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### Course: Electrical engineering > Unit 2

Lesson 3: DC circuit analysis- Circuit analysis overview
- Kirchhoff's current law
- Kirchhoff's voltage law
- Kirchhoff's laws
- Labeling voltages
- Application of the fundamental laws (setup)
- Application of the fundamental laws (solve)
- Application of the fundamental laws
- Node voltage method (steps 1 to 4)
- Node voltage method (step 5)
- Node voltage method
- Mesh current method (steps 1 to 3)
- Mesh current method (step 4)
- Mesh current method
- Loop current method
- Number of required equations
- Linearity
- Superposition

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# Mesh current method (steps 1 to 3)

We solve a circuit by writing Kirchhoff's Voltage Law in terms of "mesh currents." This video covers the first three steps out of four. Created by Willy McAllister.

## Want to join the conversation?

- Is it possible to eliminate current source from the circuit by converting it to voltage source or vice-versa ?(7 votes)
- Transforming sources and the theory of Thevenin and Norton are covered here: https://spinningnumbers.org/t/topic-dc-analysis-special.html(2 votes)

- At7:32, why you define iR2 as (i1 - i2) and not the other way around?You are writing KVL for the second mesh. Thanks.(6 votes)
- Hello Asal,

Yes, we could also have written -R2(i2-i1) and substituted this into the video's equation.

We could have simplified things even more and written:

0 = R2(i2-i1) + R3(i2) + R4(i2-i3)

It all depends on how you look at things and what you define as a voltage rise and a voltage drop. Personally I try to make the math as simple as possible. It is so easy to miss a sign...

Regards,

APD(8 votes)

- I don't understand why i3 only equals to -I, why isn't that -I-i3=0?(4 votes)
- Hello! One question: why is i3 = -I and we don't include R4(i3-i2) also?(4 votes)
- Mesh current i3 has to be equal to -I because it is the only current flowing through the current source. When you look at R4, however, the resistor current is the combination of (superposition of) two mesh currents, i3 and i2. You nearly got the expression for the
*voltage*across R4 correct. It equals R4(i2 - i3).

Mesh currents are kind of strange little animals. Sometimes they exist in a wire or component by themselves, as i3 does in the right side of the circuit. Another example is i2 flowing in R3. Other times mesh currents exist only in combination with other mesh currents, as the current in R4 or R2. I really like this weirdness. Mesh Current is my favorite analysis method, just for the entertainment value.(4 votes)

- At8:03, why don't we account for the current coming in from the current source, I, since they're on the same node?

I'm sure you're right, but I don't understand why we wouldn't have our 2nd equation as:

R2(i1-i2) - R3(i2+I) - R4(i2-i3) = 0

^(2 votes)- You are anticipating the path of I (the current from the current source on the right) before letting the math do its job. I does not necessarily flow where you think it does. When it reaches the junction where R3 and R4 connect, it has the option of splitting between those two resistors. So not all of it (necessarily) flows through R3.

The Mesh Current method is based on**mesh currents**(i1, i2, and i3, but not big I ). When you write the mesh equations you do it in terms of mesh currents. That means you won't mention**element**currents like I when you set the equations up.

The Mesh Current method is kind of fun because you are asked to put your faith in these made-up mesh currents as you set up the problem. Everything works out in the end.(4 votes)

- @5:10Why don't you include R4 into your i3 calculation?

Sal says: i3 = -I

why not: i3 = -I + R4(i2-i3)(3 votes)- i3 is completely defined by the current source. You don't need any other information.

Take a look at the last term in your second equation. The units of R4(i2-i3) is what? That expression is in volts. It is the voltage across R4. The element current through R4 is the combination of two mesh currents, (i2-i3).(2 votes)

- Instead of having i3 going clockwise and equaling -(current soruce), could you have made i3 go counter-clockwise and equal +(current source)?(2 votes)
- Hello Emily,

Yes but...

The direction of the loops is completely arbitrary. If you changed i3 you would need to go back to8:08and change your R4 calculations.

For your own sanity may I recommend consistency. Make all loops in one direction this way you can quickly use a robotic process to generate the equations.

