 Hello everyone. In this video we will discuss how to solve a linear circuit which is sourced by a DC source having only independent sources. Here is the learning outcome by the end of this session students will be able to identify the various aspects of the circuit to get the Thevenin's equivalent circuit when only independent sources are available. So, let us have a look at the problem and analyze how to solve this to finally, identify the Thevenin's equivalent and also to find the current flowing through the resistor R equals to. So, here is the question. So, where let us say that this voltage source is of a value 50 volts whereas, these values are of 5 ohms 20 and 4 and this resistance is nothing but the load resistance across which we are supposed to calculate the value that is 2 ohms. So, here is the problem. We are supposed to identify the Thevenin's equivalent circuit across this particular point I mean this particular network. So, for that let us have a look at how to solve this. So, basically I would say like we have three steps and the first step is to identify the Thevenin's voltage then we calculate the Thevenin's resistance and then finally in step three we tend to draw that Thevenin's equivalent circuit and identify the rest of the what we call the requirements like maybe the current across the 2 ohm resistor or maybe the voltage across whatever is desired in the question. So, let us have a look at the step one where so, I am going to roughly prepare a template that would ease us to identify how we are going to proceed for solution of this problem. So, let me lay down the template. So, step number one is going to identify Vth. Similarly, we are going to identify the value of Rth which stands for the Thevenin's resistance and then finally we have step number three which involves drawing the equivalent circuit that is the Thevenin's equivalent circuit ok. And finally we are going to find the value of Rl and the current that is actually flowing through that. So, let me write the KVL equation for identification of Vth following step number one where we will be writing the KVL for this loop. So, what we have here is we are going to first of all disconnect the load. So, this is very important step the first step that you are going to follow is disconnect this part ok. So, step one involves disconnecting the load and then redrawing the diagram. So, let me redraw the diagram by disconnect the load. So, this is a important step that will ease you to identify what exactly we are performing and what part of the circuit is important and what part of the circuit can be neglected for the moment we are trying to analyze the circuit. So, here we have 50 volts this is going to be 20 and this is going to be 4 ohms ok. Remember that we have eliminated this and whatever voltage that will be developed across this particular part is going to be referred as Thevenin's voltage and that is what we will be identifying in a short while. Now, for this we have to apply KVL. So, I am going to assume a current I flowing through this and hence apply KVL which will give me 50 minus 25 times I is equal to 0 and thus on simplification this will give me 50 by 25 that is 2 amps ok. So, but this is not what we are interested in. We are interested in the voltage developed across this one. Now, the voltage developed across the output port is going to be merrily the voltage developed across 20 volts plus the voltage developed across the 4 ohms resistance, but as you can see in the given circuit that the voltage developed across 4 ohms is going to be 0 volts. The reason is we do not have any sort of current flowing through this one and hence though Vth is equal to voltage drop across 4 ohms plus 20 ohms. Now, this just remains like Vth is equal to the voltage developed across 20 ohm resistor only since this is 0. Therefore, we can follow that Vth is equal to nothing, but the resistance R into the current flowing that what that is what we have just identified that is 2 amperes. So, I would not mention the unit here. Hence, the value of Vth is that we are able to obtain is 40 volts. So, step number 1 is successfully accomplished. Similarly, we have step number 2 where we are going to eliminate the source and redraw the circuit for the sake of step 2 again. So, here I am having the voltage source. This voltage source is short circuit at this time and of course, RL is still in a removed state. So, whatever you look into the circuit across these points A and B that is known as Thevenin's resistance. So, now I have 5 ohms, 20 ohms and 4 ohms back into the picture whereas, I have simply replaced this particular 50 volt source with a short circuit. So, what generally we are trying to do here is we consider that all the sources present in a given linear circuit are ideal and when we say these voltage sources which are independent are ideal, independent ideal voltage source generally have an ideal resistance of R s which is equals to 0. So, this is as good as I am replacing this voltage source with its own internal resistance which is nothing but 0 volt sorry 0 ohms. So, here is the circuit which will help us to identify the Thevenin's resistance. So, simply as you can see we have Rth value as nothing but 5 ohms which is in parallel with 20. This whole combination is in series with 4 ohms. Therefore, Rth is going to be in just some sort of simplification and finally, what we get here is 8 ohms. So, the Thevenin's resistance that we have calculated is 8 ohms hence we have we have finished calculating the step number 2. Now, we are getting into step number 3 where we are supposed to draw that Thevenin's equivalent circuit. So, let me draw this. This is our Thevenin's equivalent circuit which contains only one single voltage source Vth and one single resistance Rth across the same given points A and B and now we are going to connect the load back. Whatever we have disconnected in the step number 1 the same thing should be connected back again and of course, the value of Rth is going to be 8 ohms and the value of Vth is going to be 40 volts that is what we have calculated in step number 1. Now, let us calculate the real desired part that is we are supposed to calculate the current flowing through the resistor Rl. So, here is the solution for that the current flowing through Rl let it be I once again. So, the current I flowing in the given loop of Thevenin's equivalent is going to be simply Vth upon the value Rth plus Rl. So, this brings us to the value of 40 upon 8 plus 2 that is 10. So, simply this is nothing but 4 amps ok. So, the final answer what we are having is 4 amperes. This is how we calculate Thevenin's equivalent circuit. Here are the references. Thank you.