 Hello everyone, in this video we will be discussing how to solve for a Thevenin's equivalent circuit where a linear, okay so now let's have a look at how we can identify Thevenin's equivalent circuit for a linear network where dependent and independent both the sources are available. So let's have a look at a simple problem where we have a small network which has independent as well as dependent sources both. So as you can see we have a dependent source and this dependent source is a current source having a value of vx by 4000, okay. So as you can see this is a current source whose unit is 1 by 4000 and which is dependent on vx and since we are having an arrow inside this diamond that's a clear indication that this is a dependent current source. So to be more specific we can say like this is a voltage dependent current source. So now whenever we have dependent sources we need to be very careful and identify the quantity on which this particular dependent source is dependent on. So since this current since this voltage dependent current source is dependent on vx we need to first of all identify where this vx is in the given circuit. So that will definitely make the problem a little bit easy once you hit that unknown part of the dependency. So now let me complete the rest of the circuit. Here we have the same a and b points similar to the previous problem. These values are 2k and 3k and of course we have a dependent source whose value is mentioned as 4 volts. So let's consider the circuit and try to identify the Thevenin's equivalent circuit since it is having both dependent and independent sources and it is an important point here to hear for the students to identify. You can pause this video and identify like how you would be able to solve this problem and what things makes this easy like whether if you are going for KCL it will be easy or if we go for KBL this will be easy. So here it's a good point for the students to pause for a while and identify applying KBL is better or applying KCL is better. So if you have thought that applying KCL across this particular node called x would be easy for you people then you are definitely right because whenever we are dealing with dependent sources it's always a good way of choosing KCL if you have nodes like this. So depending upon experience it would be becoming used to this. Now let's have a look at the solution. Let me write the question that is find here we are supposed to find Thevenin's equivalent circuit. The way how we look at a problem whenever we are having only independent sources and when and in the case where we have dependent and independent sources is a little bit different. So here the problem is we can't directly move ahead and identify the values of Vth and Rth but we have to follow a little bit of different approach. We call this V open circuit voltage okay no matter whether we are having only independent sources present in the given circuit or it is a combination of independent and dependent we generally tend to identify Voc. Voc is nothing but if there was a resistor given across AB in the form of a load then we would have removed it in first step. The reason why we are removing is we are intentionally removing it so that we can identify the open circuit voltage across the given points. So anyways we are not having a load connected across these points A and B so we are good to go with identification of Voc which is nothing but Thevenin's voltage this is just an alternative name for Vth. So let's proceed and identify I'm not redrawing the diagram because we don't have this load if there was a load then we would have removed it and redrawn it so since it is given readily for us I'm not going to redraw the diagram otherwise we would have drawn it. So I will be simply applying KCL by considering these things across point Vx. So the value we are having is Vth minus 4 okay I'm writing the equation about this branch Vth minus 4 by 2k after this we have Vx by 4000 note this part Voc is nothing but Vx is nothing but Vth so I will straight away replace this Vx with Vth is equal to 0 we don't have any other branch and this is floating in air so we won't be considering that particular part. So on a little bit of simplification what I have is 1 by 2k minus 1 by 4k is equal to 4 by 2k okay so when we are trying to identify or write any sort of equation for a given circuit it is an important note that you have to make that is always try to keep the conventions same okay so if we consider this Vth as Vx here and if we consider the same thing as Vth here then that may unnecessarily confuse as if we are calculating two unknowns okay so this part is very important since this is floating and this output part is mentioned as Voc that is what we are trying to identify and that is nothing but Vth and since we are having Vx mentioned in the question itself all are one and the same and hence since we are about to identify Vth I have kept the notation of Vth everywhere throughout the KCL expressions. Now once we try to identify this Vth let me further simplify this part that is Vth I think this will be 2 minus 1 upon 4k is equal to 4 by 2k okay so this brings me to the value of Vth as okay I am getting 8 volts this closes my step number 1 now let us move ahead to calculate step number 2 where we are supposed to calculate the short circuit current okay so this is the basic difference between when we have a combination of dependent and independent sources we cannot directly calculate Vth and Rth so we try to calculate V open circuit and then we try to calculate I short circuit by replacing the output load and then we try to apply Ohm's law that is Rth is equal to Voc upon Isc that is how we tend to calculate Rth so let us have a look at how we can identify Isc for this I need to short circuit the output by keeping the load intact so this is what we call as Isc which stands for short circuit current so here again we will be keeping this independent source as it is and of course the load will be connected back into the picture so the values are 2k and 3k okay and this will still remain Vx by 4000 okay now when we are trying to calculate so when we are trying to calculate Isc let us simply apply the short circuit current that is okay so if you look at this particular diagram very keenly you will understand that the entire current is going to be flowing only through this Isc short circuit path means whatever current is so if you look at the circuit keenly you will identify that whatever current that this 4 volt is generating the entire current will be moving ahead there will be no passage of the current through this one okay so whatever current is being generated by this dependent source that is only because of this independent source and whatever current that this independent source is generating that will be flowing through this short circuit as we all are aware of the fact that current always flows through the least resistive path so this becomes redundant so let me eliminate this part okay so this is what the equivalent circuit is for the calculation of Isc so now let us write the loop equation considering this loop where the current flowing through this loop in a clockwise direction is going to be Isc so let me write the equation for this so let me write the case equation for this that is I have 4 minus I have 2k plus 3k it will be 5k into Isc so just keep an eye on the notation so here I am just mentioning it as Isc in the lower case both are one in the same and this will be equals to 0 so on simplification the value that I am getting is 4 by 5 4 by 5 milliamps okay this will help us in calculation of Rdh which is a part of over step 2 that is VoC upon Isc okay so the value of VoC is nothing but 8 volts upon what we have Isc as 4 by 5 so on calculating this what I am getting is 10k so the value of Thevenin's resistance for the given circuit is 10k this completes our step number 2 now the final step is going to be so step number 3 is going to be drawing the equivalent circuit that is we have Rdh whose value is 10k and the value of Thevenin's voltage that is 8 volts so this closes our problem here are the references thank you