Thevenin's Theorem Explained (With examples of Independent and Dependent Sources)

In this video, we will learn about the Thevenin's Theorem in the electrical circuits.
Thevenin's Theorem is a very useful technique in solving the circuit problems when your entire focus is on a particular element in the network and you are not bothered much about the remaining network or circuit.
So, using this Thevenin's Theorem we can replace the entire circuit by the Thevenin's equivalent voltage source in series with Thevenin's equivalent resistance across it's terminal.
So, in this video, we will see that how to find this Thevenin's equivalent voltage and Thevenin's equivalent resistance.
Now, there are three different possible cases of the circuit for which we can find the Thevenin's equivalent circuit.
1) Circuit contains only independent sources.
2) Circuit contains independent and dependent sources.
3) Circuit contains only dependent sources.
So, in this video, we will see that, how to find Thevenin's equivalent circuit for all these cases with an example for each case.
At the end of the video, we will see Norton's Theorem which is dual of Thevenin's theorem and will see that how to convert any Thevenin's equivalent circuit into Norton's equivalent circuit using the source transformation.
The time-stamped link for the different topics is given below:
0:20 Usefulness of Thevenin's Theorem
4:04 Thevenin's theorem for the circuit which contains only independent sources.
6:58 Thevenin's theorem for the circuits which contains independent and dependent sources.
11:49 Thevenin's theorem for the circuits which contains only dependent sources.
14:42 Norton's theorem
Link for the other useful videos:
Classification of Electrical Network:
• Classification of Elec...
Source Transformation:
• Source transformation ...
This video will be helpful to all the students of science and engineering in understanding the Thevenin's Theorem in the electrical circuits.
#TheveninTheorem
#TheveninExamples
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[@ALLABOUTELECTRONICS]

Minor correction: 5:30 In Case 1, the value of Vth will be 15 V instead of 5V.

[@rahuljaiswal4409]

it is -15V

[@danieleyitoyo8089]

10-10(-0.5) -10(-0.5)=+5 so, 10+5=5

[@alihajjhassan84]

So helpful thank you!!

[@gaddamful]

can u please mention this in the description instead. also good idea wioth the time stamps

[@ALLABOUTELECTRONICS]

I have already Pinned this comment.

[@unnimaya5681]

I just came here to see what to do with the dependent sources. But I got to learn why this theorem is needed. Thanks for that.

[@bingebites1_9]

anybody who is confused about voltage Vab in case 1 see below: just apply KVL at node A like this: 0=(Va-10)/10 + (Va-20)/10 SO 2Va = 30 Va = 15v Note: B is taken as ground so Vab is now only Va as Vab= Va-Vb

[@madhur7590]

You've applied kcl not kvl

[@TriscuitFTW]

@@madhur7590 he did node voltage to prove there's a mistake. i literally looked to the comments because i was certain that first mistake was an error

[@sivaarun4351]

Wow thanks....I needed this😁

[@piyush501]

Thanku brother' my mind was blowing regarding this before seeing your comment ❤️🙌

[@nitingautam6762]

Yes. Vab in first case is 15 volt

[@aminulislamishmam5607]

you're the boss. you explain the theory well and also the practical applications. Thank you

[@AJ-et3vf]

Awesome video sir! Thank you! Exactly what I need!

[@Gaffari2000]

Very helpful, thank you!

[@deepakdhuriya76]

Best explanation i have ever seen on KZbin ... Thanku sir for helping students like us

[@GLu-tb1pb]

review: Only independent sources: remove selected resistance, calculated voltage via mesh analysis and resistance via removing sources Independent and dependent: Apply KCL on dependent sources and short A and B to find volt and resistance only dependent: apply 1v/1A source from B to A to find the resistance, as it is equal to Va (Volt = 0)

[@KavyaRam]

One doubt...why to make independent sources zero while calculating thevenin's resistance? What is the significance of this point?? Also what does thevenin's voltage actually mean

[@ahmadfarris8308]

This video helps me a lot. Clear explanation + good examples. Thanks @All About Electronics.

