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## Electrical engineering

### 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 (step 4)

We complete the solution of a circuit using the mesh current method. Fourth step (solve) out of four. Created by Willy McAllister.

## Want to join the conversation?

- When you talk about scaling the equations down (6:24), wouldn't that affect the 5 Volts on the other side as well? Does it have to do with the currents being expected to be in mA? Thank you in advance.(5 votes)
- The scaling up/down by 1000 happens on just the left side of both equations. Every term on the left side is current x resistance. If all currents are in mA, and all resistance is in kOhms, then 10^-3 from the mA cancels out the 10^+3 from the kOhms, and you are just left with the integer parts. In Ohm's Law, the units cancel like this: kOhms x mA = 1. And for big resistors: MegOhms x uA = 1.(12 votes)

- I'm a bit confused by the final results. I understand the algebra and where all the figures come from but-

How are the values for i_1 i_2 and i_3 related? Shouldn't KCL have something to say here?

Plus, is there some relationship between i_r2 i_r4 and I? In other words: what do 1.8mA, 1.6mA and 2mA have to do with one another?

Thanks!(4 votes) - I may be wrong, but at 3 : 40 when -i3 is factored out by -R4 to make +RI, this would make RI positive and then subtracted on the other side to make it -RI. On the video it's positive.(2 votes)
- There is an extra - sign that comes from i3 = -I, mentioned at3:46. So there are three - signs on the left side, resulting in +RI on the right side.(6 votes)

- I'm 14 and I'm wondering whether this is university work or not.Thnaks(3 votes)
- This is definitely university level work. But if you have an understanding of simultaneous equations from your high school algebra class, this circuit analysis method is within reach.(3 votes)

- why don't you just show us solving it with matrices instead of making it longer and longer..(2 votes)
- You are correct that this problem could be solved with matrix methods. However, I don't assume the reader has taken a class in Linear Algebra. These systems of equations are fairly small, so solving by ordinary algebra is not too onerous, and is accessible to anyone who can solve simultaneous equations.

However, you've made me think about adding an addendum showing a matrix solution.(5 votes)

- Ugh.

So I use the alternative rule set for walking around the loop (add when encountering +, subtract when encountering -), which meant I was doing the inverse for all of the calculations. Everything went fine until you gave the hard values for the variables, then suddenly my calculations kind of got thrown out of whack. I got the inverse values for for i_1 and i_2 (which was expected, given that I had to multiply by -4 instead of 4 for the cancellation of i_2 to work at7:04), and I got the inverse for i_R2 (also expected), but i_R4 is coming out as 2.4mA instead of the expected -1.6mA. I noticed that everything checks out if I tweak the schematic slightly by inverting either i_3 or I, but I'm not sure which is allowed, if either.

So I guess my question is if using the alternative rule set when following these demonstrations is allowed, and if so, what aspects should I be changing in the given values for the math to still work? Sorry for being a pain, but the alternative rule set just works better for me and I don't want to go against what feels natural if I don't have to.(3 votes)- Your description suggests you have a sign error somewhere along the way.

The only thing you are doing different than the video is to change the sign of the terms in the KVL equations. The mesh currents still flow clockwise, right? When I did the problem I got these two equations,

3k i1 - 2k i2 = 5

-2k i1 + 8k i2 = -6

These are the same as the video's equations, with every sign flipped. So they should solve down to the same answer.

Where you got iR4 = 2.4mA, that points to a sign error in the second equation where you move the term +3k i3 = +3k (-2mA) over to the right side. It should end up -6 on the right.

Sign errors like this are the hardest part of manual circuit analysis. I have problems with this all the time. My habit is to triple check frequently as I go along, (and use a circuit simulator to help check my own calculations. http://spinningnumbers.org/a/circuit-sandbox.html)(3 votes)

- So at7:10on the video why would multiply the equation by 4? So far that is my only problem that I have.(2 votes)
- As I say this I'm looking closely at the two simultaneous equations, trying to think of a way to eliminate either i_1 or i_2. The thing that caught my eye was the +2 and -8 coefficients of i_2. I can turn the 2 into a +8 by multiplying the first equation by 4. That sets up the cancellation of i_2 when the equations are added.(2 votes)

- Could the combination of R1,R2,R3,R4,V and I be such that the resulting current flowing through R2 or/and R4 be zero? (for example if i1=i2 or/and i2=i3)(2 votes)
- Kind of a fun question.

There is the degenerate case where you set V and I to zero. Then no current flows anywhere. That's not so interesting.

To get I_R2 = 0 or I_R4 = 0 you have to arrange for both of resistor terminals to be at the same voltage. Assume the bottom of both resistors is connected to ground (V = 0). Then you have to figure out how many ways you can get their upper node to be 0.

One way I can think of is to make V a negative value while I is still a positive value (current flowing up).

I bet you would have a lot of fun trying to figure this out using this circuit simulator: https://spinningnumbers.org/a/circuit-sandbox.html(1 vote)

- I'm having trouble figuring out how the algebra was worked out starting at2:05. He mentioned how I1 is mentioned twice. How does that turn into -(R1 + R2)I1 ?(1 vote)
- i1 appears twice in the top equation (the one right next to the Circled-3).

The two terms are -R1 i1 and -R2 i1.

You can factor out i1 to get ( -R1 -R2 ) i1 which is the same as - ( R1 + R2 ) i1.(3 votes)

- Why can't we use ohm's law on the resistors? 5 volts (from the source) over 3000 ohms = about 2.5 mA(1 vote)
- Where did you get 3000 ohms from?

This isn't a simple 1-resistor problem for Ohm's Law.

