Norton's Theorem Explained (With examples of Independent and Dependent Sources)
ฝัง
- เผยแพร่เมื่อ 31 ก.ค. 2024
- In this video, Norton's Theorem is explained with the Solved Examples.
The timestamps for the different topics covered in the video :
0:00 Introduction
0:20 Norton's Theorem and Steps to find the Norton's equivalent Circuit
3:33 Example 1 (When the circuit contains only independent sources)
7:02 Example 2 (When the circuits contain independent and dependent sources)
11:28 Example 3 (When circuits contain only dependent sources)
Norton's Theorem is a very useful theorem in solving circuit problems.
Norton's Theorem :
Any linear network can be replaced across the two terminals by the equivalent network which contains the current source in parallel with the impedance.
The equivalent current is known as Norton's equivalent current and can be found by finding the short circuit current across the two terminals.
The equivalent impedance is known as Norton's equivalent impedance (or Norton's equivalent resistance ) and it is the equivalent impedance which is seen through the two terminals across which one wants to find the equivalent circuit.
Norton's Theorem is dual of Thevenin's theorem and using the Source transformation, it is possible to convert the Thevenin's equivalent circuit into Norton's equivalent circuit.
In this video, the three different cases have been considered through 3 different examples.
1) Circuit contains only independent sources.
2) Circuit contains independent and dependent sources.
3) Circuit contains only dependent sources.
And for each case, Norton's equivalent circuit has been found.
Link for the other useful videos:
Classification of Electrical Network:
• Classification of Elec...
Source Transformation:
• Source transformation ...
Thevenin's Theorem
• Thevenin's Theorem Exp...
This video will be helpful to all the students of science and engineering in understanding the Norton's Theorem in the electrical circuits.
#Norton'sTheorem
#NortonsExamples
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Music Credit: www.bensound.com - วิทยาศาสตร์และเทคโนโลยี
The timestamps for the different topics covered in the video :
0:20 Norton's Theorem and Steps to find the Norton's equivalent Circuit
3:33 Example 1 (When the circuit contains only independent sources)
7:02 Example 2 (When the circuits contain independent and dependent sources)
11:28 Example 3 (When circuits contain only dependent sources)
Please keep backup of all your videos till date. Coz you know very well,Neso Academy channel got hacked. So much of hard work all gone. You should keep backup of every videos you uploaded.
Neso academy created the app in play store everything is available there.
Thanks , I like your channel and the content you provide. Is there a real example for a circuit that has only dependent sources ?
Awesome video sir! Thank you!
great channel a help to the college students
I like your way of teaching 👍👍👍 and found al most all the topics easily and subscribed 🎊🎊🎊🎊😳
Thanks a Ton ... Sir
It helped me solving today's Quiz..
Sir in example 1 could we have found out Isc using superposition theorem instead of mesh analysis?
In the last example ,When applying kcl at two nodes, for both nodes we assumed current to be in different directions for that 50 ohm resistor...can we take it like this and also why??
Sir, i have a request to you, that please make a video series on 8051 microcontroller, please sir.
Best👌🏻👌🏻
Very good..
Thank you so much♡
Osm sir.....!
Good teacher
Hello again, At 14:14, I simplify this to (10Vx+Vy+4Vy-4Vx)/200=0 which gives 6Vx + 5Vy = 0. Please tell me where I have gone wrong. Thank you.
Awesome
In this circuit diagram
When kcl applied at node Vy... Why will Vx/10 will come ?
Can you please tell me about this
Anyone please reply me about this doubt
I am confused at 8:00. In figure shouldn't 4 ohm resistor be a dummy resistor (ie no current flowing through 4 ohm resistor) since it is parallel to 10V source?
And also how is yellow part of circuit parallel to blue part of circuit? If possible can you send link of redrawn circuit
It will be a dummy resistor when the voltage at both ends of the resistor is same. Here, at one end, the voltage is 10V, and at the other end of the 4-ohm resistor, the voltage is 0V. Hence, the current will flow through it. Since it is parallel to the 10V voltage source, the voltage across 4-ohm resistor will be 10V.
Regarding your second question, if you see the one end of 5-ohm resistor (yellow part of the circuit) is connected to the 10V (Blue part) while through the short circuit, the other end is connected to the negative terminal of 10V. So, it is connected in parallel to the blue part.
For simplicity, just imagine the entire parallel combination by equivalent single element for both blue and yellow part. Then you will be able to see that.
I hope it will clear your doubt.
here at 6:39 you consider 4ohm(i2-i1) why?I cannot undersand
Sir will you please explain about 8051 microcontroller interfacing topics...
I will cover. But it will take some time.
Sir can u plz tell me the names of Institutes where M.SC in electronics can be done. I am currently a student of b.sc final year physics honours. Subject combination Physics , mathematics, chemistry.
I will be highly grateful to u.
in nit warangal
Which current should I consider while dividing Voc to find resistance, the current across A and B or the current leaving the voltage source?
The short circuit current flowing through A-B.
@@ALLABOUTELECTRONICS thanks
Here in the question 1 equivalent resistance at the terminal AB, not through 5ohm , then why we are considering 5ohm as load??
