Mesh current method (step 4 solve)

We're working on the mesh current 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, uh, the current on this side. This current source constrains I3 to be a fixed value, so we already know I3.

Then we created two equations from the two independent variables, the two currents here, to describe how the circuit works. In this video, we're going to solve these two equations. One thing to remind ourselves when we define these mesh currents, all of them went in the same direction. It helps to make them all go in the same direction, all clockwise in this case. That's so that when you have this little pattern down here of I1 minus I2 or I2 minus I3, 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 the 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. Now we begin step four, which is to solve the equations. What I want to do first is just to tidy up a little bit. I'm going to take the constants and I'm going to put them on the right-hand side, and I'm going to gather together the terms for I1 and the terms for I2 in separate places.

So if we go to 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. For the I2 term, let's see if I2 shows up. I2 just shows up once here, so it's going to be plus R2 I2, and that's going to 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. I1 terms—this is the only I1 term here, so it's plus R2 I1. Oops, plus R2 I2. That's the only I1 term. Now, if we do I2 terms, we're going to 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 going to 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 -I3 is positive R4 I3, or it's negative R4 I. When we take it to the other side, it'll be plus R4 I. Let me check that again: minus R4 times I3 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, right? 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 going to get kind of complicated and it doesn't help with our understanding. So I'm going to switch over to a specific circuit.

So what I did was I filled in some resistor values here, and I put 5 volts on the voltage source. 1 k ohm is for R1, 2 k ohms is R2, 3 k ohms is R3, 3 k ohms is R4, and there's 2 milliamps over here. So I'm just going to write these values under the equation here so I don't mess up: 1 k, 2 k, 2 k— that's 5 volts; that's 2 k ohms; 2 k ohms; 3 k ohms; and R4 was also three. R4 is three, and I is 2 milliamps.

I'm going to 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 going to be minus 1 plus 2 is 3 I1 plus 2 I2 equals -5. We know all the resistors are k ohms, and so we can actually scale our equations down by a factor of a thousand, and you just use the simple numbers without the k's.

So our second equation is 2 I1 minus 2 plus 3 plus 3 is 8 I2, and that equals 3 k ohms times 2 milliamps is +6. So this is getting close to something we can solve. 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 4 times -3 is -12, and we add it, and we're going to get 4 times 2 is 8 minus 8, and I2 drops out. That was the purpose of the four, and that equals 4 times -5 is -20 plus 6 is -14, so I1 equals -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, and I1 is 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 2 times I1, which is 1.4 milliamps, minus 8 I2 equals 6, and now we have to figure out I2. So let's solve this: -8 I2 = 6 minus 2 times 1.4 is 2.8. I2 equals 3.2 over -8, or I2 equals -0.4 milliamps. 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 -0.4 milliamps. 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 as the positive current, and from before we already knew I3.

I3 equals -2 milliamps. So let's do one last thing here; let's figure out the currents in the elements. Let's figure out a couple of them— that one and that one. So that is I R2, and this one is I R4. I2 equals I2 minus, oops, I1 minus I2 equals I1 is 1.4 minus -0.4, and so that current equals 1.8 milliamps.

We'll do one last one; we'll do I R4. That is this R. This is R4 here, and I4 is I2 minus I3, so that equals I2 is -0.4 minus I3. I3 is -2 milliamps, so I4 equals 2 minus -0.4, or 1.6 milliamps.

And there you have it; we used the mesh current method to solve our circuit.