The current coming out of the 5V source splits after the 1k resistor. You don't know ahead of time how much current goes down through the 2k resistor. The ratio of the split is determined by everything else in the circuit.

This video sequence on Mesh Current is one way to work out how it all happens.(2 votes)

## Video transcript

- [Voiceover] We're working
on the mesh current method, a method of analyzing circuits. And in the previous video
we set up our circuit, we set up our mesh currents
flowing around these loops within the circuit, and we
solved for the easy currents. That was the current on this side, this current source constrains
i3 to be a fixed value, so we already know i3. And then we created two equations from the two independent
variables, the two currents here, to describe how the circuit works. And in this video we're going
to solve these two equations. One thing to remind ourselves, when we defined these mesh currents all of them went in the same direction. It helps to make them all
go in the same direction, but all clockwise in this case. And that's so that when you have this little pattern down here of i1 minus i2, or i2 minus i3, that it becomes a familiar pattern, and it helps to get the signs right. The other thing to point out is the reason we're using
meshes and not any other loop in this circuit, not overlapping loops like we could have gone all the way around the outside in a loop. The reason that meshes, in particular the open
areas of the circuit, are a good way to do this method is because it produces exactly the right number of equations
to solve the circuit. You don't have too few equations, and you don't have too many equations. So let me move the screen up and we'll go to work on these equations. So now we begin step four, which is to solve the equations. And what I want to do first is, just to tidy up a little bit, I'm gonna take the constants and I'm gonna put them
on the right hand side, and I'm gonna gather
together the terms for i1 and the terms for i2 in separate places. So if we go at the first
equation and we look for i1, it's mentioned here twice, so we get minus and it's a combination of r1 and plus r2 times i1. And for the i2 term,
let's see if i2 shows up, i2 just shows up once here. So it's gonna be plus r2 i2. And that's gonna be equal to, let's move V over to the other side, so V becomes minus V over here. So that's our first equation. So now let's tidy up the second equation. The i1 terms, this is
the only i1 term here. So it's plus r2 i2, oops, plus r2 i1, that's the only i1 term. Now if we do i2 terms, we're gonna have minus r2 i2,
minus r3 i2, and minus r4, i2. So let's do that this way, r2 plus r3 plus r4 times i2. Now we have an i3 term in here, but we know what i3 is,
i3 we know from this. So I'm gonna write that down right here, that equals minus I, there's
lots of minus signs here, so let's see if we can get this right. Minus r4 times minus i3 is positive r4 i3, or it's negative r4 I, and when we take it to the
other side it'll be plus r4 I. Let me check that again. Minus r4 times minus
i3 is a positive sign, I gives it a negative
sign, and when we take it to the other side it gets a positive sign. Good. Alright? So this is what our
equations look like now. Let's keep going. I can continue using symbols
for the resistor values, but it's gonna get kind of complicated and it doesn't help
with our understanding, so I'm gonna switch over
to a specific circuit. So what I did was I filled
in some resistor values here and I put five volts
on the voltage source, one k ohms for r1, 2k ohms is r2, 3k ohms is r3, 3k ohms is r4, and there's two milliamps over here. So I'm just gonna write these values onto the equation here so I don't mess up. One k, two k, two k, that's five volts. That's two k ohms, two
k ohms, three k ohms, and r4 was also three, is three, and I is two milliamps. I'm gonna move the circuit down
out of the way a little bit. Now we have room for our equation. Let's fill it out again
with the real values. So it's gonna be minus
one plus two is three i1, plus two i2 equals minus five. We know all the resistors are k ohms and so we can actually
scale our equations down by a factor of 1,000 and just use the simple numbers without the ks. So our second equation is two i1 minus two plus three
plus three is eight i2, and that equals three k
ohms times two milliamps, is plus six. So this is getting close
to something we can solve. So I think the way I want to solve this is I'm going to take this equation here and multiply it by four, and then add it to this equation here. And let's see what we get. So four times minus three is minus 12, and we add it and we're gonna get. Four times two is eight,
minus eight, and i2 drops out. That was the purpose of the four. And that equals four times
minus five is minus 20, plus six is minus 14. So i1 equals minus 14 over 10, or i1 equals 1.4 milliamps, and that's cool. We get to put a box around that. So as a reminder, we defined
i1 to be the mesh current flowing around like this in i1. 1.4 milliamps. That's one of our independent variables. Now that we know i1, let's just choose one of our original equations here
and figure out what i2 is. We'll plug in two times
i1, which is 1.4 milliamps, minus eight i2 equals six. And now we have to figure out i2. So let's solve this. Minus eight i2 equals six minus two times 1.4 is 2.8. i2 equals 3.2 divided by minus eight, or i2 equals minus 0.4 milliamps. And we can draw a box around that too. So let's put that mesh current in. And our mesh current was. That was i2, and that
equals minus 0.4 milliamps. And what that means is
that the current is flowing in the negative direction, so it's actually flowing this
way through these resistors. It's actually flowing in the
counterclockwise direction is the positive current. And from before we already knew i3. i3 equals minus two milliamps. So let's do one last thing here, let's figure out the
currents in the elements. Let's just figure out a couple of them, that one and that one. So that is ir2 and this one is ir4. ir2 equals i2 minus, oops, i1 minus i2 equals, i1 is 1.4 minus negative .4, and so that current equals 1.8 milliamps. And we'll do one last one,
we'll do ir4, this is r4 here. And ir4 is i2 minus i3. So that equals, i2 is minus .4, minus i3. i3 is minus two milliamps. So ir4 equals two minus .4, or 1.6 milliamps. And there you have it, we used the mesh current
method to solve our circuit.