okay so this example problem we're gonna use mesh analysis again I mean we could use nodal analysis to solve this problem as well actually if you use nodal analysis to solve this problem it's a little bit easier you only have two equations and two unknowns but these videos here we're talking about mesh analysis so we want to make sure we do some examples using mesh analysis now the reason I point out nodal or mesh is that on the quizzes it's not going to specify necessarily to use nodal or mesh if it does specify using nodal or using mesh obviously you make sure to use the correct technique to solve the problem now since we're doing mesh analysis I'm gonna draw some loops here and again you can draw any loops you want but typically I draw what are the obvious loops and the circuit so for instance there could be a loop like this and I'll label that remember these loops are currents these are loop currents and that's I 1 now I'll purposely draw I 2 in the counterclockwise direction just so we have something that's slightly different as said normally I draw everything in the clockwise direction but I'll purposely draw this one in the counterclockwise just to show that it doesn't matter which way we draw these things and then I'll draw another one in the clockwise position I'll call this one I 2 and this one I 3 and again you're labeling doesn't matter either you know I could have labeled this one I 1 and this one I 3 it doesn't matter you're labeling it's just gonna you know change your equations slightly you'll have the same numbers and the same answers obviously these are just variables all right so now our ultimate goal is to find the power delivered by the 80 volt power supply so power for the 80 volt power supply is voltage times current there is no other short equation because a power supply isn't a resistor so I can't do like V squared over R or R squared times I so I have to do V times I the other option would be to find the power to I paid it in every single resistor but that seems like a lot of extra work because if I look at this how many branches are touching this power supply and when I say branches I should say how many loops are touching this power supply well there's only one loop I want so the current through this 80 volt power supply is just I 1 or that the current being delivered is simply I 1 so actually if we find I 1 will be able to find that power of the voltage source really quickly really easily because we know what voltages it's 80 volts and we know what I is we'll be able to determine that power similarly here I - that's the only loop touching this 8 ohm resistor so the power for the 8 ohm resistor again we could do voltage times current but since we're gonna be finding currents it would make sense to use the current square times resistance all right so now that's just kind of setting up what we're looking for of course now we have to write down mesh equations and unlike the video previously I'm not gonna write down 80 volts the voltage of the 5 ohm the voltage of the 26 we're gonna go ahead and apply Ohm's law right away but let's go ahead and start on loop 1 here so loop 1 well first off I'm gonna notice that this arrow comes into the negative side of the voltage source so that's minus 80 now you can start anywhere in this loop you could start with the 5 of them I always just tend to start with the power supplies so I come into the negative side it's minus 50 now remember the loops you're in are always the positive so I'm gonna have current times resistance I don't write the resistance first so it's 5 ohms now what's the current through this 5 ohm resistor well there are two loops that are touching that 5 ohm so let's draw those here I 1 is always the positive direction because I'm in loop 1 so I ones going to the right and what is I 3 doing I 3 is going to the left so there are in opposite directions so this would be I 1 minus I 3 then we have this 26 ohm resistor so we need the voltage across the 26 well that's going to be 26 ohms times the current now again we're in loop 1 so we know that loop 1 i1 is going to be the positive now let's see here I 1 is going down through this 26 ohm resistor and what is i2 doing I 2 is also going down through that 26 ohm resistor so this is going to be plus i2 because they're both going in the same direction and then this is equal to 0 and then of course we would combine like terms and simplify things and I'm going to go ahead and combine like terms and simplify it and just write it off up here to the side here without doing too much else here so we'd have 80 volts because I'm gonna move that to the other side is equal to well 5 plus 26 is 31 home's not 31 ohms 31 I 1 and then I'm gonna have plus 26 I 2 minus 5 I 3 so that's what I'd have for loop 1 and again of course I have three unknowns but I have three loops so we should be fine now if I look here at loop two I'm gonna erase this here again maybe just to give us some more room so again I do this so I don't have to keep scrolling back and forth between the circuit diagram alright that's good enough so for loop - I'll go ahead and do that one in blue loop - is this one right here and remember when we're doing the resistors it's always the positive so I'll