3
\$\begingroup\$

I'm having trouble understanding some concepts in simple electrical circuits and I was hoping to clarify it because it has been the my source (pun intended) of wrong answers.

Lets say I have this circuit (which I just created for the purpose of investigating the voltage/current behaviour and asking this question - it is not homework).

  1. How can I calculate the voltage at node 1? I cannot apply Ohm's Law with I1 because I don't know the resistance the component I1 has - I only know that it imposes a current of 1A and articulates the voltage necessary to do so.
  2. When doing KVL equations to the left mesh how could I specify the I1 component? Would it be I1*R? But I don't know R...

Thank you in advance!

enter image description here

\$\endgroup\$
3
  • 2
    \$\begingroup\$ ”I don't know the resistance the component I1 has” It has no resistance. You need to apply both KVL and KCL to solve it. \$\endgroup\$
    – winny
    Commented Aug 19, 2022 at 8:00
  • 1
    \$\begingroup\$ For all intents and purposes, the only impact of R1 is to alter the voltage at the indicated node. Since a current source has, itself, infinite impedance it follows that R1 in series with it doesn't impact the rest of the circuit and can be removed for analysis of the rest of the circuit. But, of course, R1 does affect that internal node's value. But it doesn't impact anything besides it. \$\endgroup\$
    – jonk
    Commented Aug 19, 2022 at 8:02
  • \$\begingroup\$ @jonk thank you very much for your reply! I mentioned your answer in other thread but then I tested in the circuit simulator and understood exactly what you explained. That point (which I dumped form the original question) is very clear to me now! \$\endgroup\$
    – ludicrous
    Commented Aug 19, 2022 at 8:05

4 Answers 4

6
\$\begingroup\$

A simple re-drawing of your schematic makes it easier to see:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the current entering node \$N_2\$ isn't impacted by \$R_1\$'s value. It's the same, regardless. So the voltage value you'd find for \$N_2\$ is the exact same, no matter what value you assign to \$R_1\$.

The only difference will be the voltage value at \$N_1\$, which will obviously be \$I_1\cdot R_1\$ above \$N_2\$. So \$R_1\$ does adjust the voltage value at \$N_1\$. But it has no impact on the voltage value at \$N_2\$.

\$\endgroup\$
5
  • \$\begingroup\$ It is clear now! So I guess a way to get analytically the voltage at N1 would be, not through KCL in that mesh, but come from N2 "upwards" through the Resistor (from which I know the current I1). So it is never possible to write KVL in a mesh whenever the mesh has a current source? \$\endgroup\$
    – ludicrous
    Commented Aug 19, 2022 at 8:15
  • \$\begingroup\$ @ludicrous That's perhaps the easier way to see it. Still, the general analytical techniques work fine. It's just that they don't provide insight. They come up with the right answers. But it's just a bunch of algebra. Not so much on insight. Resistance in series with current sources can, broadly speaking, be removed before further analysis. Just as resistors directly in parallel with a voltage source can also, broadly speaking, be removed before further analysis. (You can always go back and add them back, after that analysis to pick up on those loose details.) \$\endgroup\$
    – jonk
    Commented Aug 19, 2022 at 8:18
  • \$\begingroup\$ "So it is never possible to write KVL in a mesh whenever the mesh has a current source?" Invariably, nodal analysis is better than mesh analysis - it deals with current sources and minimises the number of unknowns. \$\endgroup\$
    – Chu
    Commented Aug 19, 2022 at 14:55
  • \$\begingroup\$ This is not a redrawing, since you deleted some wires \$\endgroup\$ Commented Aug 19, 2022 at 18:58
  • \$\begingroup\$ @user253751 Define "redrawing" as you see it. It meets my definition of one. \$\endgroup\$
    – jonk
    Commented Aug 20, 2022 at 3:37
3
\$\begingroup\$

Using superposition is a good way to solve this. First open the current source and find the voltage at node 1. By inspection, it is +0.5V since the R2/R3 form a divider and R1 has no effect (being in series).

Then short the voltage source and find the voltage at node 1. It is 1A into 100 ohms + R2||R3 = 150 ohms, so +150V.

The total voltage will then be +150V + 0.5V = +150.5V.

Even if you have to use more formal methods to solve, it's a good and quick double check.

\$\endgroup\$
2
\$\begingroup\$

The voltage across source I1 is an unknown at this point, but you give it a name, and polarity, and its value comes out with all the others, in the analysis.

So let's analyse.

Note: Below, I've tried to explain my thinking for for every one of my choices for current directions and voltage polarities, to avoid any ambiguity. However, my approach below is unnecessarily pedantic, and in reality you would probably just add up voltages around each loop when applying KVL, assuming every potential change to be positive, and relying entirely on the signs of the results to reveal actual polarity/direction. This latter methodology is necessary, in fact, when performing mesh current analysis, where voltage polarity is not explicity "chosen", rather it's decided implicitly by the chosen loop current direction. Regardless, the sign of the result always reveals actual voltage polarity and current direction.

Let me redraw your circuit with some labels to help build the equations:

schematic

simulate this circuit – Schematic created using CircuitLab

I don't know the polarity of voltage across the current source I1 yet, so I'll guess, and for voltage source V2 I have a similar problem with current direction, so I'll guess that too.