[@nikhilkapoor8868]

In case 1..there is a mistake...Vth is 15V not 5V

[@ALLABOUTELECTRONICS]

Yes, you are correct Nikhil. Thanks for mentioning it.

[@gaddamful]

do u start reading comments before commenting

[@saumeenphansalkar6722]

@@gaddamful no you moron he commented after readding

[@rajeshbandi3070]

Why in the second case, didn't short ckt the voltage source

[@fouzanbaig1963]

This video clear my all concepts of dependent and independent sources

[@ParneetKaurChemistry]

Thank you.. I understood it well😊

[@mnada72]

5:22 The minus in the last equation should be replaced by plus Thank you for the video, it's really of value

[@jayaprasadb2044]

Sir suppose if we take 20v in order to find out vth then value of vth will be change

[@anshpratapsingh5279]

TOO GOOD. THANK YOU!

[@engsherif4657]

Great thanks for you Dr sherif ahmed from egypt

[@ashvinisalunke2775]

Hi please make video on Transfer Function of Electrical Circuits. Thank you, your videos are helpful.

[@pooyaimani8331]

Your explanations are so cool 💯

[@jaydipprajapati5763]

Very blnice video and easily understood that.

[@rishiprusty3166]

When u solve the problem if u hints the eqn very helpful 2 us....btw ur vedios is awesm and it is very helpful me in my study...tq

[@ashar6118]

ckt contain dependent src, i got value of vth is 15v if am applying kcl..is dis value rgt?

[@successzone4034]

thx sir for good explanation

[@muhammadfaizanahmedkhan6371]

Could you please tell me the applications of Thevenin's , Norton's, Reciprocity, superposition and maximum ower transfer theorem.

[@physicsdiva1871]

Nice explanation

[@Cheyansh_Tiger]

in case 2 ,if dependent source value replace by 20i then how will write node eqn

[@Akash-hq5xf]

very nice section i liked it bro

[@prachiaggarwal433]

amazing explanation

[@new-jj5il]

Simply superb

[@universe1072]

really very helpful

[@adnanbajwa27]

In case 2 if the independent voltage source was present in the place dependent voltage source and vice versa than after short circuiting does the independent voltage source gets short circuit or not

[@shuvadipbhowmick7610]

you are great sir

[@mayankraj9618]

in # case 2 why don't we remove 1kohm resistance across AB..(we removed it in case 1)

[@mohamedabubacker.m831]

I didn't understand how to to find Vab in case 1 pls explain it....

[@ALLABOUTELECTRONICS]

Here node B is the reference node. (i.e Vb = 0V) So, Vab = Va - Vb = Va. So, we just need to find the voltage at node A. And it is equal to 10V - (voltage drop across 10-ohm resistor) I hope it will clear your doubt.

[@mohamedabubacker.m831]

@@ALLABOUTELECTRONICS thanks a lot

[@2023noone]

@@ALLABOUTELECTRONICS 1. is not Vab=Vb-Va, not Va-Vb Can you apply KVL in open circuit, since it is not a loop as point A and B are disconnected?

[@rahulmeena3269]

Sir if a circuit contain only independent then can we apply 2nd method i.e the method of short circuit and finding Isc

[@Natyler03]

simple , clear and right explanation! Thanks! Greetings from Paraguay

[@uygarbaran8515]

I’m so confused about the 2nd case when he shorted the resistance and the dependent source. How?

[@analysisdigwd8721]

yeah ,me too.Why is it that the dependent source is alsoshort circuited as the terminal A and B is short circiuted

[@anglotenglish6001]

He doesn't explain it well. The reason is that when he applies the short circuit across the terminals A and B, the voltage across those terminals is zero, thus the voltage of the dependant source (3.VAB) is now also equal to zero, so can be replaced with a short circuit.

