω resistor is \$1 - 0.5i_1\$.

I'd multiply \$i_1\$ by 6 to get \$V_{O}\$ and, equate this to \$1 - 0.5i_1\$ multiplied by 4 (also equals \$V_O\$): -

\$6\cdot i_1 = 4(1 - 0.5i_1)\$

Do a few lines of rearranging to find that \$i_1\$ equals 0.5 amps.

Sanity check in microcap: -

enter image description here

Note that I haven't used KVL or KCL.

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If you don't want to apply KVL or KCL, if the network is in the right configuration (as in this case) then apply Millman's theorem:

Circuit

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