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EDIT: I am not removing the linked question and redoing the question.

How do you formulate an equation on a node with two voltage sources connected in it, regardless if it is a supernode or not?

(You may ignore the statements below this.)


Though this question has been answered here, I still have doubts in mind.

Consider the circuit below. In the linked question above, both voltage sources have the same value for voltage. The circuit below has 30 V and 20 V. Will it still be zero just like the one in the linked question? If not, how should you create the nodal equation for node 1? Usually, if there is only one voltage source connected to the node, and if it is connected to the ground, you just equate V to the value of voltage source. (Let's say below, if there is no 2 kΩ resistor below node 2, V1 = 30 V.)

An additional question as well. What if there are two voltage sources, but the other one is not independent but a VCVS or CCVS? How do you formulate the equation at that certain node with those branches connected?

enter image description here

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  • \$\begingroup\$ "Will it still be zero..." Will what still be zero? \$\endgroup\$ – Alfred Centauri Nov 15 '13 at 14:14
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    \$\begingroup\$ The linked question has nothing to do with this question other than it concerns voltages at nodes, but which nodes I wonder? \$\endgroup\$ – Andy aka Nov 15 '13 at 14:27
  • \$\begingroup\$ It's not the same topology. Look at the Voltage sources, they are reversed in the question you linked to, and not in this schematics. Thus no, Vo (I suspect that Vo is the thing that would be zero) is not equal to zero here. Just by staring at the schematics. \$\endgroup\$ – Blup1980 Nov 15 '13 at 14:29
  • \$\begingroup\$ I looked more seriously and as @Andy aka just said: The linked question has nothing to do with this question. What do you want to compute ? \$\endgroup\$ – Blup1980 Nov 15 '13 at 14:31
  • \$\begingroup\$ I think you are confused with nodal analysis concepts (in particular with supernodes and references). The circuit you linked is not related with this one. I suggest you check these concepts again :) \$\endgroup\$ – Chuz Nov 15 '13 at 15:46
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In general, you don't create an equation for node 1. You take nodes 1, 2 and 3 together as a single node, called a generalized node, aka supernode, for current summing purposes, and create an equation for that. Like this: Let the voltages at nodes N1, N2 and N3 be called \$v_1\$, \$v_2\$ and \$v_3\$, respectively. You can sum the currents out of the supernode: $$ \frac{v_2}{ 2\mathrm{k\Omega}} + \frac{v_3}{ 5\mathrm{k\Omega} }+\frac{v_1}{ 4\mathrm{k\Omega} }= 0 $$ You now have three unknowns and one equation, but you know that \$ v_2 = v_1 - 30 \$ and \$v3 = v1 - 20\$, so you can back-substitute into the previous equation and solve it for \$v_1\$

Dependant sources are handled in the same manner.

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It's difficult to ascertain precisely what your question is so this is just a stab at an answer.

For node 1, the KCL equation is:

$$\dfrac{v_O - 30V}{2k\Omega} + \dfrac{v_O - 20V}{5k \Omega} + \dfrac{v_O}{4k\Omega} = 0 $$

The equations for the voltages at nodes 2 and 3 are self-evident.

If, say, the 30V source were a controlled source, e.g, \$(k \cdot v_O)\$, the first term in the above equation becomes:

$$\dfrac{v_O - k\cdot v_O}{2k\Omega} = \dfrac{v_O(1-k)}{2k\Omega}$$

If this doesn't address your question, then follow-up in the comments or with an edit to your question.

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How do you formulate an equation on a node with two voltage sources connected in it, regardless if it is a supernode or not?

Supernodes are not pre-defined as part of the circuit. They are something you define when doing the nodal analysis.

When you find a voltage source in your circuit, it is up to you to define a new supernode that encompasses the two nodes connected to the source.

After you have defined supernodes that encompass all the voltage sources in your circuit, all of the voltage sources will be "hidden" within supernodes and there will be no need to write equations for nodes with voltage sources connected to them.

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