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I am trying to find the voltage across R4 in the following circuit:

enter image description here

It confuses me that there are 2 voltage sources. I am not sure how to approach it, so I tried this:

First, I use Kirchoff's voltage law for the upper right loop, $$-5(V) +20(\Omega)I_2+10(\Omega)I4=0$$ and applying the current law for the upper node, $$I_1+I_{fromVs2}=I2$$

After that, I don't know how to proceed.

How can I find the current that comes from Vs2? If I knew its resistance I could use Ohm's Law, but I don't know it. I (think that I) can find the current accross V1 from Ohm's Law, but I am not sure how to approach it. Any help is appreciated

EDIT: This is what the currents look like:

enter image description here

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    \$\begingroup\$ Superposition theorem. \$\endgroup\$ – Phil G May 31 at 15:05
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    \$\begingroup\$ You need to learn about "mesh nodal analysis". \$\endgroup\$ – scorpdaddy May 31 at 15:06
  • \$\begingroup\$ Think about this: what is the current through R2? Is it I1? Is it I2? Something else? How is I2 related to IVS2? What is the current through R5? \$\endgroup\$ – Elliot Alderson May 31 at 15:33
  • \$\begingroup\$ Welcome to EE.SE! This appears to be a homework question. As such, you need to show us your work so far, and explain which part of the question you're having trouble with. For future reference: Homework questions on EE.SE enjoy/suffer a special treatment. We don't provide complete answers, we only provide hints or Socratic questions, and only when you have demonstrated sufficient effort of your own. Otherwise, we would be doing you a disservice, and getting swamped by homework questions at the same time. See also here. \$\endgroup\$ – Dave Tweed May 31 at 15:54
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Your picture is fairly clear with respect to the part numbers and I appreciate that. But there is a schematic editor you can use (at least on a desktop PC.) And it's a little bit better, I think, if you use it when asking questions here. (One of the reasons is that it makes it easy for me, at least, to copy it and paste it into an answer I might consider making and then modify it per my own thoughts. It makes me happy.)

schematic

simulate this circuit – Schematic created using CircuitLab

Note that I've chosen a "ground" for the schematic. You can pick any node in a schematic and label it as ground. But you only get to do that, once. So when on your own, just pick a convenient node. I've picked one I consider convenient for me.

Also in the above schematic, I've indicated some proposed current loops. These are all running clockwise. There is no particular reason why they need to be clockwise. Or counter-clockwise. You can set them up in any jumbled way you please. You merely have to remain consistent to your choices, is all. But just picking clockwise (or the opposite) and sticking with it is often an easy and equally good way.

I've also indicated three voltage nodes that are "unknown" in the circuit. You could also choose to analyze the circuit with those, instead of the current loops.

As you can tell, there are several different approaches to take. Two very common ones are called "mesh" and "nodal." Mesh uses KVL (and those current loops) and nodal uses KCL (and those labeled nodes.)

Nothing is ever perfect, so as you encounter increasingly more difficult situations, you'll find that sometimes you need to combine KCL and KVL in a kind of hybrid solution. Regardless, these methods require a "simultaneous equation solution." So unless you have a really simple case, you need to have basic skills and/or software tools to help with the solution aspect. And sometimes, those software tools aren't handy and for those times you should make sure to memorize how to apply Cramer's Rule (or various elimination methods.) Putting these into your head and not 100% depending upon software means that you always have a way to solve something (unless you lose your head.)

When setting up your mesh equations, you need to take into account all the loop currents (where applicable.) for example, resistor \$R_2\$ has two loop currents: \$I_1\$ and \$I_2\$. They are in opposite directions, so one must be subtracted from the other when applying it to \$R_2\$ in your equation development.

Just start anywhere in the loop and walk around it using KVL as a guiding principle. So, for example, here's the first loop with me mentally starting in the lower-left corner of the \$I_1\$ current loop indicated in the above schematic:

$$0\:\text{V}+12\:\text{V}-R_1\cdot I_1 - R_2\cdot\left(I_1-I_2\right)-R_3\cdot\left(I_1-I_3\right)=0\:\text{V}$$

You can follow a very similar process for the remaining two current loops. When you have all three of them, you can solve for all three "virtual" currents. Do remember that for some devices (these are all two-terminal devices in your schematic) the actual net current will be the sum or difference of multiple loop currents. So again, for example, the current in \$R_2\$ will not be either \$I_1\$ or \$I_2\$ but instead the magnitude will be absolute value of the difference between them. The direction of that current will be determined by the sign you get and which of the two currents you chose as the "forward" direction. So if you decided to compute the current in \$R_2\$ as \$I_1-I_2\$ then this means you've chosen \$I_1\$ as the "tentative direction" and if the resulting sign is positive then the net current direction follows the arrow for \$I_1\$. But if the resulting sign is negative then the net current direction is opposite that assumed arrow direction.

Can you handle the other two loops and the simultaneous solution?

The other way to do this would be using nodal analysis and KCL. But I think you are more familiar at this point with mesh. So I won't dwell on nodal. I'll just leave you with a thought about it to consider.

For example, you might focus on \$V_\text{B}\$ and develop an equation for it:

$$\frac{V_\text{B}}{R_2}+\frac{V_\text{B}}{R_3}+\frac{V_\text{B}}{R_4}=\frac{V_\text{A}}{R_2}+\frac{0\:\text{V}}{R_3}+\frac{V_\text{C}}{R_4}$$

Here, I just imagine myself standing at the node labeled \$V_\text{B}\$ and develop the left-hand side of the equation as being all of the currents "spilling away" from the node through those three resistors. The right-hand side of the equation are the currents that are "spilling into" the node from the surrounding nodes. It's pretty simple to set that one up.

But don't dwell on nodal. You are just focused on mesh analysis for now. Let me know if there's something else I can add to help out.

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I think this can be solved using Thevenin's theorem. It's getting pretty long to solve and having weird values because of the question being made is not for short calculation but get you lengthy. For solving systems having 2 or more active component normally Thevenin's theorem or Norton's theorem is used.

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