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

[@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

[@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)

[@deepakdhuriya76]

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

[@aminulislamishmam5607]

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

[@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

[@ahmadfarris8308]

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

[@AJ-et3vf]

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

[@mnada72]

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

[@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.

[@harpyb3951]

bhai kya sahi banda hai gay for him

[@fouzanbaig1963]

This video clear my all concepts of dependent and independent sources

[@engsherif4657]

Great thanks for you Dr sherif ahmed from egypt

[@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

[@saigowtam2253]

Please don't apply direct steps. It will be difficult to understand. If we have that capacity to understand that direct steps we don't need see your videos, we can do on our own.

[@santoshpal8612]

Thanks God u understand that direct steps

[@pooyaimani8331]

Your explanations are so cool 💯

[@Natyler03]

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

[@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

[@physicsdiva1871]

Nice explanation

[@uygarbaran8515]

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

[@digwd8721]

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.

[@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!

[@Akash-hq5xf]

very nice section i liked it bro

[@prachiaggarwal433]

amazing explanation

[@rajeshbandi3070]

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

[@anglotenglish6001]

For those confused about the shorting of the dependent source in part 2: 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.

[@ParneetKaurChemistry]

Thank you.. I understood it well😊

[@universe1072]

really very helpful

[@student6830]

You are so smart dude

[@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?

[@vipulsingla4486]

your lecture is very usefull and easy to understand🤓🤓

[@BentHestad]

Sir, Your channel is great!

[@jaydipprajapati5763]

Very blnice video and easily understood that.

[@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

[@Gaffari2000]

Very helpful, thank you!

[@new-jj5il]

Simply superb

[@sreerams8660]

its 15V and not 5V in Case :1

[@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.

[@arindam_03]

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

[@amoghprabhu106]

If possible I may have made multi likes.

[@ashvinisalunke2775]

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

[@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

[@coster1983]

I think this guy should be an auctioneer. Where is the KZbin slow down button

[@anshpratapsingh5279]

TOO GOOD. THANK YOU!

[@jayaprasadb2044]

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

[@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.

[@tejasrahulrokade8398]

very difficult ...waste of time

[@utkarshjain861]

Lol 😂😂

[@akshayvhatkar3382]

Please add one more video about Norton's theorem

[@muhammadfaizanahmedkhan6371]

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

[@successzone4034]

thx sir for good explanation

[@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?

[@mayankraj9618]

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

[@aakashgoel2931]

4:10 7:01 11:52

[@tedharasta]

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

[@codiator7316]

please correct your video its take my half an hour in clearing my concept before watching that comment and its not a minor mistake it is a blunder

[@noweare1]

Take it easy buddy

[@sid573238]

Basic info is not available

[@NoXYPLaysYt]

Voltage between A and B should be 15 V

[@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.

[@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.

[@bhagyashrees854]

How to find vth 5:16 Why we are not considering 20 v

[@ALLABOUTELECTRONICS]

You also need to consider the drop across the 10-ohm resistor. So, Vab will be Voltage across 10-ohm resistor + 20 V. In this case, since Voltage across 10 ohm resistor is -5V, so Vab = - 5 + 20 = 15V. I hope it will clear your doubt.

[@bhagyashrees854]

@@ALLABOUTELECTRONICS we can consider like this ah Vab=20+(10*-0.5) =15 So we can consider either 10 v or 20 v

[@ALLABOUTELECTRONICS]

@@bhagyashrees854 Yes, since Vab is intermediate voltage, so from either side it can be calculated.

[@bhagyashrees854]

@@ALLABOUTELECTRONICS ok thank u understood it👍

[@sumitbisht4161]

A video for Norton's too!!!

[@sankalpwavekar2865]

Make a video about electronic 11th standard exam

[@splinter1817]

lol there are many sources go find them

[@КевинКоршунов]

I assume the answer is {-5V}, given the direction of the current I = -0.5A therefore in that direction it goes through the negative terminal of the voltage source making it -10 - 10(-0.5)=Vab using(KVL) Vab= -5V, I assume that is the right answer...

