 Hello everyone, welcome to a video lecture on Kirchhoff's Laws. Myself, K. R. Biradar, Assistant Professor, Department of Electronics and Telecommunication Engineering, Walchin Institute of Technology, Sulapur. Let us start with the learning outcomes first. At the end of this session, you will be able to demonstrate how Kirchhoff's Laws can be used to find voltage and currents in a circuit or a network. Introduction. Network elements can be either active or passive type. Network elements may include like registers, capacitors, inductors, etcetera. Even sometimes voltage sources and current sources. Any electrical circuit or network contains one of these two types of network. Elements are a combination of both. Kirchhoff is a scientist who gave two laws to identify currents and voltages in a network. These are Kirchhoff's current law which is also called KCL. Second one, Kirchhoff's voltage law which is also called KVL. Kirchhoff's current law or KCL, Kirchhoff's current law states that the algebraic sum of currents leaving or entering a node is equal to 0. Mathematically, it can be expressed as summation small m equal to 1 to capital M I suffix small m which is equal to 0 where I suffix small m is the mth branch current leaving the node. Where capital M is the number of branches that are connected to a node. For example, we can see this figure there is a node P and currents I1, I2, I3 and I4 are connected to the node either it may enter to the node or leaving the node. In the figure the branch currents I1 and I2, I3 are entering at node P. So, we need to consider negative signs for these currents. Similarly, the branch currents I4 and I5 are leaving from the node P. So, these are considered as positive signs. Therefore, when you apply the Kirchhoff's current law equation at node P which will be I1, I2, I3 are entering to the node those are considered negative sign. Therefore, minus I1 minus I2 and minus I3 similarly I4 and I5 are leaving the node we need to consider positive sign therefore, plus I4 plus I5 which is equal to 0. If I simplify this becomes I1 plus I2 plus I3 equal to I4 plus I5. Kirchhoff's voltage law which is a second law which states that the algebraic sum of voltages around a loop or mess is equal to 0. A loop is a path that terminates at the same node where it started from. In contrast a mess is a loop that does not contain any other loop inside it. Let us verify the KVL with the help of example in the next slide. Sign conventions to be followed while applying the Kirchhoff's voltage law. When current flows through a resistance the voltage drop occurs across the resistance. The polarity of this voltage drop always depends on direction of the current. The current always flows from higher potential to lower potential. Consider the figure A the current flows from B to A therefore, B has higher potential than A. Similarly, in the figure B the current flows from A to B therefore, A has higher potential than B. In short we can say in resistors current will flow from plus to minus. The circuit diagram consists of a voltage source Vs in series with two resistors R1 and R2. The voltage drop across the resistors R1 and R2 are V1 and V2 respectively. Now you apply the KVL equation around the loop of the above circuit Vs is equal to because Vs is from minus to plus that is why you need to consider plus sign whereas, V1 is from plus to minus goes from higher potential to lower potential we need to consider minus V1. Similarly V2 is across R2 is from plus to minus that is from higher potential to lower potential we need to consider minus V2 is equal to 0. If you simplify this Vs is equal to V1 plus V2. Let us see one problem determine the current in all resistors for the circuit shown. There are three resistors one 2 ohm 1 ohm and 5 ohm which are connected in parallel having a current source of 50 ampere. We need to find the currents flowing across R1 which is I1 and across R2 which is I2 and across R3 which is I3. Now apply KCL we have a node. So, the current is flowing from this current source in this direction and it divides as I1, I2 and I3. The total current coming from this is equal to summation of these three currents I1 plus I2 plus I3. But according to Ohm's law I1 equal to V by 2 that is I equal to V by R here R equal to 2 ohm. Similarly I2 equal to V by R2 that is R2 is equal to 1 ohm. Similarly I3 equal to V by R3 R3 equal to 5 ohm. If I substitute I1, I2, I3 in the I equation therefore I becomes equal to V by 2 plus V by 1 plus V by 5. Therefore, I which is given is 50 ampere which is equal to V you need to take common V into 1 by 2 plus 1 by 1 plus 1 by 5. So, this is equal to V into 1.7. If I simplify further V becomes equal to 29.41 volt this is a voltage across all the resistors. Once we know the voltage we can find currents in each element I1 equal to V by R1. So, we already we have calculated 29.41. Divide by R1 is 2 ohm which is 14.705 ampere. Similarly, I2 equal to V by R2 which is equal to 29.41 divided by 1 ohm which comes out to be 29.41 ampere. Similarly, I3 equal to V by R3 V given 29.41 divided by R3 is 5 ohm which is equal to 5.88 ampere. What is the relation between currents in the figure below? We can see this circuit diagram having a single node and there are 5 currents I1, I2, I3, I4 and I5 are either entered to this node or leaving to this node. So, I1, I3, I4 and I5 are entered to this node and I2 is leaving this node. We need to find the relationship between the various currents. The options are first one I2 equal to I1 plus I3 plus I4 plus I5, second one I2 minus I1 equal to I3 minus I4 plus I5, third one I3 plus I4 equal to I1 plus I2 plus I5, fourth one I1 plus I5 equal to I2 plus I3 plus I4. Pass the video and try to think and try to find the answer. I hope you might have got the answer. The answer for this question is only one current is leaving all the currents are entered to this node. Therefore, I2 can take one side and remaining currents in the other side. Therefore, I2 equal to I1 plus I3 plus I4 plus I5 answer first one is right. These are the references used to prepare the above PPT. Thank you.