so for this lesson we're going to be talking about current division and doing an example with current division now current division deals with parallel circuits and the reason that deals with parallel circuits we know that circuits in parallel have the voltage all being the same but the current we said is typically not the same and so we're wondering well how does that current divide down amongst the different components so let's again just kind of look at a figure here I'm sorry that shouldn't be a voltage this should be current because we're looking at current division let's say we want to find I won well we have a current source here I'll call this I that's coming up and then it's dividing amongst these three resistive branches we want to know what the current i1 is well remember again we said this is in parallel so the voltage we know is the same across any of these components the voltage that I measure and I'll even put one more here these are all the voltage V so I do know that V is going to be i1 times r1 I mean we also know I'll put it off to the side here that V would be equal to i2 r2 and that V would also be equal to I 3 R 3 and we have one more here that V has to be equal to I times R equivalent so it's the current that's coming up through this current supply times the equivalent resistance here so of course we're wanting to find I one well actually right here this is going to be pretty quick we know what the current I is that's what we're given and I can see that I could substitute then for V or set these equations equal to each other since the voltages are the same I know that I times R equivalent is going to equal I 1 times R 1 which of course means that I 1 is just simply equal to R equivalent over R 1 times my current now this is the formula I like to remember and I like it because it's really just kind of the reciprocal of the voltage division formula remember the voltage division formula I'll scroll back here voltage division formula was RN over r eq times voltage well now for current we're getting req over r 1 so if we wanted to write for a general current I and would be our eq / our n times the current supply now there are different ways of writing this if we use in mittens we know that immittance is y is equal to 1 over R so I could write then that this would be I n is equal to 1 over Y eq / 1 / y1 times i which this simplifies to i n is equal to y1 / y eq times i now this honestly is the reason i think some people like to use admittance is then if we look at this formula this formula is in the same format as the voltage division except instead of using resistances you're using admittances so i n is not y1 it would be yn or yeah yn here there should be a yn it's in the same form as the voltage division however I will emphasize this is what I will use now that does not mean you have to use this one I'm a big big advocate that if what you're doing is correct as far as formulas are correct and your math is correct and it's not the way I'm doing it but it's still a valid approach I'm not going to mark your wrong because it's not about doing it my way it's about understanding how to analyze circuits and if you do it a correct way that's different than quote the way I would do it well then I should still give you full credit because the point isn't about doing it my way it's about understanding how to analyze circuits all right so let's look at an example here of this okay so I want to know here I'll just label them here I 1 I 2 and I 3 I want to know what all three of these currents are so find I 1 I 2 and I 3 so just remember here I 1 for instance would be our EQ over r1 times my current I 2 would be our EQ over r2 times my current and I 3 would of course be our EQ over r3 times my current so really there's not a lot to do in this problem I just have to find our EQ and req would simply be 1 over 10 plus 1 over 6 plus 1 over 18 all to the negative 1 power and let me do that on my calculator real quickly yeah we get something that's not necessarily pretty number I get that this is 3 point 1 0 3 ohms which if you I mean that's the way I write it but if you wanted to some calculators like to store them or you can have them stored as a fraction and if you did it as a fraction this would be 90 over 29 but I really don't think that's necessary to do as long as we have this to a few decimal spots that will be good enough for us now when you get into measurements and significant digits and the things of like that things like that can be important but again even if you're talking about significant digits you really wouldn't need 90 over 29 you just need an extra digit compared to what your significant digits are so we're just going to leave it a three point one zero three ohms so then to get i1 it's just 3.10 3 ohms divided by 10 times 10 amps which would be 3 point 1 0 3 amps I 2 is going to be 6 ohms divided by 10 ohm I'm sorry 3 point 1 0 3 ohms divided by 6 ohms times 10 amps and if I do that we get five point 1 7 2 amps and then for I 3 it be divided by 18 ohms times 10 amps and you'd get one point seven to four amps and so these would be the three currents now let's go ahead and do the same thing we did with the voltages let's sum them up and so let's do three point one zero three plus five point one seven two plus one point seven two four and if we add all of those up here we wind up with nine point nine nine nine amps which of course is just ten amps now technically it should add up to exactly ten amps but we're adding up to nine point nine nine nine hopefully we can all see why this would be the case that we're not getting exactly ten amps and the reason is of course because we approximated what the equivalent resistance is because we use the approximation of 3.1 3.8 when t9 but even when you use approximate ones you can see that when you sum the currents up you know you're getting that the current I'm going through those resistors is equal to the current being delivered now the other thing we should point out here is that the smaller the resistance the more current we get which really shouldn't surprise us at all I mean talking about voltage and current here and resistance and we've said that hey you can think of resistance as like resisting current flow so we shouldn't be surprised that we get more current the smaller the resistance is so you know that's what we get which again shouldn't really surprise us alright let's end this video here the next thing we'll be talking about is something called nodal and mass now a lot of textbooks like to make this a lot more complicated than it really is we're gonna go through that really quickly and then do a couple examples and then we'll start getting into some applications so no to a mesh analysis is kind of the last you know analysis technique we need for this course before we can start getting into some applications