hello everyone just a quick recap of the kirchoff's current law using nodal analysis now killed off skyrim clone or KCl is a very popular tool used in circuit analysis what it basically says is that the algebraic sum of currents into a node is zero so what this means is that given a node stay a that has several branches connected to it each of these are also nodes say this is r1 r2 r3 and r4 let the currents flowing in each of these branches be i1 i2 i3 and i-4 what KCl says is that the algebraic sum of these currents that are entering a node is zero by algebraic we mean that we are taking into account the direction of the flow of current by considering a particular sign associated with that direction so as a convention let plus be the sign denoting the current that is entering a node okay after the negative sign denote the current exiting the node now if the algebraic sum of currents into a node is to be zero this basically implies that considering the direction of currents this summation should be zero okay so I 1 is flowing into the node and we take a plus reference for entering the node so we write I 1 I 2 is exiting the node so minus I 2 I 3 is also exiting the node so minus I 3 I 4 is entering the node so plus I for this is equal to zero this is another way of saying that I 1 plus I 4 is equal to I 2 plus I 3 that means that currents or rather summation of current entering a node is equal to the summation of current exiting the node this follows very simply from conservation of charge that whatever current is entering the given node the same must be exiting that node okay now KCl can very simply be implemented using nodal analysis nodal analysis basically refers to finding these currents using node voltages all right so if node voltage say this is node a and I associate a voltage V a to this node these are other nodes let me associate say v1 v2 v3 and v4 these are the node voltages now one must understand that node voltages are measure three with respect to ground okay and not with respect to each other a node voltage refers to the voltage of that particular node with respect to ground now for us to be able to write currents in terms of node voltages it's important to understand that when we say that there is a current flowing through a resistor in this direction I we assumed that the terminal from which current was entering is at a higher potential than the terminal through which the current was exiting that is why we say there is a potential drop across this resistor based on what is the direction that you have assumed for a particular branch you need to write the current equation accordingly so what does that tell us I'll rewrite the expression of what KCl basically states what it said was current entering the node should be equal to current exiting the node so current entering node a on i1 and i-4 and current exiting this node is i2 and i3 what is I 1 I 1 is the current flowing in this direction that is what our assumption is since current flows from a node of higher potential to a node of lower potential we can say the current is basically the drop across this potential upon the resistance the drop across this potential because it's flowing in this direction we would assume that v1 is at a higher potential than VA and then I 1 therefore becomes v1 minus 3ei upon r1 similarly I 2 this is the direction of I 2 flowing from VA to v2 so i2 becomes V ay minus v2 because the direction is from a to node 2 upon R 2 similarly I 3 is in this direction flowing from a to another node 3 so I 3 is VA minus v3 upon R 3 and in the same way our I 4 flowing in this direction refers to our assumption that V 4 would be at a higher potential than VA so I 4 is V 4 minus VA upon artwork now when you plug in these expressions of current into the actual KCl and solve it in terms of node voltages we may in the end find that the direction of current that we originally assumed is not actually the correct direction of current because then that value would come out to be negative let us just demonstrate that using a very simple example now the question is if I want to use nodal analysis I need to assign node voltages to each node so let me assign this node a node voltage V eight which is unknown and this is the other node say VB and since this is grounded VB is equal to 0 let me assume some direction of current say this is i1 and this is purely an assumption to start with so let this be I to let this be i3 KCl says that summation of I into a node should be equal to 0 or the current entering a node should be equal to the current exiting the node what does this tell me that what is the current entering the node I 1 is entering I 2 is entering I 3 is also entering and nothing is exiting so this is equal to 0 how can I write each of I 1 and I 2 I 1 is flowing from node B to node V a via the 2 ohm resistor so I 1 can be written as 32 minus VA upon 2 this is because I initially assumed that the direction of i1 is this I 2 in the same way is flowing from node B to VA my resistor a to node B is that a voltage V 0 so 0 minus VA upon 8 because this is my assumed direction of current i3 is in this direction so I can write 20 - VA because the positive terminal of 20 volt voltage sources actually towards node V so 20 minus VA upon for this is what is I 3 if I solve this if I plug this into this equation 1 I would get 32 minus VA upon 2 plus minus VA account 8 plus 20 minus VA upon 4 this is equal to 0 if I solve this further should get eight is the LCM I will get 4 into 32 minus VA minus V ay plus 2 into 20 minus VA is equal to zero solving it further would mean 168 is equal to 7 VA which implies that V a would come out to be 24 volts but this is not what was asked in the question this is what is going to help us find each of these currents because we just defined each of these currents in terms of the node voltage VA right so once we know VA all we have to do is just plug VA into these expressions and we will have the values of a current that gives us a 1 is 32 minus 24 upon 2 which is 4 ampere I 2 is minus 24 upon 8 which is minus 3 ampere and I 3 is 20 minus 24 upon 4 just minus 1 ampere what this basically tells us is that when the sign of the current comes out to be negative it means that the direction that we initially assumed that is not the direction in which the current is flowing is flowing in the opposite direction so if I was to rewrite what are the actual currents in this circuit I could write I 1 that is correct this is the direction and the current that flows is four amperes I too came out to be minus three that means I two does not really flow in this direction it flows downwards and has a value 3 ampere and I three does not really flow in this direction it flows in the opposite direction and has a magnitude one ampere this is how you use nodal analysis and KCl to find currents through a given circuit now in this question we only had voltage sources if we were to have current sources for instance how are we going to solve this question let us see using another very simple example so the question is calculate the current through the 8 ohm resistor which I am referring to as i8 on a dome is in subscript to show what is the current I through in ohm resistor the circuit is this this is a 2 ohm resistor a 4 ohm resistor an 8 ohm resistor you've got a voltage source of 20 volts on this time and you've got a current source of 1 ampere it's an ideal current source of 1 ampere on the side this side is ground this node is grounded calculate I 8 if I say using nodal analysis if you wanted to use Kirchhoff's voltage law it would be difficult at first to contemplate what is the voltage drop across this current source because it is an ideal current source the basic definition of a current source is that it should provide you with the constant current irrespective of what is the drop across it that means the drop across it can be anything so a viable way of proceeding would be to use nodal analysis and KCl in such questions as before I define this node as a having a nodal voltage V and this is this node is grounded so it has not voltage zero right I assume some direction of current say this is i1 this is I 2 and this is I 3 now one must understand that assuming the direction of current in the beginning is up to us all right it is after solving will come from whether the direction was correct or it was actually the opposite direction now I could as well choose this direction as I 3 in this direction okay in that case what could I write I would write KCl as i1 plus i2 is equal to I 3 so it completely depends on how you are defining how you are assuming the direction of current in the beginning so suppose this is the direction of currents I assume so I can write I 1 plus I 2 is equal to I 3 by KCl at node a I won is one ampere because we have an ideal current source that provides one ampere current system resistor is in series with it with this current source it's the same amount of current that flows through this resistor so the current in this branch is basically 1 ampere what is I 2 I 2 is the current in this branch since it is flowing from this side to this side I am assuming that 20 volt is higher than VA and I write this expression this I 2 as 20 minus VA upon for this should be equal to I 3 I 3 is flowing from node a to the ground so I write it as VA minus 0 upon 8 solving this I will get VA is equal to 16 volt and please cross-check this with VA is equal to 16 volt the current through the 8 ohm resistor which is actually I 3 is simply VA minus 0 upon it that means 16 upon 8 that is 2 ampere this is the answer this is how you use KCl in circuits that have current sources as well we'll continue this discussion with more questions thank you for now