I have to be careful with resistors though, current in a resistor always travels from the end with higher potential to the end with lower potential, and all my voltage polarity and current direction labels for resistors must conform with that.

First KCL at branch node A, simple enough:

$$ I_1 + I_2 = I_3 $$

Since we know \$I_1 = 1A\$, this immediately simplifies to:

$$ 1 + I_2 = I_3 $$

KVL for the left loop, starting at node A, going clockwise. First I cross over R2, and this incurs a fall in potential of amount \$V_{R2}\$ according to my polarity labels there. Then I jump upwards across R1, where I encounter a rise in potential (again, I'm referring to my labels) of amount \$V_{R1}\$. Lastly by traversing source I1 potential falls by \$V_1\$, and I arrive back where I started:

$$ -V_{R2} + V_{R1} - V_1 = 0 $$

I might as well incorporate Ohm's law at this stage, so where there are resistors, I'll replace \$V\$ with \$I \times R\$, remembering that \$I_1 = 1A\$:

$$ -I_2R_2 + 1R_1 - V_1 = 0 $$

Going clockwise from A around the right hand loop, first I see a drop in potential across source V2, then a rise across R3, and finally a rise across R2:

$$ -V_2 + I_3R_3 + I_2R_2 = 0 $$

Now we have three unknowns, \$I_2\$, \$I_3\$ and \$V_1\$, and three independent equations, ready to solve. We could solve all these by substitution, but I'm lazy, and I'll use an online matrix inverter to solve this. First I need the equations in a form to help me construct the matrix equation:

$$ \begin{aligned} I_2 &&-I_3 && &&= &&-1 \\ -R_2I_2 && &&-V_1 &&= &&-R_1 \\ R_2I_2 &&+R_3I_3 && &&= &&V_2 \\ \end{aligned} $$

Plugging in all the known resistances and voltages, here's the matrix equation:

$$ \begin{bmatrix} 1 & -1& 0 \\ -100 & 0 & -1 \\ 100 & 100 & 0 \end{bmatrix} \begin{bmatrix} I_2 \\ I_3 \\ V_1 \end{bmatrix} = \begin{bmatrix} -1 \\ -100 \\ 1 \end{bmatrix} $$

Rearranged to make the matrix of unknowns the subject:

$$ \begin{bmatrix} I_2 \\ I_3 \\ V_1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -100 & 0 & -1 \\ 100 & 100 & 0 \end{bmatrix}^{-1} \begin{bmatrix} -1 \\ -100 \\ 1 \end{bmatrix} $$

Find the inverse and solve:

$$ \begin{aligned} \begin{bmatrix} I_2 \\ I_3 \\ V_1 \end{bmatrix} &=\begin{bmatrix} 0.5 & 0 & 0.005 \\ -0.5 & 0 & 0.005 \\ -50 & -1 & -0.5 \end{bmatrix} \begin{bmatrix} -1 \\ -100 \\ 1 \end{bmatrix} \\ \\ &=\begin{bmatrix} -0.495A \\ +0.505A \\ +149.5V \end{bmatrix} \end{aligned} $$

Now for some corrections to our guesses. Negative results tell us we guessed voltage polarity wrong, or current direction wrong. As you can see, \$I_2=-0.495A\$, and this answer is perfectly correct. We marked it on the schematic as a current flowing downwards, and that current turns out to be negative. This is the same as a positive current 0.495A flowing upwards.

I got lucky, and correctly guessed the the direction of \$I_3\$ and polarity of \$V_1\$.

I almost forgot, your question is "what is \$V_1\$, the voltage across current source I1?". The answer is \$V_1=149.5V\$, its lower terminal having the higher potential.

\$\endgroup\$
2
  • \$\begingroup\$ I haven't quoted any absolute potentials anywhere, only the voltages across things. The potential at the bottom of the current source is 150.5V, due to the 1V offset from source V2. \$\endgroup\$ Commented Aug 20, 2022 at 7:42
  • \$\begingroup\$ Now I see you've marked V1 to be across the current source, I took it to mean Node1, as in the OP. Ma' bad. \$\endgroup\$ Commented Aug 20, 2022 at 8:25
0
\$\begingroup\$

By Ohm's law, the voltage at Node1 is \$100\:V\$ higher than the voltage at Node2, so it's convenient to find \$V_{Node2}\$ first.

Using nodal analysis, the sum of currents away from any node = 0

Thus, at Node2,

\$\large\frac{V_{Node2} - V_{2}}{100}+\frac{V_{Node2} - 0}{100}\small-1=0\$

\$\large\frac{V_{Node2} - 1}{100}+\frac{V_{Node2} - 0}{100}\small -1=0\$

\$2V_{Node2}-1-100=0\$

\$2V_{Node2}=101 \$

\$V_{Node2}=50.5\:\:V\$

Hence,

\$V_{Node1}=150.5\:V\$

\$\endgroup\$
1
  • \$\begingroup\$ Why did this get a downvote? \$\endgroup\$ Commented Aug 20, 2022 at 6:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.