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Can we apply nodal analysis to find out the current in each branch for this circuit:

Circuit

I am having trouble to find the correct reference point where I can take V = 0.

(If you are down voting the question, please mention the reason.)

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    \$\begingroup\$ Please add identifiers to each element, with which the referencing of an element is easier. The 10 V source and the 10 kOhm resistor (which one ;-) ?) can be swapped, then you get a resistor triangle, with a Y-Δ (Wye to triangle) conversion you can get started. \$\endgroup\$ Jan 7 at 7:04
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    \$\begingroup\$ Previous comment was a practical approach (more engineer like than mathematical). With nodal analysis, you can choose any node as 0 V! I would use the minus of all voltage sources as 0V. \$\endgroup\$ Jan 7 at 7:11
  • \$\begingroup\$ I am an absolute beginner in electronics. Can you please answer this explicitly, just using KCL or KVL. I don't know about Y- $ \Delta $ . I only know KCL and KVL. \$\endgroup\$ Jan 7 at 7:11
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    \$\begingroup\$ @SagarK For nodal, you must select a node to be 0 V. (There's a deeper, non-electronics reason due to the Rank-Nullity Theorem.) Other than that, it just works. You are allowed to swap around 2-terminal devices in a branch. So if the voltage source is at an inconvenient location and the only problem is a resistor in the way (in series) then just swap the two. Your circuit allows this for the 10 V and 10 k Ohm devices, for example. Just swap them if it helps you. \$\endgroup\$ Jan 7 at 10:26
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    \$\begingroup\$ @SagarK.Biswal A branch has everywhere the same current -- it must. So you can swap around 2-terminal devices in the chain. This works for two voltage sources. But it cannot work for two current sources (unless they magically have the same current.) \$\endgroup\$ Jan 7 at 11:58

4 Answers 4

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No zero-volt "ground" is ever necessary.

To apply KVL, you traverse each element in a loop, summing potential differences as you go. When you get back to where you started in the loop, you must necessarily end up at the same potential you started with. That means the sum of potential differences around a loop is zero, not that any particular node has an absolute potential of zero, or any absolute potential for that matter.

All that matters is how potential changes (increases or decreases) as you go around the loop, not any absolute potential of any node along that path. All you can ever say about the potential of any node in a circuit is how different it is from the potential at some other node in the same circuit. If you call something 0V, that's just for convenience, not because it is actually zero. There's no such thing as zero volts, anywhere. All there is is potential differences.

schematic

simulate this circuit – Schematic created using CircuitLab

You can guess current directions, but make sure annotated voltages across resistances conform with the principle that current through a resistor always enters the end of a resistor with the higher potential, and leaves from the lower potential terminal.

KCL applied here is trivial:

$$ I_1 + I_2 = I_3 $$

KVL around the left loop goes like this: start anywhere, and go in any direction. I'll go clockwise, starting at node X. First jump over R1, which incurs a fall in potential (according to my annotated polarity), so subtract \$V_A\$. Cross over V3, another fall in potential, another subtraction. Traverse R3, again subtracting \$V_C\$ in accordance with the marked polarity. Finally jump over V1, this time adding, since this incurs a rise in potential. Back at X the potential is still whatever it was when we started, so the total change in potential around the loop is zero:

$$ -V_A -V_3 -V_C + V_1 = 0 $$

We didn't refer to any absolute potentials at all, just potential differences, or changes.

It's probably easiest at this point to invoke Ohm's law, and substitute known potential differences, which I could have done from the start:

$$ -I_1R_1 - 10 - I_3R_3 +20 = 0 $$

Do the same for the right hand loop. This time I'll start at Y, going clockwise. It doesn't matter where you start, or which direction you go:

$$ -5 + I_3R_3 + 10 + I_2R_2 = 0 $$

These, then, are your three equations:

$$ \begin{aligned} I_1 + I_2 - I_3 &= 0 \\ -I_1R_1 - 10 - I_3R_3 +20 &= 0 \\ -5 + I_3R_3 + 10 + I_2R_2 &= 0 \\ \end{aligned} $$

You have three independent simultaneous equations, and three unknowns, \$I_1\$, \$I_2\$ and \$I_3\$, so this can be solved.

I think I'll be criticised for doing the work for you, but the point I made about there being no such thing as zero volts requires an example of why it's not necessary. Below is the same schematic with the voltages \$V_A\$, \$V_B\$ and \$V_C\$ shown, and node X arbitrarily designated as ground (0V). I've also included a couple of voltmeters showing the potentials of Y and Z relative to ground:

schematic

simulate this circuit

Notice how the solution yielded a negative voltage across R2. That simply means I guessed the direction of current \$I_2\$ incorrectly, and therefore also the polarity of voltage across R2. I could correct this now by swapping R2's "+" and "−" symbols and marking the voltage "+8V" instead. It means the same thing.

Now the same circuit again, with that correction, and this time with node Z set as the zero volt reference:

schematic

simulate this circuit

I doesn't matter where I call "zero volts", the potential differences don't change anywhere. For example, the difference \$V_X - V_Y = +15V\$ in both cases.

Most notably, I didn't have to start with any idea of where zero volts might be, because:

  1. zero volts doesn't really exist, and

  2. KVL is all about potential differences, not absolutes.

This means that you can place "ground", a zero volt reference point, anywhere you like, it doesn't change anything about this circuit at all, except perhaps how you would describe it to another engineer.

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Sure, you can apply the nodal analysis to that circuit.

The more obvious node to use as reference is shown below.

enter image description here

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  • \$\begingroup\$ If you wrote a little more I'd +1 your answer. It's a bit oversimplistic as it is. (But not wrong.) More of a comment to my eyes. Each to their own. \$\endgroup\$ Jan 7 at 14:56
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Explicitly done based on comments: I have swapped the elements in the middle branch. enter image description here

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    \$\begingroup\$ Nice catch! I'll +1! \$\endgroup\$ Jan 7 at 13:34
  • \$\begingroup\$ @periblepsis thank you :) \$\endgroup\$ Jan 7 at 13:45
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yes of course < 1st you can convert the voltage source to current source < then the comon node will be the reference node .

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  • \$\begingroup\$ AFAIK no voltage source can be converted always to a current source in theoretical terms. Therefore I downvote that answer. \$\endgroup\$ Jan 7 at 7:06
  • \$\begingroup\$ @BitLauncher It is routinely done in Spice programs. In fact, you can find that specific recommendation in the LTspice help documents and also in the groups.io forum. Spice is more stable with current sources and experts prefer them over voltage sources in models they develop. I have personally been repeatedly reminded by the developer of LTspice, Mike, of this when I experienced excessive runtime. Regardless, here the author of this answer isn't necessarily wrong. It's certainly valid in this case. So I'll upvote to compensate and encourage the author. \$\endgroup\$ Jan 7 at 13:32
  • \$\begingroup\$ @periblepsis: I do not know Spice program (anymore - used it 25 years ago). But I am curious how you would model above circuit with current sources. Guessing: and then current is voltage and voltage is current and Ohm values are Siemens values? \$\endgroup\$ Jan 8 at 0:01

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