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My professor told me one method, a well known and trusted youtuber told another method and our assignment solutions have another method for nodal analysis so I can't figure out which method is actually correct.

enter image description here

I'm getting different answers for different methods and I just can't understand what's actually correct and what's wrong.

Edit: Okay since some people have asked for the equations, here you go:

I edited the circuit a little and removed the 5 ohm resistor and replaced the 30 ohm one with 20 ohm for easy calculations.

enter image description here

and this is the equation for the first circuit that I got enter image description here

and this is the equation for the second circuit

enter image description here

I'm kind of surprised that I got the same answer. My professor told me to read the theory again when I suggested him the second method

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  • \$\begingroup\$ Label the currents, write equation(s), then we can tell you where you went wrong. \$\endgroup\$
    – Mattman944
    Nov 7, 2023 at 13:10
  • \$\begingroup\$ Yes I just edited the question, check it out \$\endgroup\$ Nov 7, 2023 at 13:29

2 Answers 2

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Learning to draw

From your earlier question and this one I think I'd recommend gaining some practice redrawing schematics in different ways. Some schematics are the logical equivalent of a Gordian Knot and desperately need to be redrawn. (Hopefully, you encounter these later on when you are more experienced in redrawing.) Others can be made simpler in order to help you understand them for analysis, but are terrible for figuring out how to wire them up on a breadboard.

There are different audiences for schematics. And not everyone who draws them are equally good at the task. So get good at this process, yourself. It's worth the effort and will pay off well over time. And knowing how to draw schematics takes you more than halfway towards understanding how to analyze them.

Here's your schematic shown in a schematic targeting someone without any electronics experience but with some parts, wire, and a power supply:

enter image description here

This is not a very good schematic for analysis, though. It is purely designed to make construction easy to follow and where the least amount of wiring is involved.

For nodal analysis, one of these would be much better:

schematic

simulate this circuit – Schematic created using CircuitLab

But neither of those are all that helpful for mesh analysis. The following would be better for mesh as the loops are clear:

schematic

simulate this circuit

Practice redrawing schematics until it feels fairly natural to you.

Nodal analysis

The current direction isn't important. In fact, you can mostly ignore that question when faced only with resistors.

Let's ignore current direction and just do nodal analysis on your schematic:

schematic

simulate this circuit

There are just two unknown nodes. (Ground and \$+15\:\text{V}\$ are known.) So let's completely avoid assigning a specific current direction and instead say that the current goes both ways at once. (This isn't unlike mesh analysis, which can often claim that a single device carries two loop currents at the same time.) By deciding it goes both ways, we don't have to make a choice.

Let's start with node \$v_1\$:

$$\begin{align*} \frac{v_1}{R_1}+\frac{v_1}{R_2}+\frac{v_1}{R_3}&=\frac{+15\:\text{V}}{R_1}+\frac{0\:\text{V}}{R_2}+\frac{v_2}{R_3} \end{align*}$$

To read the above, imagine that the left side has all the current arrows pointing away from node \$v_1\$ towards the other nodes. And that the right side has all the current arrows pointing into node \$v_1\$ from the other nodes. These of course must balance out.

So, there's no need to be forced to make a choice. Instead, choose both ways at once! You can't go wrong when you treat both directions as simultaneously true!!!

So, the other equation would then be:

$$\begin{align*} \frac{v_2}{R_3}+\frac{v_2}{R_4}&=\frac{v_1}{R_3}+\frac{0\:\text{V}}{R_4} \end{align*}$$

Using freely available SymPy, SageMath, and Python:

for i in list(solve([
    Eq(v1/r1+v1/r2+v1/r3,15/r1+0/r2+v2/r3),
    Eq(v2/r3+v2/r4,v1/r3+0/r4)],[v1,v2]).values()):
        i.subs({r1:10,r2:40,r3:30,r4:5})
420/43  # v1 = 9.76744186046512
60/43   # v2 = 1.39534883720930

And a separate LTspice run provides similar results:

V(v1):   9.76744     voltage
V(v2):   1.39535     voltage

The point here is that you can save yourself some grief and just not worry about current directions when dealing with resistors.

Both directions are correct!

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Normally, for DC (or instantaneous), current is assumed to be flowing from higher potential to lower potential.

But for analysis, there's no "correct flowing direction". You just pick a direction and label them (e.g. entering to node is +, exiting the node is –), do your calculations and depending on the result's sign you'll see the actual direction.

The signs might be different than your initial assumptions, but the magnitudes cannot be different.


I'll do the first one for you.

First, I picked a reference point (ground) and put the ground symbol (all the voltages will be with respect to this point) then I put the current directions:

schematic

simulate this circuit – Schematic created using CircuitLab

So let's take the currents entering to the node as +, and exiting ones as –. For V1 node we can write:

$$ (-i_1)+(-i_2)+(-i_3)=0 $$

And, since currents flow from higher to lower potential, with the directions we selected we can write the following for the currents:-

$$ i_1=\frac{V_1-V}{R_1} \\ i_2=\frac{V_1-0}{R_2} \\ i_3=\frac{V_1-0}{R_3+R_4} $$

Note that we don't know if V1 is greater than ground or V yet. For the directions we picked we assume V1 is greater than both V, and ground (0V).

Now put these into the first equation:

$$ -\Big(\frac{V_1-V}{R_1}\Big) + \Big(-\frac{V_1-0}{R_2}\Big)+\Big(-\frac{V_1-0}{R_3+R_4}\Big)= 0 \\ -\Big(\frac{V_1-15}{10}\Big) + \Big(-\frac{V_1}{40}\Big)+\Big(-\frac{V_1}{35}\Big)= 0 $$

Solve this for \$V_1\$, and calculate \$i_1\$ from its equation given above. You'll see the result is negative. This means that the actual direction is different than its initial i.e. we took it as exiting but it must be entering.

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  • \$\begingroup\$ I updated the question, can you check it? \$\endgroup\$ Nov 7, 2023 at 13:28
  • \$\begingroup\$ @SpaciousCoder78 see updated answer (I was updating already before your comment). \$\endgroup\$ Nov 7, 2023 at 13:43
  • \$\begingroup\$ Saw yours but I just want to know if my 2nd equation's method is correct or not \$\endgroup\$ Nov 7, 2023 at 14:03
  • \$\begingroup\$ @SpaciousCoder78 I can't check your attempt at the moment, sorry, but you can re-write taking my approach above and compare. \$\endgroup\$ Nov 7, 2023 at 14:47

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