WEBVTT 00:00:01.780 --> 00:00:05.330 In this video, we're gonna demonstrate the super mesh technique. 00:00:05.330 --> 00:00:06.760 As you look at this, 00:00:06.760 --> 00:00:11.450 you'll notice that there are three meshes, this one here to the left. 00:00:11.450 --> 00:00:15.400 And we'll label the current in that mesh current I1. 00:00:15.400 --> 00:00:18.850 We have a mesh up here. 00:00:18.850 --> 00:00:21.790 And we'll call this mesh current I2, and 00:00:21.790 --> 00:00:26.370 we have a mesh here with a mesh current that we'll label I3. 00:00:27.830 --> 00:00:32.530 And further before we even get started we'll notice that separating mesh one from 00:00:32.530 --> 00:00:36.370 mesh three is a branch that contains a current source. 00:00:36.370 --> 00:00:39.180 And as we've mentioned in the past, we don't have a. 00:00:39.180 --> 00:00:44.250 A numerical or mathematical relationship between the voltage across 00:00:44.250 --> 00:00:46.240 a current source and the current itself. 00:00:46.240 --> 00:00:51.160 That will then require us to do a super mesh which we will 00:00:52.710 --> 00:00:55.560 Be writing around there. 00:00:55.560 --> 00:00:58.790 Let's get started first of all though by writing the mesh 00:00:58.790 --> 00:01:02.000 current equation around this mesh number one. 00:01:02.000 --> 00:01:06.410 Starting down here in the lower left hand corner, we have going minus to plus. 00:01:06.410 --> 00:01:08.169 It will be a negative 12 volts. 00:01:09.410 --> 00:01:15.920 And then across this resistor of a voltage drop of five resistance five times current 00:01:15.920 --> 00:01:22.590 rate through that which is I1-12 00:01:22.590 --> 00:01:27.690 plus the volts struck across this film resistor which is going to be 3 00:01:27.690 --> 00:01:32.950 times I 1- I3 and 00:01:32.950 --> 00:01:37.810 the sum of those voltage drop must equal 0. 00:01:37.810 --> 00:01:43.220 Now, the Super Mesh starting here at this point going up and around the 15 ohm, and 00:01:43.220 --> 00:01:50.280 when we're across the 15 ohm resistor now the voltage drop there being 15 times I2. 00:01:50.280 --> 00:01:54.860 Now continuing on down here, 00:01:54.860 --> 00:01:59.840 not going around this branch here because that's where the current source is. 00:01:59.840 --> 00:02:04.817 Continuing on down here across the 10 ohm resistor we have a voltage drop 00:02:04.817 --> 00:02:09.229 of +10 x (I3) Plus, 00:02:09.229 --> 00:02:13.720 now continuing around here, across the 3 ohm resistor, +3 times. 00:02:13.720 --> 00:02:19.904 Now the current going in this direction is I3-I1. 00:02:19.904 --> 00:02:24.980 I3-I1+ the voltage 00:02:24.980 --> 00:02:30.040 drop across that 5 ohm resistor Is five times. 00:02:30.040 --> 00:02:32.170 And again, because we're going from right to left, 00:02:32.170 --> 00:02:35.370 the current going in that direction will be I2 minus I1. 00:02:35.370 --> 00:02:42.110 I2 minus I1, the sum of those terms must equal 0. 00:02:42.110 --> 00:02:44.420 And finally the third equation we need for 00:02:44.420 --> 00:02:49.700 these three unknowns comes from the relationship between I2, I3. 00:02:49.700 --> 00:02:50.740 And this current source. 00:02:51.940 --> 00:02:56.060 In this case, I2 is going in the direction of the current source, so 00:02:56.060 --> 00:03:02.290 we're going to have I2 minus I3, 00:03:02.290 --> 00:03:08.530 which is the current going through that branch in terms of the mesh currents. 00:03:08.530 --> 00:03:11.040 It's going to equal 2 Amps. 00:03:12.690 --> 00:03:15.620 So we have our three equations and three unknowns. 00:03:15.620 --> 00:03:17.240 Let's go ahead and combine terms. 00:03:17.240 --> 00:03:24.276 Taking the first equation we have I1 times, 00:03:24.276 --> 00:03:29.089 we got 5 there plus 3 is 8 plus 00:03:29.089 --> 00:03:33.730 I2 times We have a negative 5 00:03:36.352 --> 00:03:41.636 Plus I3, we have a negative 3. 00:03:44.377 --> 00:03:48.687 And then we have this negative 12 volts on the left-hand side we bring to the other 00:03:48.687 --> 00:03:50.720 side as a positive 12 volts. 00:03:50.720 --> 00:03:51.998 And there is our first equation. 00:03:54.119 --> 00:03:57.260 Second equation combining like terms, we have I1. 00:03:57.260 --> 00:04:02.508 And in this we have -3 and a -5. 