[@SoumyajitMondal]

@@anglotenglish6001 oo yes yes. Thanks for your explanation

[@dianaalothman9831]

@@anglotenglish6001 no he was mistaken when he short circuited the dependent voltage source. what you said is only true if the dependent voltage source was connected in PARALLEL to a and b terminals then that would be a short circuit. the dependent source actually becomes connected in parallel to the 2k ohm resistor

[@prithvigupta8215]

@@anglotenglish6001 thanks man .you saved me from confusion.

[@student6830]

You are so smart dude

[@Abhishek-hi9tx]

Best💥💥💥🔥🔥🔥🔥🔥

[@akshayvhatkar3382]

Please add one more video about Norton's theorem

[@arindam_03]

Please explain more easily and give regular examples for B. Tech students

[@sudhakiran495]

Please upload this video again after modifying the vth value in case 1

[@spelunkerd]

I"m still trying to intuitively understand why we short circuit a voltage source but open a current source when calculating Thevenin's resistance. Why the difference in how these two are manipulated? I guess voltage makes sense. After all, there is no resistance in an ideal battery. The best I can come up with as far as a current source is the way it is usually in parallel to a resistance, and so opening the source, which will never change depending on the load characteristics, is fair. Anybody have other ideas?

[@-ShraddhaP]

how did you consider Vab at 5:17

[@harshvardhangavandi7733]

In case 3, is the 3i dependent source a current-controlled voltage source, if not then how did you subtract 3i from Va in case it is not a voltage source (that would be like subtracting current from voltage otherwise).

[@kumudk4kk93]

Yes, I think it is a CcVs , and the symbol of a dependent voltage source is also there. If it was a current source, an arrow(symbol of a dependent current source) would have been used.

[@daiyaan3427]

how did he get 10-10i for v in case 1, can someone explain?

[@shanthankasarla4920]

Yes I am talking about case 1 @ Vab

[@sumitbisht4161]

A video for Norton's too!!!

[@poojashah6183]

👌👌👍

[@manibalan2141]

Why was I1 = -1A in third case at 13.20. plz help

[@sreerams8660]

its 15V and not 5V in Case :1

[@sumanth9929]

Sir, @ 10:59 you've eliminated the 2 ohm parallel 2 ohm resistances to 2 ohm resistance. But without making parallel formula how would it be 2 ohm? It might be : 2*2/(2+2) = 1 ohm ??

[@ALLABOUTELECTRONICS]

If you closely observe, the 2 K resistor is in parallel with AB terminals (which are short-circuited). That means their effective resistance will become zero. And in the equivalent circuit, there will be only 5V voltage source in series with 2 K resistor. I hope it will clear your doubt.

[@chillacombs8790]

hi am a CE student....my lecturer wasn't that clear especially on obtaining the thevenin resistance....but lol...you made it simple sir..thumbs up!

[@harpyb3951]

bhai kya sahi banda hai gay for him

[@vipulsingla4486]

your lecture is very usefull and easy to understand🤓🤓

[@manojsinghrawat8125]

Veryyyy niceeee videooooooooooooo

[@student6830]

He made a small mistake in example 1, but he already corrected it, I don't know why 99% of the comments are about that mistake? Let's just appreciate his effort instead.

[@BentHestad]

Sir, Your channel is great!

[@sandipsadhukhan9287]

Which software use to create video ?

[@abkrikot]

At 8:45 in case #2 while explaining Super node this stumped me Vx - Vth = 3Vth , You Jumped from Vx-Vth = 3Vab and equated Vab to Vth

[@maramsaritharani4180]

Doubt about case1 Vab calculation

[@tedharasta]

I was confused but you did it wrong. Can you please change the value of Vab from 5v to 15v

[@abhishekchaudhri691]

i didn't understand why and how 1 ohm resistor got short circuited when A and B r shorted

[@ALLABOUTELECTRONICS]

When A and B are short-circuited, it offers zero resistance. So, anything in parallel to it will also offer zero resistance. Because all current will flow through the short circuit. And that's why 1K resistor will also get short-circuited. I hope it will clear your doubt.