Regards,

APD(3 votes)

- Can we use thevenin' s law to analyze this circuit?(2 votes)
- Hello Varney,

Absolutely 💯. The circuit may be simplified by converting Thevenin circuits to their equivalent Norton circuits and vise versa.

Do whatever makes the work easy - or whatever your instructor is looking for...

Regards,

APD(2 votes)

- sir,

how do we know that there is voltage rise or voltage drop... as the teacher says that there is always voltage drop accross the resistor but how can we judge it by + and - signs?(1 vote)- The voltage drops in the direction the current passes through the resistor. The current goes from high potential to low.(3 votes)

- Hello,

This question is for the working done in7:32.

I am quite connfused as to how the second equation written for Mesh 2 is not R2(i2-i1) - R3i2 - R4(i2-i3). Particularly, why the voltage across R2 changes from -R2(i1-i2) in M1 to R2(i1-i2) in M2? This is what I did after assuming that going from negative to positive terminal gives a positive voltage:

I wrote the equation for M1 as:

5 - R1i1 -R2i1 + R2i2 = 0

5 - R1i1 -R2(i1-i2) = 0

M2:

R2i2 - R2i1 - R3i2 -R4i2 + R4i3 = 0

R2(i2-i1) - R3i2 - R4(i2-i3) = 0

-R2(i1-i2) - R3i2 - R4(i2-i3) = 0

I attempted to take the voltage with respect to each current loop. This way I thought that the voltages will remain same (and they are the same), but your working is different? Is it because that you take voltages with respect to the mesh instead of the loop? As you can see, I am quite confused. So, please help me in understanding the real process.

Thank you in advance.(1 vote)- We are talking about the sign of the first voltage term. The video has -R2(i1-i2) for the first mesh, but uses +R2(i1-i2) for the second mesh. You are asking why we don't use the same expression, -R2(i1-i2) (or equivalently +R2(i2-i1)), for both meshes.

The reason is how you encounter R2's voltage polarity signs as you travel around the two different KVL arcs.

Assuming you start your KVL trip in the lower left corner of each mesh... For mesh 1 when you get to R2 you 'enter' it through its + voltage sign and exit through its - sign. In that direction of travel, that voltage is a*drop*and the overall term gets a - sign.

When you do the trip for the second mesh the first thing you encounter is the - sign of R2's voltage polarity. That means as you go through R2 you are getting a voltage*rise*. That's why this term gets a + sign in the second mesh equation.

A point to emphasize... When you have a resistor with two mesh currents, like R2 (and R4), be careful with which mesh current goes first in the subtraction, (i1-i2) is correct, whereas (i2-i1) is not. The first mesh current in the subtraction is always the one that flows into the + voltage terminal of the resistor. That is true for both mesh 1 and mesh 2.

Now that you have the order of the subtraction always right, the + and - signs in the KVL equation depend on the order of travel as discussed above.(2 votes)

