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Why in the following solution the 6Ω resistor is not considered?

circuit

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    \$\begingroup\$ I suggest you do the calculation twice: 1) without the 6 ohms resistor in place (replace it with a wire) 2) with the 6 ohm resistor in place. Then compare the answers. \$\endgroup\$ – Bimpelrekkie Feb 8 '16 at 15:56
  • \$\begingroup\$ Your sample calculation is wrong. KCL sums currents into a node. The value 2v<sub>0</sub> is not a current term. \$\endgroup\$ – Michael Karas Feb 8 '16 at 16:25
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    \$\begingroup\$ @MichaelKaras - if you bothered to look at the diagram you would see that there is a voltage dependent current source whose value is 2Vo, so there is nothing wrong with its use in the equation \$\endgroup\$ – I. Wolfe Feb 8 '16 at 16:30
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    \$\begingroup\$ @MichaelKaras Yes it is. Look to the right of the schematic. Don't confuse him/her more. \$\endgroup\$ – Asmyldof Feb 8 '16 at 16:31
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    \$\begingroup\$ You found out that in order to calculate V0, the 6 ohm resistor does nothing. Why is that so ? It is because what you're interested in is the current through the 4 ohm resistor. As that current is defined by 2 current sources in parallel (so you can sum their values) the value of the 6 ohm resistor is irrelevant. It could be any value because that would not change the current as that current is set by the 2 sources. If Vo is the voltage across the 6 ohm resistor then indeed the value of the 4 ohm resistor does not matter for the same reasons. \$\endgroup\$ – Bimpelrekkie Feb 9 '16 at 9:53
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The amount of current flowing from the 6-ohm branch into the node can be determined without considering the 6-ohm resistor in the circuit. The current in that branch is easily obtained by dividing the potential difference across any component in the branch by the impedance of that component.

So the current in the 6-ohm branch is simply \$I = v_0/4\$. And this can be added to other currents flowing into the node (\$10 A\$ and \$2v_0\$) to get a node equation according to KCL.

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  • \$\begingroup\$ So if Vo were the voltage of the 6Ω resistor the solution would be this: Vo/6 + 10 + 2Vo = 0 ⇒ Vo + 60 + 12Vo = 0 ⇒ 13 Vo = -60 ⇒ Vo = -4.615. It's right? \$\endgroup\$ – ᴜsᴇʀ Feb 9 '16 at 9:25
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    \$\begingroup\$ Yes, if Vo was the voltage across the 6-ohm resistor \$\endgroup\$ – TisteAndii Feb 9 '16 at 9:30
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This is simply because the variable of interest is Vo.

There is only one current running through a branch. The current through the 6 ohm is the same current running through the 4 ohm, which is the current through the branch. Since KCL deals with currents leaving or entering a node from its branches and the current through the 4 ohm is the same as the current through the branch, we don't have to include the 6 ohm resistor and the equation solves directly for V0.

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  • \$\begingroup\$ it's not that the 6ohm resistor isn't connected to the node, that phrase is very misleading. You could very easily make V0 the voltage across both 4ohm and 6ohm resistors, which would make both relevant. the only reason 6ohm is not relevant is because V0 is defined as the voltage across the 4ohm resistor. likewise you could make V1 the voltage across the 6ohm reistor and use it, ignoring the 4ohm resistor (of course this would cause issues with the dep current source, but you see my point) \$\endgroup\$ – I. Wolfe Feb 8 '16 at 16:35
  • \$\begingroup\$ That's true, but in that case, the resistors would have to be combined, then the equivalent resistor would be connected. I guess I wasn't sure if the question was in reference to the use of KCL, or calculating V0. \$\endgroup\$ – JosephQ Feb 8 '16 at 16:40
  • \$\begingroup\$ It is most likely a school problem to calculate Vo using KCL. The 6 ohm resistance is completely dismiss-able, and OP was wondering why that is. The answer is not that the 6 ohm resistance isn't connected to the right node, as it is in series with the 4 ohm and between the same two nodes. The reason is because the problem is an arbitrary school problem, and all you care about is the current in the branch, which does not require the 6ohm to compute. I could swap the positions of the two resistors and get the same answer, making your comment of nodes irrelevant. \$\endgroup\$ – I. Wolfe Feb 8 '16 at 16:50
  • \$\begingroup\$ I see your point. Let me know if my edited answer is better. \$\endgroup\$ – JosephQ Feb 8 '16 at 17:00
  • \$\begingroup\$ Yup, much better answer now. \$\endgroup\$ – I. Wolfe Feb 8 '16 at 17:22
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KCL only depends on the sum of the currents flowing into any particular node. Taking the top node, you have three branches of current:

The 10A source and the variable 2*Vo source are pretty much given to us. In the remaining branch, we are given a 4 ohm resistor with a voltage difference across it. V=IR will allow us to find the current in that one branch, which would be the same at ANY point in that one branch.

That being said there should be ways of calculating this with the 4 ohm resistor but it would simply add extra complication for no reason.

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The 4\$\Omega\$ resistor is in series with a current source (actually a parallel combination of two current sources).
Therefore the current through the 4\$\Omega\$ resistor must be the same as the current through the current source (actually sum currents through both current souces), no matter what other components (6\$\Omega\$ resistors or any other resistor or voltage source) are also in series.

If you care only for currents you can ignore resistors in series with current sources just as you can ignore resistors in parallel with voltage sources if you care only for voltages (the dual case).

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Let's take a look at this. I'll name the currents to make the math clearer:

schematic

simulate this circuit – Schematic created using CircuitLab

\$I_i\$ and \$I_d\$ both flow from ground to node A. \$I_0\$ flows from node A to ground through the resistors. So our KCL equation for node A is:

$$I_i + I_d = I_0$$

\$I_i\$ is an independent current source, and we already know its current, so we're done there. \$I_d\$ is a voltage-controlled current source. Its controlling voltage is the voltage across the \$4\ \Omega\$ resistor. Note that I've (arbitrarily) made \$I_0\$ flow into the negative terminal of this resistor. The dependent source's current is thus given by:

$$I_d = 2 \cdot -I_0 \cdot 4 \Omega = -8I_0$$

Now we have two equations with two unknown variables:

$$I_0 = I_d + 10 \mathrm A$$ $$I_d = -8I_0$$

Note that the \$6\ \Omega\$ resistor does not appear in either of these equations. That's because our sources are all current sources, and the one control voltage depends only on the sum of the two source currents. The absolute voltage is irrelevant.

Let's finish this off. Substituting the second equation into the first gives:

$$I_0 = -8I_0 + 10 \mathrm A$$ $$9I_0 = 10 \mathrm A$$ $$I_0 = \frac {10} {9} \mathrm A \approx 1.111 \mathrm A$$

Since we know \$I_0\$, we can calculate \$I_d\$:

$$I_d = -8I_0 = -\frac {80} 9 \mathrm A \approx -8.889 \mathrm A$$

The current through \$I_d\$ is negative, which makes sense given the polarity of \$V_0\$. Let's double-check our result using the original KVL equation:

$$10 \mathrm A - 8.889 \mathrm A = 1.111 \mathrm A$$ $$10 \mathrm A = 10 \mathrm A$$

Success! Feel free to post a comment if you have any questions.

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