[@yec-praneethKondeti]

bro akkada dependent source sum lo ab terminals short circuit cheste 1kohm ,dependent source zero enduku aindi

[@jinshikami7525]

Hey could you try this software? Bump into: 'Circuit Solver' by Phasor Systems on Google Play.

[@Niha1399]

Sir I didnt understand how did u get Vab=10-(10i)

[@ALLABOUTELECTRONICS]

To find, Vab, KVL has been applied in the loop. Now, Vab = Va- Vb ( the Potential difference between Node A and B). As node B is reference node, so the voltage at that node will be zero. Now, the voltage at Node A will be 10 - (10I) (that is 10V - voltage drop across 10-ohm resistor, using KVL) And hence, Vab = 10 -(10I) I hope it will clear your doubt. If you still have any doubt then do let me know here.

[@othoiothoi4299]

sir how is node b a reference node

[@ALLABOUTELECTRONICS]

See, node B is at zero potential. Or in another way, if say, it is a ground of the circuit. And in the circuit generally, the node with zero potential is considered as the reference node. I hope, it will clear your doubt. If you still have any question then do let me know here.

[@inspiratex]

what if both resistors are not equal

[@saimanimorri8942]

Sir voltage at node A 10-(10i) ala aindii

[@AdmMusicc]

In case 1, since the points A and B aren't connected, why didn't we take the 10 ohm resistances in series?

[@ALLABOUTELECTRONICS]

Because we want to find the equivalent resistance across terminal A and B. What you are saying would be true if we want to find the resistance at some other point. Let's say the points where the 10V voltage source is connected there we want to find the resistance. In that case, both resistors would come in series. I hope it will clear your doubt.

[@yankeefan031]

In case 2, why is the V used in the equivalent TH circuit 5 and not .5 as calculated at the beginning of the problem

[@ALLABOUTELECTRONICS]

It is 0.5V, Please check at 11:41 in the video.

[@rahiabrar8400]

I have a doubt. If the V in the final reduced circuit is thevenin voltage then how come the potential difference across A&B is thevenin voltage because as there is a resistance in series with equivalent voltage of circuit the voltage across terminals will not be the same as the equivalent V. Can you please elaborate?

[@ALLABOUTELECTRONICS]

It is the representation of the linear circuit. Or I would say using Thevenin's theorem you are getting an equivalent circuit, which has Thevenin's voltage source in seires with the Thevenin's equivalent resistance. And across that circuit generally load is connected. (Across terminal A and B of the circuit). So, when you turn off all the independent sources and find the voltage across AB that's circuits equivalent voltage. It's not the voltage that you get across the load. Of course, it will be different. And it will be RL*Vth/(RL + Rth) . I think you got confused with it. Isn't it?

[@shreyasuresh8291]

I have a doubt in the case 1 to find R th u short circuited the voltage sources, but in the case 2 why you dint short circuit the 5 v voltage source??? Plz clear my doubt

[@ALLABOUTELECTRONICS]

Because in case 2, the circuit contains the dependent source. So, there is a different procedure for finding the Thevenin's theorem in that case. Please go through the example, and after that if you have any doubt then do let me know here.

[@sharminmonika8887]

Nice tutorial. ,By the way .....should we always use 1 A as test current in the 3rd case ?Actually,I found the value of Thevenin's resistance varied for other test currents.

[@ALLABOUTELECTRONICS]

Any value of current source can be used as test current. Based on the value, voltage across the current source will change. But ratio of V/I should remain the same. The advantage of choosing 1A current source is that, denominator is 1. So, calculation becomes easy.

