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Current time:0:00Total duration:14:43

Kirchhoff's law application: 2-loop circuit solving

Video transcript

suppose we are given a circuit with some resistors and some batteries we are asked to find out the current through each branch current through each of these wires is what we need to figure out so how do we do this well usually when we have a circuit the first thing i look for is if we can reduce the circuit by using series and parallel formula but over here you can't do that if you don't trust me go ahead give it a try you can't reduce the circuit using series or parallel so in general what do we do when we can't do that when we have more complicated circuits that's where we use kirchhoff's laws and so that's what we're going to do over here as well you use keyshaft's laws to figure this out but before i solve this problem let me give you a general overview of how to solve any problem um any any complicated step how do you solve any complicated circuit like this so think of it mathematically since i want to calculate current through each branch and there are three branches over here as you can see colored differently i have three unknowns and to figure out three unknowns i need a three equations and it's kirchhoff's laws they're going to help me build these equations the first equation can be built using the kirchhoff's current law which is applied to nodes where three or more wires get connected and the remaining two equations can be figured out by using kirchhoff's voltage law kvl which is applied to closed loops like a loop over here or over here or this big loop over here now i'm assuming that you have already practiced keshav's laws otherwise this is this video is going to be pretty confusing so if you aren't familiar with this i highly recommend to go back and watch videos and do practice on our website on these laws and then come back over here but if you have well then let's go ahead and solve this bad boy all right where do i start well since i want to calculate the current we can start by assuming some current in each of these branches so let's say in this branch yellow branch over here the current is moving to the right let me call that value as i1 and and i've colored this over here just to indicate that the current through this entire branch will be the same so over here will be i1 here will be i1 here will be i1 because the current has nowhere else to go and you could ask me why am i choosing it go to to go towards the right well i've randomly chosen that you could have also chosen it to go backwards for example you know you're free feel free to choose whichever direction you want similarly in this branch let's assume the current is i don't know maybe going upwards so let's call that current as i2 and again now here also you are free to choose whatever you want but since it's so nice i can see that you know there's i1 going in and there is i2 going in i like to think of it as okay these two currents are merging and coming out from here so if i1 goes in and i2 goes in then what is the current that comes out you can say i3 but that i3 has to be i1 plus i2 right right i1 and i2 go in and the current that goes out comes out over here must be i1 plus i3 i2 sorry and in effect what i have now done is applied kcl this is kirchhoff's current law which says that whatever current enters a node should exactly equal the current exiting the node and you can see that the same applies here as well i1 plus i2 come in comes in over here i2 goes out i1 goes out over here everything works nice and well so because i've applied kcl already i only have two variables now which means i only need two more equations does that make sense so kcl is done now to build two more equations i'll have to use kvl so let me write that down so next step would be kvl and this is where things get usually a little tricky with the signs and everything but if you you know if you do this conceptually then there will be no problem whatsoever all right so how do you apply kvl you choose a closed loop first so for example i can choose this closed loop over here and the whole idea is you start from some random point let's say i start from a point somewhere over here and let me just give that point a name let's call it point as a and then you figure out what the voltage at this point is and whatever that value is and then you keep track of the changes in the voltage at every point and then you come back all the way if you do that you will build an equation and the whole idea behind kvl is energy conservation okay and again like mentioned before this is something i've been explaining great detail in a previous video so if you need more details feel free to go back and watch that and of course please practice as well this is definitely requires practice all right having said that let's quickly start this so the way i do this is i start at this point and i say let's say that the voltage over here is va okay some voltage i'm going to call that va and now i keep walking and every time i cross a battery or across a resistor i will figure out what the new voltage is so let's say when i walk walk walk and i cross this battery notice some voltage gets added or subtracted so the voltage over here is going to be va plus or minus something so now the question is what is the voltage at this point well it's going to be first question is am i going up in potential or am i going down in potential the way to figure that out at least for a battery would be we'll look at the terminals this is the negative terminal of the battery this is the negative terminal of the battery this is the positive negative means low positive means high so i'm going from low to high meaning when i'm going from here to here i'm going uphill it's like climbing a mountain i'm going uphill all right so when i go from here to here voltage gets added so the voltage over here must be va plus something plus how much 2 volts it's given in the battery it's given over here in the question itself so the voltage at this point is va plus 2. all right let's continue walking walk all the way until we come back now when i when i cross this resistor again there's going to be a potential drop remember ohm's law whenever a current flows through a resistor there will be an ir drop now the question is when i go from here to here am i going up in potential or am i going down in potential how do i figure it out over here well notice i'm going in the same direction as the current current is moving this way right and remember through a resistor current always flows from higher potential to lower potential think of positive charges they will always move from higher to lower potential so i can again say this has to be positive this side and this side has to be negative and basically i'm saying that this chart should be higher than this one because current is moving this way and since i'm also going in the same direction i'm also walking i can now say that i'm going down high to low and so when i come over here the potential drops by some value so negative by how much when i go from here to here potential drops by the whatever is the voltage over