[@ALLABOUTELECTRONICS]

You have applied KVL incorrectly. If you are considering 10V voltage source as negative (Negative to the Positive terminal) then drop across the 10-ohm resistor should be positive. So, if you consider negative to positive going direction of the voltage as positive then KVL equation would like 10 - (10)(-0.5)- Vab =0 , And if you simplify it then Vab= 15V. I hope it will clear your confusion.

[@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.

[@infinitx1330]

could have made the explanation more lengthy and elaborate

[@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

[@homeshwarideotale7550]

Plz do minor corrections, in 1st case V=15 not 5....it take our lot of time to understand ur mistake and for a while we doubt our application.... So plz change this asap

[@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.

[@hemantsawant9624]

There is a mistake according to me in case 2 while you consider the dependent current source and 2 nodes as a super node and applying KCL you forgot to keep in mind the current going in and out and mentioned all the currents to be incoming current thus all positive signs in the equation kindly verify and respond.

[@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.

[@shanthankasarla4920]

Yes I am talking about case 1 @ Vab

[@daiyaan3427]

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

[@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.

[@ornilhasan6845]

For you subtitle circuit Couldn't see clearly

[@ashar6118]

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

[@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.

[@Abhishek-hi9tx]

Best💥💥💥🔥🔥🔥🔥🔥

[@manibalan2141]

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

[@manojsinghrawat8125]

Veryyyy niceeee videooooooooooooo

[@shellysingh1786]

Please make a video on foster and cauer form also.

[@DILSHADKHAN-eg8bv]

Nice sir

[@DILSHADKHAN-eg8bv]

Hi

[@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.

[@EATW]

norton theorem with dependent source

[@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

[@sudhakiran495]

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

[@hemenchamuah2244]

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

[@noweare1]

It can be a negative voltage.

[@zaheer561]

many points are not well explained

[@funofun6317]

Please upload clearly for students sir . To find Vth we using delta method. But here is more different. No one use this please upload as useful

[@pravinkumarsharma200]

Vth in first case will be 15V. Do review and update.

[@kerensobol1645]

there is also a problem in case 2, you sayd that to find Rth we need to shorten all sources and you didnt shorted the 5V source.

[@ALLABOUTELECTRONICS]

Well, that's the case when all the sources are independent. When the circuit contains a dependent source then the problem will be treated differently. (which is the case in the second example, as 3Vab is the dependent source) I hope it will clear your doubt.

[@vandana6813]

In thevinen's theorem case 1 thevenin voltage is 15

[@vandana6813]

Good

[@dipenprajapati7469]

Sir please make video on Norton's theorem

[@rahulmeena3269]

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

[@hitenarya4987]

There was a calculation mistake while writing thevenins voltage Vab should be 15

[@ALLABOUTELECTRONICS]

Yes, that's right. I have already mention that in the comments.

[@hitenarya4987]

Sorry i didn't see it

[@ALLABOUTELECTRONICS]

Just see the pinned comment.

[@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

[@zaheer561]

it's 15v sir

[@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.

[@kollasowjanya8285]

Tq u brother 🥰

[@kollasowjanya8285]

Enti brother symbol pettaru

[@rushikeshgawade1288]

good

[@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.

[@EteRnaL_OP56]

In case 1 Taking right branch instead of left we get -15V WHY SHOULDN'T WE TAKE THIS AS A THEVENIN'S VOLTAGE

[@ALLABOUTELECTRONICS]

By every way, you will get Vth = 15V. If you take, 10 - (10x (-0.5)) , then Vth = 15V Or if you consider other branch, then applying KVL, Vth - 10x I -20 = 0 Therefore, Vth = 10X I + 20 And here Vth = -0.5A So, Vth = 20 - 5 =15V. so, whatever way you calculate, Vth is 15V. I hope it will clear your doubt.