00:04:02.508 --> 00:04:07.290 That makes -8 times I1 + I2 times, 00:04:07.290 --> 00:04:11.964 we got 15 00:04:13.660 --> 00:04:20.029 plus five is 20 times i2 plus i3, we've 00:04:21.260 --> 00:04:26.780 got 10i3 plus 3 is 13i3 equals, and 00:04:28.720 --> 00:04:32.740 there are no constants there So 00:04:32.740 --> 00:04:36.720 the sum of those terms equals 0. 00:04:36.720 --> 00:04:40.992 Finally, our third equation is I2, 00:04:43.895 --> 00:04:52.421 Times 1 + I3(-1) = 2. 00:04:54.060 --> 00:04:56.170 And we have our three equations and three unknowns so 00:04:56.170 --> 00:05:00.880 get ready with your calculator or whatever you choose to solve through your systems 00:05:01.940 --> 00:05:07.700 through your equations and three unknowns. 00:05:07.700 --> 00:05:11.205 And when you do we get the following. 00:05:11.205 --> 00:05:12.230 I1=2.03 amps. 00:05:13.730 --> 00:05:20.170 I two is equal to one point two eight amps 00:05:20.170 --> 00:05:24.918 and I three is equal to negative 00:05:24.918 --> 00:05:29.432 point seven two amps. 00:05:31.020 --> 00:05:34.100 LEts do a couple of consistency checks here just to make sure we didn't make any 00:05:34.100 --> 00:05:39.210 mistakes not to prove that our work was correct but at least to look at a couple 00:05:39.210 --> 00:05:43.580 of calculations that would be consistent with what we know to be true. 00:05:43.580 --> 00:05:48.565 For example we know that i two minus i three equals two amps. 00:05:48.565 --> 00:05:53.590 So I2, that is 1.28 00:05:53.590 --> 00:05:57.690 minus I3, well I3 is a -0.72 amps. 00:05:57.690 --> 00:06:03.220 The sum of those, they're supposed to equal 2 amps, 00:06:03.220 --> 00:06:04.800 and sure enough that works there. 00:06:06.800 --> 00:06:10.520 And of course we could plug those values into any one of those equations And 00:06:10.520 --> 00:06:12.280 proves that our solutions are correct. 00:06:13.570 --> 00:06:16.060 But now that we know what I1, I2, and I3 are. 00:06:16.060 --> 00:06:18.840 Let's use them to say for 00:06:18.840 --> 00:06:24.480 example now determine what the voltages drop is across this current source. 00:06:25.700 --> 00:06:28.530 We said that we couldn't numerically calculate it based upon the current going 00:06:28.530 --> 00:06:33.260 through it, but now that we know the currents around it, we can write a KVL 00:06:33.260 --> 00:06:38.720 once again around that loop, and solve for the voltage across the current. 00:06:38.720 --> 00:06:44.130 Starting here and going clockwise, across that 3 ohm resistor, 00:06:44.130 --> 00:06:49.890 we're going to have Three times I three minus I one. 00:06:51.420 --> 00:06:56.270 Now, going across this current source with the voltage as referenced plus to minus. 00:06:56.270 --> 00:07:00.080 That represents a voltage drop, so it would be plus v. 00:07:01.870 --> 00:07:05.704 Now, coming down across here, that would be plus 10. 00:07:06.880 --> 00:07:12.680 Oops, plus 10 times I3 00:07:14.740 --> 00:07:19.730 equals that's where we started from, 0. 00:07:19.730 --> 00:07:25.108 Now I plug in our values for I1, I2, 00:07:25.108 --> 00:07:30.151 and I3 and we have 3 times I [SOUND] So 00:07:30.151 --> 00:07:36.876 3 times I3, well, I3 is a -0.72 amps, 00:07:36.876 --> 00:07:41.247 -0.72,- I1, which is 00:07:41.247 --> 00:07:46.796 2.03 amps, + V + 10 times I3, 00:07:46.796 --> 00:07:51.868 which is a negative 0.72 amps. 00:07:51.868 --> 00:07:54.968 The sum of those things has to equal 0. 00:07:54.968 --> 00:08:00.061 Solving this for V gives us 00:08:00.061 --> 00:08:04.697 V = 15.45 volts. 00:08:07.588 --> 00:08:12.358 All righty, now looking at this source, we see that the current is reference into 00:08:12.358 --> 00:08:16.660 the negative terminal of our voltage and leaving the positive terminal. 00:08:16.660 --> 00:08:18.040 So it's acting as a source. 00:08:18.040 --> 00:08:24.220 And we can then say that the power being generated by that source is equal to 00:08:24.220 --> 00:08:29.300 negaitive i times v Is equal to negative i 00:08:29.300 --> 00:08:34.200 which is two times v which is 15.45 and the power 00:08:36.100 --> 00:08:40.025 then would be equal to an negative what is that 00:08:40.025 --> 00:08:45.930 30.9 watts the fact that 00:08:45.930 --> 00:08:50.230 it's negative means that it's energy being put into the circuit from the source. 00:08:51.540 --> 00:08:54.625 So once again, once we know all three of the mass/f currents, 00:08:54.625 --> 00:08:57.829 we can determine any voltage or current or for that matter power or 00:08:57.829 --> 00:09:01.113 any other quantity that we might want to know within this circuit.