 Hello everyone. In this video, we will be discussing how to extract Norton's Equivalent Circuit where a linear circuit is operated by the presence of independent as well as dependent sources. Here is the learning outcome. By the end of this session, students would be able to draw a Norton's Equivalent Circuit provided that there are independent and dependent sources both available in the given linear network. Okay, so let's have a look at another problem where we have the presence of an independent voltage source as well as a dependent source. So if you observe properly, this dependent source is a voltage source and that is dependent on 2000 I mean that is dependent on some factor multiplied by a current that is IA and IA is nothing but the first loop current. So to be more specific and more precise, I would say like this is a current dependent voltage source which is dependent on an independent source whose value is 12 volts. So once you identify this part, we can track down the question like we are about to find the current IB which is nothing but the current flowing through the load. So this is going to be the IB. So in cases where there are linear circuits and we are about to apply some Norton's kind of theorem, then we have to first of all deactivate the load that is this one 667 ohms and then short circuit that and identify I short circuit. So the steps that we need to follow for the calculation of IB or the Norton's Equivalent First of All is to eliminate the load like what we have already done for the presence of independent and dependent source based Thevenin's based problems. So whether we are trying to calculate a Thevenin's Equivalent or Norton's Equivalent wherever we have a linear circuit where there is a presence of independent as well as dependent sources, we need to first of all eliminate the load and then identify the value of current IAC and then identify the value of VOC by open circuiting the terminals and then we are going to simply apply ohms law for the calculation of RTH that is RTH can be equated with VOC upon IC and finally identify the value of RTH. So let us have a look at how to proceed with this problem. First steps first so let us have a look at the solution first. Step number one is to calculate the short circuit current so for that we need to redraw the diagram by eliminating this particular load. So let me quickly draw that so we have so we have 6000 ohms here so we have 6000 ohms here then we also have this dependent source then we have one more resistor and then we also have this independent voltage source active but as soon as we try to short circuit this particular path by eliminating the 667 resistance then definitely this resistor becomes redundant. I mean whatever current that is passing from this entire source is going to pass through the short circuit again it would be a good moment for the students to pass the video and identify why this becomes redundant. So if you have identified it correctly like current since current has the property of flowing always through the least resistive path there would be zero amps of current flowing through this one and hence whether this is a part of the circuit or not this becomes redundant since there is not going to be any sort of current passes through this resistor and hence there would be zero volts of voltage drop across this resistor. So this circuit can further be simplified and redrawn like this so since there is a short circuit here the voltage drop across this is going to be zero and the current flowing through this path is also going to be zero which was mentioned to be IA and if you look in more detailed weave and if you look and if you have a more detailed look then you will identify that we have this dependent source values 2000 times IA. So if this IA becomes zero then this entire dependent source is also going to be eliminated and hence we need to redraw the diagram once again where we have only this independent 12 volt source in series with 6000 ohms dependent source is gone and the 1000 ohm resistor is also gone and we have already replaced the load with a short circuit and hence now we are only supposed to calculate this short circuit current flowing through a simple Thevenin's equivalent. This is as good as a Thevenin's equivalent hence we are having 12 volts and 6000 ohms. Therefore, we have a direct formula here by simply applying the Ohm's law that is ISC is equal to 12 volts upon 6000 this comes to 2 milliamps. This completes our first step of calculation of ISC further let us move to step number 2 where we are going to calculate. So in step number 2 we are going to calculate VoC that is open circuit voltage. So to find open circuit voltage again we have to redraw the circuit and let us identify how the things work. So we have this 12 volt source we also have this 6000 ohm resistance we have this dependent source that is 2000 times IA and IA is flowing through the 1000 ohm resistance and of course this time we need to leave these points X and Y open and the voltage that we are going to calculate across this is going to be VoC. We can calculate the value of VoC. So let us identify the value of VoC by applying KVL. So what I am having here is we have 12 volts minus 6000 times IA minus 2000 times IA and of course one more element we are having that is 1000 times IA and on the RHS we are going to have a 0. So on simplification so on simplifying what I have here is 12 6000 times IA minus 2000 times IA. So on simplifying the value that I am getting here is 9000 times IA is simply equals to 12. So the value of IA is nothing but 12 by 9000 here which can be calculated which can be simplified further to 4 upon 3000 amps. So this is our value of IA that will give us the value for calculation of VoC. So just by observing this part of the circuit we can easily calculate the value of VoC as 1000 times IA. Therefore by substituting the value of IA into this one we get the value of VoC as 1000 times 4 by 3000 which brings me to the value of VoC is equal to 4 by 3 volts. So this closes our step 2. Now the final step is to draw the Norton's equivalent circuit and find the value of Rn. So step 3 is to draw the Norton's equivalent by calculating Rn. Here the value of Rn is going to be simply VoC by IAC which we have calculated in our previous two steps. So the value of VoC is going to be 4 by 3 and the value of IAC is going to be 2 milli amperes. So the value of Rn that I am getting is 667 ohms okay. So coincidentally as you can observe that so coincidentally as you can observe we have the value of Rn and the value of Rn that we have calculated as the same. So this doesn't end up the problem. We were supposed to calculate the current IB which was nothing but the current flowing through the load. Now let us redraw the Norton's equivalent circuit that is the Norton's current whose value is 2 milli ampere and then it is connected in parallel with the Norton's resistance that is 667 ohms. This is our Norton's equivalent circuit. Now I am going to connect the load back so this load will be connected back to the Norton's equivalent circuit which was also coincidentally identified to be 667 ohms only. Now this 2 ampere is going to flow through two similar resistances that is 667 and 667. So even without applying any sort of current division rule we can easily identify that one ampere is going to flow through this and one ampere is going to flow through this. So I am not going to apply any sort of current division rule. I am going to directly write like the value of IB as mentioned in the problem here and this is going to be the value of IB. The value of IB is going to be 1 milli ampere. So this is how we generally focus on the real part of the problem like what we want to identify. The only thing that comes into picture in the meanwhile unlike our mesh and nodal analysis problems is we need to focus on first of all calculating the Thevenin's or Norton's equivalent. After that we need to focus on the real desired part of the problem and that is how we end up with a solution. Here are the references. Thank you.