 Dear members, welcome you again to one of the experiment based on verification of some parameters. This experiment is said to be verification of KCL. You are very familiar with the rule Kichop's current law, which is very much resembling with the parameter or you can say with the theory called current user rule. If you know one, you can easily do the other because circuit diagram is also same for both. Let us look into the circuit diagram of the KCL. This is the circuit diagram we have in the KCL. Here you can see we have power supply. We have one multimeter which is used as an emitter. We have another multimeter which is used as an emitter. Again one more multimeter is used as an emitter and one more multimeter is used as an emitter. We have resistances R1, R2, R3 of values 3.2 kilo ohm, 2.2 kilo ohm and 5.6 kilo ohm. So this is how we see the circuit diagram to verify the current law. So this circuit which we have discussed now has been seen over here on the platform. This is the circuit diagram. You can see this part is going up to worry about because it is for another experiment. So you see concentrated over here. These are three resistances as we have discussed in the circuit diagram. Those are 3.3, 2.2 and 5.6. We have connected this with a source. Now there are four different multimeters. Now here also I will show you. See this is one multimeter. This will calculate the total current in the circuit. This is the second multimeter that is going to calculate the current through the second branch or you can say through the resistance of 3.3 kilo ohm. This is the third multimeter which is used as an emitter. It is going to calculate the current through the third branch or through 2.2 kilo ohm resistance. This is the fourth multimeter which is going to be calculating the current through the last resistance that is 5.6 kilo ohm. And to supply, to give supply to this circuit we are using a dual bar supply. We are using one of the more of the dual bar supply to give supply off. We are using 12 volts. So here from this 12 volts supply we will give 12 volts to the circuit. Let us on the circuit. This is the one. Here you can see it is around 11.9 let us make it exactly 12. It is 12.0 volts. This is the voltage we are giving to the circuit. Due to this voltage there will be obviously flow of current. Now that flow of current is first. What is the total flow of current is being measured by this voltage. So you are seeing here the total is how much I have kept to measure again. I am reminding you back. To measure the current we have to keep this volt from half to this range of 8 meter. You can see this is 8 meter. Here are various ranges. 200 micro ohm milli, 20 milli and 200 milli. I am keeping it at 20 milli. And it is giving a value that will be stable. It is giving around 11.09 if what? Or you can say 11.1 which is in the range of milli. So this is 11.1 milli ampere. Similarly, this is the total. Let us go to the circuit. I have made it manually. I will show you. In that circuit everything was connected. Not many multimeter and all were connected. I have simplified it to a small circuit. This is the voltage source. This is the first branch, second branch, third branch. Now what is case here? Case here says that in a node it is the sum of all the incoming currents and the outgoing currents. Here if we say this entire is one single node then incoming current is only i and all these currents are outgoing. So I am writing what? Incoming current that is i is equal to i1 plus i2 plus i3 that is sum of all the outgoing currents. Which in a node sum of all the incoming currents is equal to sum of all the outgoing currents. So this is what case here. Now we have to verify what? That this current should be equal to the sum of all. This we can verify practically using the circuit. Let us see in the circuit in the first reading we are going to calculate this i. Now we are going to come to this first emitter. What is the total current? We are saying it is 11.11. So I am writing this suppose 11.11 in this over here. Here we are writing it to be suppose this i is how much 11.11 milli ampere. So this is equal to be something. Now again go back to the emitter. First second emitter. In this second emitter I am using the same mode at 220 milli ampere. And I am getting my second i1 current as to be 3.54. What? Unique milli ampere. So again from here I am writing in the i2, in this i1 is equal to how much? 3.54 milli ampere. This is what I have written. 3.54. First current in the branch of 3.3 kilo. Again coming back to the second branch. That is through R2. What is the amount of current flowing? It is showing 5.38 milli ampere is the amount of current flowing. So again that thing we will write down here. 5.38 milli ampere plus. So we will return total is equal to first plus second and plus third. So third I am not going to show you over there. I am directly writing by saying it is 2.13. So I am writing as 2.13 milli ampere. So if I sum up this 3.54, 5.38 and 2.13. I am total getting what? 5 plus 3 is almost 8, 9, 10. Then this 0.5, 0.3 become 0.8. Then 8, 4, 0.9, 11.9 is again coming 11.1. So you are getting what? 11.1 as your incoming current. 1 milli ampere and you are getting your outgoing current is also equal to 11.1. So this proves your pitch of flow. All incoming current is equal to all your outgoing current. And how this again resembles to current deviation rule? You can say in the current deviation rule what is happening here in this circuit? This is total current coming. The current is divided here, here and here. So now your total current has to be divided into this, this and this. So what will be your I1? What will be I2? What will be your I3? So this I is equal to I1 plus I2 plus I3. This is how your KCL is verified. All incoming current that is only one I is equal to all outgoing current that is I1 plus I2 plus I3. That we have used this circuit using this multimeters and all we have verified it. So thank you for watching this video.