The circuit diagram is given below, I keep getting the wrong answer for the mesh currents, I always get -1 for first loop and -7 for second loop. Therefor Ix comes out to be 6 which is wrong. It satisfies my two equations as well so it seems to me that my equations are wrong. Correct answer should be 2, which I can easily get by nodal analysis. It'd be awesome if you can arrive at the correct answer with mesh analysis as my main goal is to learn mesh analysis


I've tried several tutorials but nothing seems to be wrong in my approach, I'm posting my attempt below as well.

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    \$\begingroup\$ Can you show the work you did to get -1 and -7? For example, if loop 1 is on the left and if both loops point clockwise, then I would get 0+24-I1*12-(I1-I2)*6=0. \$\endgroup\$ Commented Jul 27, 2023 at 12:21
  • \$\begingroup\$ @periblepsis It was on paper 😁, but basically I assume both currents in clockwise direction and apply KVL in each loop \$\endgroup\$ Commented Jul 27, 2023 at 12:23
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    \$\begingroup\$ Then type in your work. No need to show a picture. Just typing as I show you. \$\endgroup\$ Commented Jul 27, 2023 at 12:24
  • \$\begingroup\$ Could you please just post a solution, I can tally it with my work to find my mistake \$\endgroup\$ Commented Jul 27, 2023 at 12:27
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    \$\begingroup\$ You need to post your attempt at a solution so we can see that you are not trying to cheat an exam. \$\endgroup\$
    – Andy aka
    Commented Jul 27, 2023 at 13:07

3 Answers 3



simulate this circuit – Schematic created using CircuitLab

KVL applied to the left-hand loop, travelling clockwise (in the direction of current \$I_1\$) from A. Potential rises as we traverse V1, so we add. Potential always drops as we traverse a resistance in the direction of current, so we subtract for R1. There is then another drop in potential as we jump over R3, but we must also account for \$I_2\$ flowing "backwards" with respect to our assumed direction of \$I_1\$. Since we are now back where we started, at the potential we started with, the total change in potential is zero:

$$ \begin{aligned} +24 - 12I_1 -6(I_1 - I_2) &= 0 \\ \\ -18I_1 +6I_2 &= -24 \\ \\ 3I_1 - I_2 &= 4 \\ \\ \end{aligned} $$

This agrees with your own equation for the left loop.

KVL applied to the right-hand loop, travelling clockwise (in the direction of current \$I_2\$) from B. First we encounter a fall in potential as we traverse V2, so subtract that. Then there's a fall in potential across R3. Don't concern yourself with the polarity we assumed from our treatment of the left-hand loop, here all terms must conform to our supposed direction of \$I_2\$, so we subtract, while still accounting for \$I_1\$ in the other direction. Then there's another drop in potential to subtract as we jump over R2:

$$ \begin{aligned} -15 - 6(I_2 - I_1) - 3I_2 &= 0 \\ \\ -9I_2 +6I_1 &= 15 \\ \\ 2I_1 - 3I_2 &= 5 \end{aligned} $$

Here our equations differ. You have a different sign for the voltage across R2.

Solving these gave me:

$$ \begin{aligned} I_1 = +1A \\ \\ I_2 = -1A \\ \\ \end{aligned} $$

Applying KCL at node C:

$$ \begin{aligned} I_1 &= I_2 + I_X \\ \\ I_X &= I_1 - I_2 \\ \\ &= 1A - (-1A) \\ \\ &= +2A \end{aligned} $$


Thanks for taking a moment to help out. Appreciated.

enter image description here


$$\begin{align*} 0\:\text{V}+24\:\text{V}-12\:\Omega\cdot I_1-6\:\Omega\cdot\left(I_1-I_2\right)&=0\:\text{V} \\\\ 0\:\text{V}-6\:\Omega\cdot\left(I_2-I_1\right)-3\:\Omega\cdot I_2 - 15\:\text{V}&=0\:\text{V} \end{align*}$$

The SymPy/Sage/Python solver says:

{i1: 1, i2: -1}

From the perspective of the net current going from top to bottom through the \$6\:\Omega\$ resistor, the total should be \$I_1-I_2\$ or \$1\:\text{A}-\left(-1\:\text{A}\right)=2\:\text{A}\$.

You wrote in comments:

I can tally it with my work to find my mistake

So I'll leave it to you. And again, thanks for taking a moment. You made it a lot easier for me to see what you were trying to do and that made me more confident I could offer some thoughts that might be useful.

  • \$\begingroup\$ please avoid using Vx since that's not the approach of Mesh Analysis, I was able to reach at correct answer by Vx too through nodal analysis but as I said earlier my goal isn't the correct answer, my goal is to find it by mesh analysis to improve my understanding of the concept. Secondly, When you applied KVL your first line was same as mine (if I assume the loop current in opposite direction) but from there the second line makes no sense. It'd be helpful if you can find the mistake in my solution. Thanks for the help 🙂 \$\endgroup\$ Commented Jul 28, 2023 at 10:58
  • \$\begingroup\$ @RizwanLiaqat I'll delete anything to do with Vx. My apologies that it caused you any trouble. I'll reduce this to KVL only. The 2nd KVL line is perfect and gets the right answer. You already promised (and wrote) that "I can tally it with my work to find my mistake". Were you incorrect when you wrote that? Because it now sounds like you cannot find your own mistake, just reading this comment now. Would that be a new correct assessment? \$\endgroup\$ Commented Jul 28, 2023 at 11:16
  • \$\begingroup\$ I posted a question for assistance, let's stay focused on that idea 😀. But if it makes you feel better, I did try to tally your solution and couldn't figure it out \$\endgroup\$ Commented Jul 28, 2023 at 11:19
  • \$\begingroup\$ @RizwanLiaqat I am exactly focused. The answer only addresses your question now and your promise that you can find things on your own given a correct answer. \$\endgroup\$ Commented Jul 28, 2023 at 11:19
  • \$\begingroup\$ @RizwanLiaqat If you find you were wrong -- that you now realize you cannot work it out on your own and need to ask questions, feel free. I'll listen. If I can understand you, I'll try and provide a thought in reply. However, trying to "find the mistake in my solution" isn't something I often do. And only then when there is good reason, in my opinion. (I don't read minds very well, unfortunately. And my crystal ball never did work right.) \$\endgroup\$ Commented Jul 28, 2023 at 11:26

Mesh based analysis of the given network. enter image description here


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