[@danieleyitoyo8089]

Then u apply kvl what about 20v

[@kollasowjanya8285]

Tq u brother 🥰

[@kollasowjanya8285]

Enti brother symbol pettaru

[@sureshtelugu5556]

Tq sir

[@srikantakumargouda9561]

All about electrical

[@hemanth3744]

How do you found Vab in first case (please respond)

[@ALLABOUTELECTRONICS]

Vab is the voltage between node A and B. Vab = Va - Vb. Node B is a reference node and is at ground potential. Therefore, Vb = 0. So, only you need to find the voltage at node A. The same has been found by applying KVL. I hope it will clear your doubt.

[@savyasanchighose5176]

sir pls upload video of Norton's theoram for lelectrical network

[@ALLABOUTELECTRONICS]

I have explained Norton's theorem at the end of the video. Please check the description for timestamps. I think it is at 14:42

[@keerthiraja9860]

At 9:58 Why A&B terminals were shorted?

[@rushalmshah]

To find the short circuit current through terminal A and B.

[@mohitkwatra9587]

Please tell that why current source is replaced by open ckt and voltage source by closed ckt In detqil you told it in vedio i cudnt understand it

[@ALLABOUTELECTRONICS]

In Thevenin's theorem, while finding Thevenin's equivalent circuit any source is replaced by its internal resistance. And voltage and current sources which are used during the examples are ideal sources. Ideal current source has infinite input impedance and ideal voltage source has zero series internal resistance. That is why current source is replaced by open circuit and voltage source by short circuit. I hope it will clear your doubt.

[@hemantsingh1253]

You have calculated wrong value of Vth in case 1 I think it would be 15 volt

[@ALLABOUTELECTRONICS]

Hey Hemant, yes I have already mentioned that in the pinned comment.

[@sankalpwavekar2865]

Make a video about electronic 11th standard exam

[@splinter1817]

lol there are many sources go find them

[@goflow6425]

Sir...u said that Vab=10-10(I)....how u give value of A as 10 & B also 10?????

[@harishhreddy5234]

Exactely i too had that dought

[@ALLABOUTELECTRONICS]

If you see the left side of the A and B. There is a 10V resistor in series with 10-ohm resistor. Here Node B is considered as a reference node. So, the voltage between the two nodes will be the 10V minus voltage drop across the 10-ohm resistor. If I is current in the circuit then Vab = 10 - (10 ohm x I) I hope it will clear your doubt.

[@shivasai8547]

Sir explain some more problems on thevenins therom plz

[@ALLABOUTELECTRONICS]

Many problems have been already covered on the second channel. Please check this playlist. kzbin.info/aero/PLH9R5x7JVXCEvlSCakgkrfdHfxBU25MYf More problems will be uploaded there.

[@ravikumar-jm7gj]

Can u displace the formula for every example for every vedio then we calculate the example like can u keep the formula in brackets before the calculation then we can understand that this formula is used in this step if u put this we easily understand that this formula is used in this step

[@ALLABOUTELECTRONICS]

Hey Ravi, that's good suggestion. Yes, I will do so from now on each video. So, that you all can easily understand.

[@amoghprabhu106]

If possible I may have made multi likes.

[@krishnakumarprathipati7186]

Sir in first sum Vab=15v sir

[@rohithivn8300]

Is the test current in case 3...1amp...Same for all circuits???? Or does it changes???

[@ALLABOUTELECTRONICS]

well, we can apply the current source of any amplitude. But to make calculations simple, 1A current source is usually applied.

[@durgasmart4639]

No need to take the current (I) amplitude as 1A. Take I Then Apply kcl then you should get the ratio if Va and I then that is thevenin resistance..... I hope my explanation was right.

[@ninhnguyen1495]

What is this app?

[@amanyadav6607]

At 10:35 why 3Vab is short circuited?

[@ALLABOUTELECTRONICS]

Vab is the voltage across the terminal A and B. When terminal A and B is short Circuited then Vab =0. And hence dependent source will also become zero. I hope it will clear your doubt.