## Video transcript

- [Voiceover] Now we're
gonna discuss the second of two popular ways to analyze circuits, and this one is called
the mesh current method. This is actually one of my favorite. There's a fun spot here
where we make up currents flowing around in circles
inside the circuit, and it's interesting. In the mesh current method,
what we do is we define mesh currents, so we
need to define that word. A mesh current. A mesh current is a current
that flows in a mesh. And a mesh is the open
windows of the circuit. Here's one of the meshes in our circuit. There's another mesh
here, it's an open window, and this third one here is another one. And this will be our
mesh current number one. We'll call it i1, and we can define these other mesh currents, like this. And I'm gonna define them all to go around the same direction. They're all going clockwise, around our circuit. Now let me take a second
to label the other components in our circuit here. We'll call this V, we'll call this R1, R2, R3, R4, and this has a current source in it. We'll call that I. Makes it a little bit more interesting, and this is one of the reasons we use the mesh current method. The word mesh actually
comes from like screen doors or window screens where
there's wires and there's spaces between the wires,
so a space in a wire network is called a mesh. That's where the name comes from. These mesh currents are
kind of interesting. In some sense they exist, and then, but it doesn't seem like they really do. How can these currents be
flowing around in circles? Let's look a little closely. It's pretty clear that the
current flowing through our voltage source, V, is
equal to mesh current i1. And also, when we get up
here to R1, mesh current i1 is flowing through R1
in this direction here. What happens when we get to R2? We have one mesh current
flowing down through R2, and another mesh current
flowing up through R2. How can this make sense? Let me go over here, and
we'll draw, redraw R2. Here's R2, and now I'll
redraw the mesh currents. So we have a mesh current,
i1 going down like that, and we have a mesh current, i2, like that. And I'll label on here, I'll
label the element current, for R2, we'll call that iR2. And what's gonna happen,
this is a really wonderful application of the
principle of superposition. i1 and i2 superimposed
inside R2 to create iR2. So if we wanted to know iR2, it would be the sum of these two currents. Now i1 is flowing in the
same direction as iR2, so this equals i1, and i2
is going in the opposite direction, so it gets a minus sign. And if we somehow knew
i1 and i2, this is how we would compute the actually current that's flowing in R2. Next, since this is a
method, I want to list out the steps of the method. The mesh current method is the first thing is to draw the meshes. And we did that over here,
when we drew i1, i2, and i3. And the second step is
to solve the easy ones. When we say an easy one, here's
an example of an easy one. The current source I, right
here, defines what i3 is. So i3 is an example of an
easy current to solve for. And then the third step in the method is to write Kirchhoff's voltage law. KVL. Write Kirchoff's voltage law equations, for the two remaining mesh currents. And the fourth step is solve. So in this video we're gonna
do the first three steps, and then in the next video,
we'll go the full solve. Okay, so let's run our
mesh current method. Number one, draw the meshes. Here's step one, happening right here. So we could check off, we did step one. Step two is solve for the easy ones. And with think, i3 counts as an easy one. i3 flows down, through the current source, and so, step number two, so step two, is i3 equals negative I, because they're pointing
in opposite directions. But that's it, that's the easy one. We've done that. Now we get to the next step. So now let's start step three. We're gonna write KVL equations
on each of these meshes. We'll start here, and we're
gonna go around this mesh in this direction and write
Kirchhoff's voltage law. So the first thing we
come to is a voltage rise. We go in the minus side and out the plus. That's a voltage rise of V volts. Next we get to R1, and R1 has
a current flowing through it, a mesh current, and it will
have a voltage on it like that, because the current's
coming in from this side. And that will give us a
voltage, we go in the plus, out the minus, so that's
gonna give us a voltage drop of... a drop of R1 times i1. Right. Now we're over to here,
we go around the corner, and we go through R2. Now, R2 is odd because it has i1 and i2 flowing into it, so what are we gonna do? Well we already figured
out this element current is written down right over here. There it is right there. So if I label the current going down here, I have a plus and a minus, this is gonna be another voltage drop going around the loop. So it's gonna have a value of R2 times this value here. i1 minus i2. And now we've made it
all the way back to home, to where we started from, and so that's our KVL equation, which we set equal to zero. All right. Next step, let's do the other mesh. We're gonna go start
here, and we're gonna go around this mesh, in the same way. First, the first term,
we're gonna see a voltage, oh look, it's a voltage rise, so I'll have a plus here. A voltage rise, and it's the same term that we had in the previous equation. It'll be plus R2 times i1 minus i2. That's the voltage rise
going from here to here. Now we go through R3. Now R3 just has one mesh current. It just has i2 flowing through it. So we could put that there. So that counts as a voltage drop, gets a minus sign, and
it's value is R3 times i2. That's that one. Now we go over to R4. R4, this is the same
thing with two currents that we had over here with R2. So we do the same thing again. This is gonna be a... we'll define this as a voltage drop, there's the element current. And it's the same form as we had here when we calculated the current for R2. So this is gonna be a voltage drop, give it a minus sign, and it's gonna be R4 times, this time it's i2 going down. i2 minus i3. And do we get home? Now we're all the way home, so our KVL equation is
all that equal to zero. So these are our two
simultaneous equations that represent our
current, and we wrote them using the mesh current method, so we get to check off step three. And in the next video, we'll
solve these two equations.