Hello guys, welcome back to Last Moment Tutions. I am Dhruvi and today we will study Thevenin's and Norton's theorem. So first let's start with Thevenin's theorem. So it states that any two terminals of a network can be replaced by an equivalent voltage source and an equivalent series resistance.
Now what does this mean? Whenever we have two terminals of a network, we can replace them with two components i.e. voltage and an equivalent series resistance. So, consider this is your network.
So, if this is your network and this is your load resistor RL, current flowing through it is IL. So, if this is your network, you can replace it. You will replace this entire network with a voltage.
So, voltage source and an equivalent series resistance. So, series resistor is here. So we call that to be Vth and Rth and then your RL current flowing through it is Il.
So this was a short explanation that what we are doing is that we are replacing the two terminals of the entire network with a voltage source and a series resistor. That is Vth and Rth. So here our voltage source Vth is the voltage across the two terminals with load if any removed. i.e. if there is any IRL, then we have to calculate the voltage across the circuit by removing it. And same for series resistance, the resistance of the network measured between two terminals with load, again we have to remove the load.
And the voltage source which we considered in the previous step, we have to use its internal resistance i.e. short circuit. Now this may seem a bit confusing. If we go directly to numerical, then it would be easy for you.
So, let's understand step by step in numerical. So, let's move on to it. So, this is our numerical. We need to find the equivalent network for the terminals A and B.
So, now we don't have any load resistance here. If there was RL, then we would have to remove it. But if it is not here, then it is not an issue. So, now our two terminals are A and B. so the voltage across this will be written as Vth because we have to calculate the value of Vth and Rth so the first step is the voltage across the two terminals is Vth fine now this is your circuit so what will you do?
you will apply KVL right? so we have only one mesh here right? that is I1 this mesh is not being made here because these are open terminals so we will write KVL so applying KVL to the mesh So, KVL's equation will be 5 minus 10 times I1 minus 10 times I1 again. Now, this is not a mesh. So, only current flowing through this 10 ohm resistor is I1 is equal to 0. So, 5 minus 20 I1 is equal to 0. 5 is equal to 2. I1 and I1 is equal to 0.25 amperes.
So this was our first step. Now next step is we will write Vth equation. So for Vth this is the loop right.
So what will be our Vth equation? We will start from here and So, I will write the Vth equation here. So, Vth equation will be 5-10I1 again.
Then, here it will be plus terminal so, plus 8I1. Then, here if there is no current flow then, 0 current. So, 0 into anything is 0. And then, we have is Vth. So, minus Vth is equal to 0. So, this is 5-10I1 plus 8I1 minus Vth is equal to 0. And on solving, we will get the value of Vth because we already know the value of I1. So, we will get the value of Vth of 4.5V.
So, again I will repeat. First, we defined Vth. After that, we solved all the meshes.
There is only one mesh being made. So, we solved this and took out the value of I1. After that, we have to calculate the equation of Vth.
So, that is outer loop. So, we wrote the equation of Vth and we calculated the value of Vth. Now, what we have to do is find the current. Because only after finding the current, we will be able to find Rth. Because Rth is Vth by current.
So, for current, we had said in the definition that we have to replace the voltage with equivalent internal resistance. So, we have to replace it with Rth. So, this is what we have to do. short circuit because voltage is equivalent to short circuit. So, the current flow from here, I am naming it as IN current flow.
Now, we have two mesh. Earlier, it was an open terminal, so there was only one mesh. Now, there are two.
So, let us consider the current flow from here, we name it as I2. Now, we have to write the equation again, KVL to mesh 1. So, what will be your KVL to mesh 1? 5 volts is 5, minus 10 I1 because current flowing through this is 10 I1.
Now we will come back here. so that is minus 10 times, now we are considering the first loop so I1 minus I2 is equal to 0 so on solving you get 20 I1 minus 10 I2 is equal to 5 so this is your first equation now similarly we have to apply in KVL mesh 2 that means in this after applying in this, how will be your equation? see this is plus 8 I1 so 8 I1 minus 1 times current flowing through this is I2 so minus 1 times I2 minus 10 times considering this loop so I2 minus I1 is equals to 0 so its equation will be 18 I1 minus 11 times I2 is equals to 0 so we have got two equations we have got two unknowns I1 and I2 if we put in calculator then we will get value of I1 1.375 amperes and I2 value will get as 2.25 amperes now we have considered the current flow through this as IN i.e. it is equal to I2 right so IN is equal to I2 so its value is 2.25 amperes right now what we have to do we need to calculate the value of RTH because in 7-in-k we have to define the voltage source and circuit in series resistance so RTH is Vth by IN we have already got the value of Vth i.e. 4.5 volts So 4.5 divided by IN is 2.25 So the value of RTH that we get is 2 Ohms Now we have got the value of RTH So always remember whenever you are doing Thevenin So your circuit representation will be Vth means 4.5V With that series resistor will be there That is RTH that is 2 Ohms And the two terminals that you considered So this is your final Thevenin equivalent network for the question. So I hope you understood this.
