This is the problem I'm working on. It's a circuit that I originally had to solve using Thevenin's theorem (which I successfully was able to do so), but now I want to solve it using the mesh current theorem.

This is my approach so far:


simulate this circuit – Schematic created using CircuitLab

I'm having problems with the third loop, only because of the fact that there are two batteries that are facing at each other; i tried to use falstad to see maybe if it'd show me the direction but it didnt so tha's why i'm asking for help from you guys! so far I know that:

loop 1: E1-R1-RL

loop 2: -E1+RL-E2+R2

loop 3: ??

node 1: IV2+IR1=IRL

node 2: ??

node 3: IE1=IR1+Ix

node 4: Ix=IR2+IE2

Thank you : )!

  • \$\begingroup\$ Why not combine E2 and E3 plus R2 and R3 into a single voltage source and series resistor as per rules a la Thevenin. Or, use Millman's theorem to solve. Kirchhoff is such a tedious and generally pointless method that teaches nothing sensible about EE (of course I have to say IMHO to avoid being savaged by Kirchhoff squadrons) \$\endgroup\$
    – Andy aka
    Commented Feb 10, 2022 at 19:28
  • 2
    \$\begingroup\$ You don't need to know a direction for any of the loops, you only need to define a direction. Add up the voltages of however many batteries there are in the loop paying attention to the signs. When you've done the sums, you'll find out which direction the current is actually flowing, it will be +ve if aligned with the direction you've defined, and -ve if opposite. Have you noticed that node3 is node2? \$\endgroup\$
    – Neil_UK
    Commented Feb 10, 2022 at 19:29
  • \$\begingroup\$ Hey, Andy! so i have actually explained this problem had to be solved using Thevenin's theorem in the brackets where it said [explain the image] but it didn't want to show up. I actually managed to solve it using Thevenin's and Nortons as well, but not really working for me using the mesh currents method! \$\endgroup\$ Commented Feb 10, 2022 at 19:30
  • \$\begingroup\$ that's right neil I'm just not really used to using a website to create any circuit sorry about that one! \$\endgroup\$ Commented Feb 10, 2022 at 19:32
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    \$\begingroup\$ Write each of your equations in the same form. In mesh analysis, the things that you are trying to solve for are the mesh currents. The equations are of the form (sums of voltages=0). So, each term in your equations should be a voltage. Notice that the voltages across the resistors always involve the mesh currents. This is how you end up with three equations in the three unknown mesh currents. \$\endgroup\$
    – user69795
    Commented Feb 10, 2022 at 23:28

1 Answer 1


Not an answer -- just something to consider for another time

Keeping the voltages positive (you did, as evidenced by your setting for \$E_1\$), the left side is a very close schematic to yours except that I swapped the series devices of \$R_2\$ and \$E_2\$ (without affecting the analysis):


simulate this circuit – Schematic created using CircuitLab

On the right side, is the equivalent circuit but drawn without busing around power and ground lines. (I'm allowed to select exactly one node and call it ground and I think you can easily see which one I picked!) It's easier to read, now. The three KVL loops are easier to see on the right: (1) start from \$-E_1\$ and work down to ground; (2) start from \$+E_2\$ and work down to ground; and (3) start from \$+E_3\$ and work down to ground.

Or, with KCL, you only have one unknown node to solve.

An answer

There are six readily available loops, but you only need the right three of them. My suggestion is to take the obvious three:


simulate this circuit

Let's assume to start in the lower left-hand corner of each, call that corner \$0\:\text{V}\$, and then walk around the loop clockwise (the direction I'd like to assume for each loop current.) The loop currents will be called \$I_1\$, \$I_2\$, and \$I_3\$.

Do note that my choice (always clockwise) isn't the choice you wrote. What matters is consistency of application, not the choice itself.

The three equations from my choice are:

$$\begin{align*} 0\:\text{V} - R_1\,I_1 -R_{_\text{L}}\left(I_1-I_2\right) + E_1 &=0\:\text{V} \\\\ 0\:\text{V} - E_1 - R_{_\text{L}}\left(I_2-I_1\right) - E_2 - R_{2}\left(I_2-I_3\right) &=0\:\text{V} \\\\ 0\:\text{V} - R_2\left(I_3-I_2\right) + E_2 - R_3\,I_3 -E_3 &=0\:\text{V} \end{align*}$$

These do solve out correctly and they specify one set of equations that are defensible.

Now, let's go over your equations:

loop 1: E1-R1-RL

Here you are mixing apples and oranges and you wrote an expression and not an equation.

If you do start at the point in between \$R_{_\text{L}}\$ and \$E_1\$, then you would write \$E_1 - R_1\,I_1 -R_{_\text{L}}\left(I_1-I_2\right)=0\:\text{V}\$. Note that by multiplying a current (or sum of currents) by a resistance you get a voltage difference. You can add voltage differences to voltage differences and you are fine. You cannot add a voltage difference to a resistance, though. That does not make any sense.

So your expression is mixing dimensions and it should become obvious to you that you cannot do this. If you ever see that happen, you know something is very wrong.

Now, you could have written:


And that would have been okay. You still would not have the currents showing there, but at least it is an accurate statement if you also understand that \$E_{R_1}=R_1\,I_1\$ and \$E_{R_{_\text{L}}}=R_{_\text{L}}\left(I_1-I_2\right)\$.

Does that more make sense, now?

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    \$\begingroup\$ then the answers would be \$ I_3=-\frac{79}{115} \ \ ; \ \ I_2=-\frac{181}{115} \ \ ; \ \ I_1=\frac{7}{23} \$ || Same as Multisim. Thank you really helpful! \$\endgroup\$ Commented Feb 13, 2022 at 13:27

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