start with the 8 ohm resistor so it's eight now the only loop touching that 8 ohm resistor is i2 so this is just simply 8 times i2 and this is my voltage again that I'm summing up because current times resistance is a voltage and now I'll just move on to the next one which is the 90 ohm resistor and I'm gonna notice here that well again I'm in loop 2 so this is gonna be I 2 and it's positive automatically but this I 2 is going to the left through the 90 ohm and what's nine this I three doing it's also going to the left through the 90 ohm so that means since they're both going to the left it's going to be plus i3 and then we've got this 26 on 1 which is 26 and we've already got the currents kind of drawn in there so we know what directions they're going so this is going to be 26 and this is going to be i2 + i1 equals 0 and then sorting this out a little bit you have 8 I 2 plus 90 i2 plus 90 I 3 plus 26 I 2 plus 26 I 1 equals 0 and then combining terms together what we'd have here is we'd have 0 is equal to 26 I 1 plus well let's do that addition I'd have 8 + 90 + 26 and it gets me 124 I 2 plus 90 I 3 so that's my second equation and then we'll go ahead and do the last one here you might want a positive again if you don't have this written down because I'm gonna erase this and we're gonna do loop three so I'll do in red and again doesn't matter where you start I often like just to start if there's one that's obviously only having one loop touching it like this thirty one resistor only has I three touching it so I'd start with thirty I three plus well then I'm going to have ninety and again these are in the same directions so this is a three plus I 2 and then I have plus five now I'm in loop three so loop three is automatically the positive one and then I one is obviously going in the opposite direction through that 5 ohm resistor right here and so this would be minus i1 equals zero and again kind of multiplying some things out we get this if we multiply everything out and then if we write our equation here we'd have zero is equal to well I have minus five and I made a mistake right there that should be i1 i1 plus 30 no not 39 t i2 plus this would be 120 plus five so that's 125 I three and I'm gonna double check my arithmetic on there and then now we have three equations and three unknowns so we would solve that so and if we solve that system of equations I wind up with I 1 is equal to 5 amps I 2 is equal to negative 2.5 amps and I 3 is equal to 2 amps now you might say well what happens if I didn't draw them all in the same direction as you do all that would happen is these would these would be negative or pluses were to change would change for instance you know let me draw it here in green if I drew a current instead clockwise there well they're just in the opposite directions so if I did it here this green current would be positive 2.5 amps now all my equations would look different because now I've changed the direction and the green current is going in the opposite direction so I'd have a bunch of minuses where I'd have pluses like all of these pluses that I had here I 3 plus I 2 they'd all become minuses so your equations would look slightly different but you'd still get the same answer and my point here is that it really doesn't matter the direction you draw these loops what matters is that you apply them correctly now we know that the problem here I'm going to erase this equation here problem was asking us for the power delivered power supply and to get the power its voltage times current right well let's see here the current i1 is going into the negative side of the power supply so my power for the 80 volt let's take it off green here the power for the 80 volt would simply be it's going to be negative eighty because I'm coming into the negative side of that power supply times and the only current going through there is five amps that's the only current I've got going through there and so if I do that you get negative 400 watts which of course this means that the power supply is delivering 400 watts because remember if you get a negative that means power is being delivered if you get a power a positive that means power is being absorbed so the power supply is delivering 400 watts and then the power for the 8 ohm resistor remember we said we're going to just do the current square times the resistance now there's only one loop that's touching that a an ohm resistor the only loop that's touching it is i2 so this would be negative two point five squared times 8 ohms which if you do that I'm just double-checking my computation here you get 50 watts and again if you had drawn the current and used the green arrow current you'd still get the same result because negative two point five squared is the same thing as positive two point five squared and so that's the result of it now the reason I did the as I said normally I always draw mine clockwise but I wanted to show that you can draw them counterclockwise and clockwise and it doesn't make a difference you still get to the correct answers in the end so it really is just what what is convenient to you now we're going to look at one more video on mesh analysis and then we're going to move on as I said to some more applications