Hello everyone, I'm Jesse Mason and in this edition of Teach Me we'll use Kirchhoff's rules to obtain three linear equations with three unknown currents. Then we'll set these equations into an augmented matrix and row reduced to determine values for each current. Finally we'll use the current values along with the resistance values to determine the voltage across and power dissipated by our resistors here. Alright, y'all ready for this? Let's begin. When using Kirchhoff's rules to analyze a circuit we always start by labeling our junctions and then our currents and then our loops. Okay, starting with our junctions we have J1, J2, J3 and we'll call this J4. Moving on to currents we'll start with the current coming out of the two-volt battery. We'll call that I1 and since any amount of current leaving a battery must necessarily re-enter the battery, we'll call this I1 as well. Now when I want to reach this junction one it splits into two currents. We'll call this current I2 and normally we'd call this current here I3. Instead we're going to invoke the junction rule up front and say I1 is equal to I2 plus I3 and if we rearrange this equation we have I3 is equal to I1 minus I2 and that is how we're going to label this current. And we're going to do the same with junction 2. Here we'll call this our new I3 and this current will be the difference between I1 minus I2 and I3. So we'll write this current as I1 minus I2 minus I3 and just to be good and thorough we'll indicate the continuation of the current down here. By the way what all of this is doing is reducing the number of current variables that we're dealing with which will in turn reduce the size of our matrix making it a little less complex. Okay, when these two currents rejoin they become I1 minus I2 and that's it for labeling currents. Now we label our loops. This would be loop A, here we have loop B and loop C. Now that our circuits labeled we can apply the loop rule. The loop rule states that the sum of the voltages around a closed loop is equal to zero. We'll start by labeling loop A. Here we're going to start in the upper left hand corner and move clockwise around our loop. So we have minus 5 I1 minus 6 I2 minus 3 I1 plus 2 equals 0. This simplifies to negative 8 I1 minus 6 I2 and since we need an entry for I3 we're going to just write 0 I3 here equals negative 2 and we'll multiply through by negative 1 to get rid of those negative signs. Okay, for loop B we have minus 2 times I1 minus I2 the current there minus 5 I3 plus 4 plus because we're going against the current 6 I2 equals 0. And the simplify will distribute combine-like terms and multiply through by negative 1 and we do we get 2 I1 minus 8 I2 plus 5 I3 equals 4. Okay, now we apply the loop rule to loop C and the 2 O and 8 O resistor I'm going to combine as a 10 O resistor. So we get minus 10 times I1 minus I2 minus I3 minus 8 plus 5 I3 equals 0. We'll distribute and combine-like terms and this equation simplifies to negative 10 I1 plus 10 I2 plus 15 I3 equals 8. Wonderful, we have three equations and three unknowns. To solve these equations we're going to set up an augmented matrix. Now I like to draw mine right on top of my equations so I can see the entries more clearly. So the first row we have 8, 6, 0 and 2, second row we have 2, negative 8, 5 and 4, third row, negative 10, 10, 15 and 8. Okay, so there's our augmented matrix. Time to flex that matrix muscle of yours and do some goussjord on elimination. Now if the following matrix maneuvers move to quickly for your taste, feel free to click the link below for a more thorough explanation of all the linear algebra that we're doing here. Okay, so the first row operation we'll do on our original matrix is multiply row 2 by 5, add row 3 to it, and replace row 3 with the result. This gives us a 0 in the first entry of row 3. Next we'll take row 1, subtract 4 times row 2, and replace row 2 with the result. This gives us a 0 in the first entry of row 2. And at this point all of our entries are even numbers. So we'll do three steps in one and multiply every row by 1 half. Next we'll take row 2 and add to it 19, 15 times row 3, and we'll replace row 3 with the result. This gives us a 0 in the second column third row. Now we'll multiply row 3 by 3, 46, and this will give us our first solution, which is why you see a little happy face here. Hey, it's the small things in life, right? Okay, now we're going to take row 2 and add to it 10 times row 3, replace row 2 with the result. This will give us a 0 in the third column of the second row. Next we'll multiply row 2 by 1 19th, and that gives us our second solution. Yeah, that's my only face. Okay, now we're going to take row 1 and subtract 3 times row 2, and replace row 1 with the result. This gives us a 0 in the second column of the first row. We're almost there. Last thing we're going to do, we're going to multiply row 1 by 1 fourth, and then our matrix is in row reduced echelon form. Okay, summarizing our results, we have I1 is equal to 0.25 amperes, I2 is equal to 0 amps, and I3 is equal to 0.7 amps. And believe it or not, I didn't rig this problem to have such gorgeous results. Matter of fact, if I told you how it came up with the initial numbers here, you'd probably call me up to see if I was pulling your leg. Okay, so the last thing we're going to do here is set up a solutions table. This table will not only allow us to organize a results, but we'll make our final calculations much quicker. So in the first column we'll have resistance, second column current through the resistor, third column voltage across the resistor, and last column power dissipated by that resistor. We'll start with resistances. First we have the 5 on the left, the 2 on the center, then the 2 on the right, then the 3 on the bottom, the 6 on the center, the 5 on the center, and then the 8 on the right. Next we move to currents. The current through the 5 on resistors is just I1, which is 0.25. The current through the 2 on resistors, I1 minus I2, that difference is also 0.25 amps. And the current through this 2 on resistor here is I1 minus I2 minus I3. That difference is actually negative. We get negative 0.45 amps. This negative sign tells us that the direction we chose to describe positive charge flow through that leg is not the right direction. It's actually going the other way. But that's part of the beauty of Kirchhoff's rules. You choose the directions arbitrarily, and in the end you find out if you've chosen wisely. Okay, so the 3 on resistor here is CsI1, which is 0.25 amps. The 6 on the center leg here is CsI2, which is 0 amps. Again, unplanned, but really cool. Then the 5 on resistor CsI3, which is 0.7 amps, and the 8 on resistor, just like the 2 on above it, sees negative 0.45 amps. Excellent. Now that we have all these currents, we can use Ohm's Law to determine the voltages across our resistors. And now you can see why we use the table. We just need to multiply columns. So this first product is equal to 1.25 volts. Product to the second two numbers here is 0.50. Here we have negative 0.90 volts. This negative sign is actually telling us that the direction of the voltage drop is right to left, not left to right like we had assumed. Our next product is going to be 0.75 volts. And the voltage drop across a resistor, which CsI0 current is not surprisingly 0. Here we have 3.5 volts, and this one is negative 3.6 volts. Okay, lastly, to determine the power dissipated across each one of these resistors, we multiply, you guessed it, the previous two columns, current and voltage. Of course, if you prefer to use the formula that has I squared R instead of IV, that's just fine too. More power to you. Actually, it's the same amount of power. Hey look, the resistor that CsI0 current dissipates no power. Good thing. Well, there you have it. Using Kirkoffs rules and conjunction with some linear algebra techniques to determine the current through, voltage across and power dissipated by a bunch of resistors. I'm Jesse Mason, I hope you found this video helpful, and until next time, happy learning. .