 Hello everyone in this video we will be discussing about how to solve for a Norton's equivalent circuit for a given linear network where only independent sources are operating. Here is the learning outcome by the end of this session students will be able to identify various parameters of the given linear network and will be successfully able to draw the Norton's equivalent circuit with only independent sources. Okay, so now let us have a look at a circuit where we will be identifying the Norton's equivalent circuit and also we will be identifying the current through 30 ohm resistor that is the load in our case by using Norton's theorem. So let us have a look at the circuit. We are having four resistors connected in the form of two adjacent meshes where the load mentioned is of 30 ohms. So let us have a look at this particular circuit. This is 10 volts. We have a couple of resistors connected in the way as shown in the circuit. These are the two points mentioned and of course the load is connected across these two points A and B where the value mentioned of the load is where the value of the load mentioned is 30 ohms. So again whenever we are trying to identify a Norton's equivalent circuit again we have the same possible ways either it has only independent source or it has a combination of dependent and independent source or it only has a dependent source. So currently if you observe at the circuit we only have independent source whose value is mentioned as 10 volts. So let us have a look at this. Similar to Thevenin's theorem we are going to solve this in three steps. So the first step is going to be the calculation of IC and then in step number 2 we are going to calculate the value of Rn and then in step number 3 we are going to draw the final equivalent circuit of Norton. So let me again make a simple template which will ease us in proceeding towards the solution of the problem easily. So let me first write the template. So again as we need for the Thevenin's theorem let me first of all design a template for this one and then we will proceed with solution of the problem. So we basically have three steps. In first step let us identify the value of IC. So I am using either the upper case or the lower case. So both are one and the same. So this is going to be the task for the first one. So for this we have to first of all disconnect the load which is similar to Thevenin's theorem type 2 where there was a combination of independent and dependent sources. So for this we need to first of all redraw the diagram and short circuit the terminals a and b by replacing the load of 30 ohms which is mentioned in the given problem. And of course we need to connect the source and show the short circuit which we refer as the short circuit current. The current flowing through the short circuit is going to be the short circuit current. Here we have 5 ohms again we here we have 10 ohms and 20 ohms 10 volt is connected back. Similarly let me first of all draw the layout of step number 2 where we are going to find. So in step number 2 we are going to find the equivalent resistance for this theorem. So for Norton's based theorems we tend to call that equivalent resistance by removing the load as Rn which stands for the Norton's resistance. So we are going to find the value of Rn in step number 2 and finally in step number 3 we are going to draw the equivalent circuit which is nothing but Norton's equivalent circuit. So now coming back to the step number 1 before that let me first of all draw the circuit for this. For step number 2 what we are supposed to do is let the load remain removed here we are going to deactivate the source by short circuiting a voltage source and open circuiting a current source. And we generally try to replace the voltage source with its own internal resistance which is 0 ohms and if this would have been a current source in place of a voltage source then we would have simply open circuited that because for a given ideal independent current source the internal resistance is considered to be infinite. So here we have replaced that voltage source with a short circuit the values remain the same. So whatever resistance that we tend to calculate by looking back into the circuit across these points A and B this is known as Rn. Now let us proceed to solve the problem by going back to step number 1. So in step number 1 as we have short circuited this one the points A, B are short circuited to find the value of Iac. So let us move ahead and calculate the value of so let me call this point as X and the voltage drop across this point is going to be Vx. So we are going to do this by applying Kcl. So let me apply Kcl at node X and identify the value of Iac. So let us do that. So here we are having the first term as Vx minus 10 the first branch upon 5 plus we have Vx by 10 I mean this branch and then finally we have Vx by 20 ohms and on RHS we are having 0. So let me simplify this. So I have taken this Vx as a common term and then what we have here is minus 10 by 5 I am pushing that on the RHS so on simplification. So after simplifying I am getting 7 by 20 and on RHS I am getting it as 2. So I can now further simplify and get the value of Vx that is 40 by 7 volts ok but I am not interested in the value of Vx alone but I am interested in calculation of Iac. So solving further I can easily calculate the value of Iac which is nothing but Vx upon the value of 20 I mean the value of resistance that is 20. So the value I am getting after putting this 40 by 7 by 20 is 2 by 7. So step number 1 to calculate the value of Iac is accomplished. Now let us move to step number 2 where we have eliminated the voltage source by replacing it with its own ideal internal resistance that is RHS is equal to 0 and hence we show that with the help of a short circuit like this and then we are going to look back into the circuit to identify the value of Rn. So the value of Rn can be easily calculated by looking at the circuit it is pretty clear that we have 5 ohms in parallel with 10 which is in series with 20 ohms. So let me calculate this I am having 5 in parallel with 10 in series with 20. So the value that I am getting is 70 by 3 ohms and this is the value that we have got as a result of step number 2. Now the final step is to draw the Norton's equivalent circuit. So Norton's equivalent circuit is basically represented by drawing current source in parallel with its own Norton's resistance and this is how it is drawn. This is the Norton's current which will be shown in parallel with Rn and of course in parallel to that whatever load value was mentioned earlier which we have eliminated in step number 1 would be connected back. So here we have a resistor of 30 ohms which is nothing but our Rl. Norton's value is 70 by 3 and the Norton's current is 2 by 7 amps. So this is our Norton's equivalent circuit as shown here but that does not end up a problem. Our problem was to calculate the current flowing through the 30 ohm resistance using the Norton's theorem or by applying Norton's theorem. So here what we have is we have 2 by 7 amps of current flowing like this. So the final so here we have 2 by 7 amps of current flowing like this that is going to split into these two branches. So let us apply the current division rule and identify the current flowing through 30 ohm resistance. So this is going to be the current flowing through 30 ohm resistance. So this is going to be the current flowing through 30 ohm resistance. The incoming current I am simply trying to apply the current division rule which says the incoming current into the opposite branch resistance that is going to be 70 upon 3 upon the desired resistance that is 30 plus the opposite branch resistance. So let me check how much value I am getting on calculation. So the value I am getting is 1 by 8 amperes. So this ends the problem. We have successfully calculated the current flowing through this resistance that comes to be 1 by 8 amperes by applying Norton's theorem. Here are the references. Thank you.