[@sid573238]

Basic info is not available

[@koyasrinath3283]

Can you tell in which case thevinin and norton circuits are not equivalent

[@ALLABOUTELECTRONICS]

Norton's equivalent circuit is just source transformation performed on Thevenin's equivalent circuit. So, any circuit for which you can find Thevenin's equivalent circuit, you can also convert it into Norton's equivalent circuit.

[@chillacombs8790]

only in complex circuits wih multilateral sources.remember it only works for bilateral and linear sources

[@hemenchamuah2244]

Nice video sir.....Sir can V(th) be negative???

[@noweare1]

It can be a negative voltage.

[@NoXYPLaysYt]

Voltage between A and B should be 15 V

[@debasishsau4323]

In dependent source circuit how to understand I1=1Amp?

[@ALLABOUTELECTRONICS]

1A current source is applied externally. If you apply KCL at node A, then I1 = -1 A.

[@jeremyyoung3343]

Can you please explain again why "short circuiting" the AB terminal in Case 2 ends up "short circuiting" the rest of the circuit as well?

[@ALLABOUTELECTRONICS]

In case 2, when we short circuit two terminals A and B, 1k resistor will be in parallel with short circuit (or ideally zero ohm resistance), hence the equivalent resistance of that combination will be zero. Now, as terminals A and B are short circuited, so potential difference between node A and B will zero or we can say that VAB=0. So, dependent voltage source 3VAB will also be zero. Now, again if you redraw the circuit then 2K resistor is again in parallel with short circuit terminal. So, it's equivalent resistance will be zero. And hence, eventually both resistors and dependent voltage source will get short circuited. And we end up with 5V voltage source in series with 2K resistor. I hope, it will clear your doubt. If you still have any question then do let me know here.

[@jeremyyoung3343]

EGG SELL LENT explanation. I understand it in the context of equivalent resistance! Thank you.

[@md.nasiruddin6608]

@@ALLABOUTELECTRONICS nice explanation

[@balamuralikrishna_d]

How u got vab man?

[@lavanyaramesh1241]

In finding vth y r u not calculating voltage at AB due to 20 voltage source?????

[@ALLABOUTELECTRONICS]

In this example, the motive is to find the voltage at node A, which is equal to the 10V minus the drop across the 10-ohm resistor. If we go further then there will be drop across one more 10-ohm resistor. But here we are not interested in the voltage after node A. If you start the loop from 20V voltage source up to node A then also you will get the same voltage. e.g 20V - (10) (0.5) = 15V. So, the point is no need to complete the loop here to find the voltage at node A. And you can start either from 10V voltage source or from the other end (20V voltage source), you will get the same result. I hope it will clear your doubt.

[@krishnakumarprathipati7186]

Sir how u have taken I1 as -1

[@ALLABOUTELECTRONICS]

The test current of 1A is applied at Node A. So, by applying KCL at node A, i1 = -1A.

[@viveksamani7924]

@ALL ABOUT ELECTRONICS at 4:30 when you just applied kvl and wrote the equantion as 10=10 I +10 I + 20 according to the arrow you dram (yellow one , clockwise)........but its wrong, the correct equation would be 10= 10 I + 10 I - 20 as we are moving from higher to lower potential of 20V source as the arrow indicated by you is clockwise........i may be wrong!

[@ALLABOUTELECTRONICS]

Please check it once again. You will get it. Just start from the 10V voltage source in the clock wise direction. Since the potential is rising (negative to positive voltage, let's consider it as positive. As per the direction of the current, we will have voltage drop across two 10 ohm resistors. And once gain for 20V voltage source, since we are moving from higher to lower potential, so there will be a voltage drop. So, three terms in the KVL equation will be negative. The equation would be like +10 - 10I - 10I - 20 = 0 Or, it can be written as 10 = 10I + 10I + 20. I hope, it will clear your doubt.