In this given circuit, the 5 ohm resistor is a part of the circuit and not the load. If we have been specifically asked in the question that 5 ohm resistor is the load and find the equivalent circuit across the 5 ohm resistor then we need to find the equivalent circuit excluding 5 ohm resistor.
I hope it will clear your doubt.
In this given circuit, the 5 ohm resistor is a part of the circuit and not the load. If we have been specifically asked in the question that 5 ohm resistor is the load and find the equivalent circuit across the 5 ohm resistor then we need to find the equivalent circuit excluding 5 ohm resistor.
I hope it will clear your doubt.
Thanku sir
sir how at 6:05 did you take I1 and I2 for both the loops individually?
Here both I1 and I2 are assumed as a mesh currents. And separate mesh equations are applied for both mesh.
at 10:20 why did you take the direction of current through 5ohm in the opposite direction?
I have just applied the KCL at node A. Incoming and outgoing current through the node should be the same. Since terminal A-B are open, there is only one-way current can go. From Va to Vx.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thanks a lot😇❤
Sir please explain Basic electronic subject
I have already made videos on it.
Please check the basic electronics playlist on the playlist section of the channel.
In 2nd question how is 4ohm resistor is in parallel with 5 ohm resistor and dependant source
If you see the one end of the 4-ohm resistor, then it is connected to the one end of 5 ohm resistor. And once we short the terminal A-B then bottom end of 4 ohm resistor is connected to the other end of 5-ohm resistor.
That's why they are in parallel.
I hope it will clear your doubt.
sir at 6:18 why are u using the opposite convention of assigning sign to potential differences
Yes, that's right, but It will not make any difference in the final result. Because, if you take the negative sign outside, then we will get the same result. So, I have directly used the second step. I hope you got it.
👍🏻
At 5:57 why you take -12 not +12 because there is a voltage gain
It depends on what sign convention you are using. Here, when there is a drop, it is considered as positive. (e.g., 8*I2). While applying the mesh equation in the second loop, as you are saying, yes there is an increase in voltage, as we go across 12V voltage source. (From negative to Positive). But here because of the opposite sign convention, it is considered as negative. I hope it will clear your doubt.
how do we know that the, 8,4&8 ohm resistors are in series?
Because the current flowing through all three resistors is same. From the A terminal, if you apply some current, then it will get divided between 8 ohm and 5 ohm, and whatever current that is flowing through 8 ohm resistor, the same current will also flow through 4 ohm and another 8 ohm resistor. (Because there is no other node in between)
In example 2 ,at any node all the entering currents equal to all the leaving currents then y did u make 2ix=va-vx/2?
I think you are referring to 10:18 right ?? Well, if so then it is 2ix = (Va - Vx) / (5)
Please check it once again.
@@ALLABOUTELECTRONICS Tq sir
how 50 0hm is leaving in the 3rd case while applying kcl at node Vy
If we assume, it is entering the node Vy, then it will be Vx - Vy/50. And the entire equation will be Vx/10 + Vy/10 = (Vx - Vy)/10
When you assume it is leaving the node Vy, then it will be Vx/10 + Vy/10 + (Vy - Vx)/10.
So, it's just a matter of direction and the sign. it won't affect the end result.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thank You
once A and B gets short circuited,why do 5ohm resistor gets short circuited?
5.14
any one please
It's connected in parallel to the node A-B. The short circuit has zero resistance. If you take the parallel combination of any resistor with zero ohm resistance, it will be zero ohm. (Short circuit)
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thankyou...i got it...understood..thankyou
Sorry, I have written my earlier comment incorrectly. At 14:07 Simplifying (20Vx+Vy+4Vy-4Vx)/200 I get 16Vx+5Vy=0. Where am I going wrong? Thank you.
You are correct. Actually, while writing, I made a mistake over there. But then at 14:17, I have written the correct expression. So, it was a mistake from my side.
@@ALLABOUTELECTRONICS Thank you so much. Not trying to catch you out....just to prevent myself going nuts. All the best.
Why there is 0A in 3rd case at final..?
Because when the circuit contains only dependent sources, then there will be no Norton's equivalent current in the equivalent circuit. We can only find Norton's equivalent impedance for such circuits.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS yes.. I understood. Thanks for replying for my question.. Your videos very high level... Hats off sir
why 0A in the last example
The circuit has only a dependent source. That means there is no excitation in the circuit. Therefore, we can only find Norton's equivalent resistance of such circuit. Without any independent source, Norton's equivalent current will be zero. I hope it will clear your doubt.
Why reference node VB taken as zero ..give some detail explaination sir please
Generally, in any circuit there has to have some reference node, and with respect to that node, the voltage at the different nodes are measured or calculated. Typically the ground node (whose voltage is zero) is considered as the reference node and with respect to that, the voltage of the different nodes are measured. But in any circuit, if the ground node is not mentioned specifically then any node can be considered as reference node (the voltage at that node is assumed to be zero) and based on that, the voltage at the different nodes is calculated. Typically the node where maximum branches are meeting is considered as a reference node. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS thanks sir
I got clarified
Binod