here and that's v equals i r and i know i is i one i know r is one so the potential drop over here is going to be i one times one and now i am over here this is the voltage at this point does that make sense the same thing continue you walk walk walk walk walk walk walk and you keep doing this all the way till you come over here so great idea for you to pause the video and see if you can just complete and see if you can build the equation all right if you're giving it a shot let's continue so i come over here now i cross this resistor this time notice i'm going in the opposite direction of the current so again same up the same rule if the current is going this way this must be the positive this must be the negative but remember i am walking in the opposite direction so notice i'm going from low to high i'm going in the opposite direction i'm going low to high meaning i'm going up the potential all right and so when i come over here my potential increases so let me use the same color increases by how much um ir 5 i2 times phi and this is where mistakes are usually made i mean when i was studying this i used to not think about it think about whether i'm going up the voltage or down the voltage i used to mug up in the direction means negative opposite the direction of the current means positive without even understanding what was going on and obviously if you try to mug up like this you are bound to forget it sometime definitely during the exams that's what used to happen to me but now i don't do that don't mug up those arbitrary um you know conventions instead think about it think about are you going up in potential or down in potential all right so finally i'm over here now i'm going from here to here when i'm going through a battery you don't look at the current because a battery is an active device through a battery current can flow either ways it can flow from higher to lower potential or from over to higher also because battery is a pump it can pump charges in the opposite direction so when you're going through the battery that's how you don't look at the current you only look at the positive and the negative terminals another place where commonly i should make mistakes so now when i'm going through a battery notice this is the negative terminal of the battery this is the positive terminal of the battery and so when i go from here to here again i'm going up in potential just by looking at the terminals all right low to high and so by the time i'm over here what's the voltage what's the potential that has to be plus 2 and now i'm over here this is the voltage at this point right but hey that point has the same voltage as point a because there is no more potential difference that means that has to equal va and ladies and gentlemen this is how we build equation using kvl okay now all i have to do is go ahead and and reduce this so va cancels out on both sides and notice what i end up with i end up with 2 plus 2 is 4 minus i 1 plus 5 i 2 equals zero one equation down one more to go to build a second equation we have to choose a different loop maybe we can choose this one or we can choose this big loop as well okay and i want you to give this a shot so since we want since i know i want us to get the same equation let's choose a particular loop let's choose this and let's use this small loop all right so let's say we start let's do it together let's say we start at this point let me call it as point b and let's walk in this direction like this let's walk this way and come back over here so i want you to give this a shot okay see if you can build an equation and then let you know then let's see if it matches with what i get all right if you've given this a shot this is what you end up with i didn't want to you know spend time again writing the whole thing but again you can pause the video and just confirm that this is what you got and if you basically reduce that you will eventually end up with this equation and of course you know if we had gone the other way around you have gotten the same equation you have gotten all positives but it's the same thing right the equation would have remained the same and of course if you have taken a different loop let's say you have chosen a different loop that this big loop then you would have gotten a different equation but eventually the final answer turns out would have to turn out to be the same all right anyways for our loop this would be the equation and again if you've got some place wrong nobody's just you know just go back and check you know what where the error was but now that we have the two equations our physics is pretty much done and now all we have to do is algebra which i believe that you would be able to take care of this so i'll just quickly brush through the algebra part so i'm just going to minimize all the physics part this is now the algebra so if i just put that equation over here i just try to match the coefficients so i'm going to match the coefficients of i1 with over this one so i'll just multiply this whole equation by 3 and that gives me a new equation all the coefficients multiplied by 3 and then i basically subtract this equation to finally give me i2 equal to negative 2 amperes i'm just brushing through this i'm pretty sure you can do this whole by yourself and again if you plug this back in you will get the value of i1 and you'll get i1 as negative 6 amperes so this is just algebra and there we have it we have solved our problem let me just get rid of this again if you need some time on this pause the video all right but that's pretty much it so we have solved our problem and now finally just to make sense of these numbers you might be wondering what does this negative sign mean well the negative sign is basically saying hey we chose the wrong direction when we assumed current to be flowing this way i1 really flows like this but it doesn't matter we got it it means that the current over here is 6 amperes flowing down i2 is also negative meaning a we got this wrong as well so this has to be 2 amperes downwards and the current over here is just i1 plus i2 so the current over here will be negative 8 i'm not writing that down and that again means that the current in this branch has to be going in the opposite direction 8 amperes this way and so the final thing is that you know you know may be asked further questions for example they could ask you what is the potential difference between let's say point a and point b how do you calculate that well you can use the same idea you can use the same krishoff's law but instead of walking through the entire loop you start from point a and you end at point b so i'm not gonna do that i'm pretty sure you can work this out yourself but you go walk you start over here va you walk walk walk walk like this and you'll get va plus or minus something and then you stop here so that will equal vb and then you do the algebra and find out what va minus vb would be and that'll be the potential difference between these two points and because you can choose any path you want you can walk like this and reach here you can walk like this and reach here in all cases va minus vb has to be the same great idea to confirm that