And now we will move on to Norton theory. So Norton theorem states that any two terminals of a network can be replaced by an equivalent current source and an equivalent parallel resistance. Now what has changed?
In Thevenin, there was equivalent voltage source and series resistance. Now there is equivalent current source and parallel resistance. If I give a diagrammatic representation of this, then it is your entire network. Similarly, you have RL here, current flowing through it is IL.
So, how will you make its Norton equivalent? Your current source will be formed because we have a current source now. And parallel to it will be your resistor.
So, we say it as Rn. Similarly, if you have RL in the final representation, then you have to show RL. Now, its calculation is also mainly same as that of Thevenin's. So, let us solve one question on the same. So, this is our question.
Find Norton's equivalent network across terminals A and B. Now, same. First, we will see our terminals A and B. Across that, we will find our voltage.
That is Vth plus minus. So, this is our first step. Now, whatever currents are there, we will give mesh current in its representation. So, my...
Consider that I have given the Ix current in the mesh. So, I2 is equal to Ix. You can see that it is in the same direction. Right? This is the open terminal.
So, we do not have to think about it that it will become a mesh. But you can see that the direction of Ix is flowing like this. That is opposite to that of I1.
So, we can say that I1 is equal to minus Ix. Fine? Now, what we will do is the same. We have one mesh.
So, we will apply KVL. to the mesh so kvl equation will be 5-10x I2 right so 10x I2 is Ix right again here minus 5 minus 5 times Ix so Ix is again I2 minus 10 times I2 now I2 is Ix again or we can write it as I2 as well so this will be 5-10x I2 Minus 5 times I2 minus 10 times I2 is equals to 0. So this will be 5 minus 10, 10, 20. This is 25 I2 is equals to 0. So 25 I2 is equals to 5. So I2 will be 1 by 5, right? That is equals to 0.2 amperes. Now I2 is your Ix. So your I1 will be minus Ix, that is minus 0.2 amperes.
So, I hope you understood this much. It was same as we did in Thevenin. Now, what will be the second step?
You have to write the equation of Vth. So, you have to consider this outer loop. So, while writing the equation of Vth, what it will be?
It will be 5, we will start from here, right? minus 10 times I2. okay minus 10 times i2 now plus 3 times i1 minus now 10 times but is there any current flowing through it no right because it is open circuit terminal so this is minus vth is equals to 0 so this will be 5 minus 10 i2 plus 3 i1 minus vth is equals to 0 so when we substitute values because we have i2 and i1 both values So, on calculating, the value of Vth is 2.4V.
I hope you understood this much. Now, what will be the next step? We will represent this with our internal resistance. We will consider the current flow as Ion current flow.
Now, we have two mesh. We will solve two mesh. Let us see that.
Now, we have two mesh. So let us say current Ix is flowing through this mesh and Iy through this. So now similarly I2 is equal to IxRi because same current is flowing.
And my value of I1 will be equal to Iy. So Iy minus Ix through the equation. So now we will write KVL to mesh 1. Similarly, starting from here, so 5 minus 10 times I2, I2 means our Ix, right? Again, minus 5 times Ix minus Iy minus 10 times I2 is equals to 0. This is our for mesh 1, that is, considering this, our equation is done.
That is, 5 minus 10 Ix minus 5 times, the current flowing here is Ix minus Iy minus 10 times I2, right? Right. because this is a given thing now next is is to solve this our simplified equation is 25 ix minus 5 iy is equals to 5 so let us say this is our equation number one similarly applying kvl to mesh 2 okay now to apply kvl to mesh 2 similarly we'll start from here right 10 i2 you can see that we have written everything in the form of I into R and this is our voltage so we are considering it as 10I2 similarly here I have written 10I2 minus 5 times this will be Iy minus Ix considering this as my main mesh plus 3 times I1 minus 10 times current flowing through this is Iy so Iy is equal to zero so on simplification we get 12 ix minus 12 iy is equal to zero so this is our second equation so now when we solve equation number one and equation number two the value of ix that we get is 0.25 amperes and the value of iy is 0.25 amperes again right we said that in is our current flow here so that is equals to iy and 0.25 amperes here fine Now, once we are done with this, our final step is to calculate the value of resistor. So, that is Vth upon In.
Right. So, we got the value of Vth is 2.4V upon In is 0.25A. So, on calculation, its value is 9.6 Ohms. Just like we made equivalent circuit in the 7-in, it is very much important that we make it in the Norton as well.
So, in Norton, we have a current source. There is a resistor in parallel to it. and if there is load then load resistor otherwise your terminal that is A and B so our resistor is 9.6 ohms Rn and our current element is 0.25 amperes so this we are done with Norton also so I hope you have understood this please like, share and subscribe to our channel and share it with your friends as much as possible for any queries you can write to us in the comment section See you in next lecture.
Till then, bye-bye.