The downloadable pdf link for the practice problems on the phasor and Series AC Circuits (RC, RL and RLC circuits ) has been updated in the description. drive.google.com/open?id=0B3FOmN0FcD6OMUJBUHFlNEw5VXM
Great video and a simple explanation. I believe there is a mistake at frame 18:22 converting to rectangular. Z = 5.776 - j3.33 (in the video it is 5.57 - j3.33). So it should be R = 5.776 ohms.
Hello sir thank you so much for your detailed explanation..it really helped me in my exams..I wish to watch more videos of yours..thank u once again sir :)
I wish you had included some examples with parallel circuits to show how to use the phasor diagram with such circuits. Is it possible to do it for the sake of completeness of the subject.
Sir, at 8:24. The phase is shown as phi=tan-1(1/wCR). Shouldn't it be phi=tan-1(-1/wCR). I mean to say the negative sign is missing. I'll be thankful if you reply fast. I have an exam tomorrow.
It is just representation of the impedance in rectangular form, as impedance we got was in polar form. So, the real part will be the resistance and imaginary part will be the reactance. I hope it is clear to you now. If you still have any doubt then do let me know here.
No, the amplitude may not be the same. It will depend on the value of R. Let's say I= 2A and R= 2 ohm, then VR = 4V. In that case, both I and Vr are in phase but the amplitude is not the same.
@@charleswolfe6849 No, that's alright. If you see the phasor diagram, the voltage leads the current. Or in another way, I can say that the current lags the voltage by some angle. That's why I = Im sin (wt - Φ). I hope it will clear your doubt.
Thanks a lot for your viedeo sir . Your viedeo was easy to understand Could you please upload so many problems related to rlc circuits and current locus
I have provided the link in the description for the more videos. Also check the community tab section of the channel, where regularly the quiz related to the different topic is posted. If you go to the earlier quiz then you will find some quiz related to RLC circuit as well.
The impedance will have a real and imaginary part. So, it can be treated as a complex number. And the algebra which is used for the complex number can be applied over here too.
I think you mean to say the impedance of the inductor right? The impedance of the inductor or reactance is 2*pi*f*L At DC, f = 0. So, in steady state condition its impedance will be zero.
Generally, in Series RL, RC and RLC circuits current is taken as reference because current that is flowing through all the components is same, but the voltage is different. And we don't know about the resultant voltage. In the given circuit suppose if you know the resultant voltage amplitude and phase, then you can take it as a reference and current will lead or lag accordingly. For parallel RC, RL and RLC circuits, as the voltage is same across all the components, so voltage is taken as reference and resultant current is found. So, in short, suppose if you take resultant voltage V as a reference then current will lead or lag accordingly. For example, in series RC circuit if you take voltage V as reference, then current will lead the voltage by some angle Phi. I hope it will clear your doubt. If you still have any confussion do let me know here.
4+ j3 is the total impedance (R + jwL) In the given example w= 300 and L = 0.01. S0, wL= 3. Hence, total impedance Z= 4+j3. I hope it will clear your doubt.
The link for the solution has been already updated. I have already pinned that comment. Here is the link: drive.google.com/open?id=1bMn2TbCP9fF0EceQqxHpi8mYr_l53ZPw
Its rectangular form to polar form conversion. If a+jb is a complex number then it can be represented in the polar form as r< theta. Where r= sqrt (a^2 + b^2) and theta = tan-1(b/a) I hope it will clear your doubt.
Let's say 1V of phasor is making an angle of 45 degrees with the reference axis. Then in rectangular form it will be 1*Cos(45) + j 1*sin(45) = 0.707 + j 0.707
If you want to solve more problems then I would suggest this book. Network Analysis and synthesis by Ravish R Singh www.amazon.in/Network-Analysis-Synthesis-Ravish-Singh/dp/1259062953/ref=sr_1_1?ie=UTF8&qid=1513183137&sr=8-1&keywords=ravish+singh+network+analysis
If XL > Xc, then VL>Vc. Because they are connected in series and the same current is flowing through XL and XC. So, in the phasor diagram, the net voltage due to Xc and XL would be VL-Vc.
The solution for the practice problems is uploaded. Please check the link
drive.google.com/file/d/1bMn2TbCP9fF0EceQqxHpi8mYr_l53ZPw/view?usp=sharing
More examples for practice will be updated in the description very soon.
The downloadable pdf link for the practice problems on the phasor and Series AC Circuits (RC, RL and RLC circuits ) has been updated in the description.
drive.google.com/open?id=0B3FOmN0FcD6OMUJBUHFlNEw5VXM
Thanks
Can u please explain me last question
U probably forgot that question but how did we seperate z
After this video..I have really understood the phasor ... Thank you sir 😊 you
the conversion of 6.67
Thanks for making this amazing tutorial, it's really very educative.
Despite your pronunciation, It was clear and understandable among many other videos on the internet! Thank you, great job
My final on electrical is going to be held next week.. TQ for teaching me😁
Khatarnak dhamaka 🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
Great video and a simple explanation.
I believe there is a mistake at frame 18:22 converting to rectangular.
Z = 5.776 - j3.33 (in the video it is 5.57 - j3.33).
So it should be R = 5.776 ohms.
Thanks a lot for making this easy to understand .
Thanks a mil for this video! Makes so much sense!🙌
Thanks bro you cleared my concepts in a single go
You explanation is in depth cool bro.support for you from our side will always be there
Hello sir thank you so much for your detailed explanation..it really helped me in my exams..I wish to watch more videos of yours..thank u once again sir :)
This channel has got a unique talent
Thanks a lot
I wish this channel will upload many more useful videos of this type
Thanks once again
I wish you had included some examples with parallel circuits to show how to use the phasor diagram with such circuits. Is it possible to do it for the sake of completeness of the subject.
thanks
....................... you made my day ...... keep it up bro
You are doing good job. Upload as much videos as you can about electronics
you are a genius among all duplicates on youtube.
You're a KING
In example 1.how you changed the impedance into polar form i.e., (283《90° =5《36.86°
At 4:32, shouldn't it be i(t)=im sin(wt) and v(t)=vm sin(wt-phi) ? It seems quite confusing.
your videos are the best 👌
amazing explanation sir, really helpful, thanks a lot!!
Thanks for your help from Al Habib Ahmed from Manipur India
Thank you ❣️
What a beauty of explanation.Thank you.Well done.
The solution of the practice problems will be uploaded very soon.
plzz plzzzz plzzz make a video on graphical representation of RC , RL and RLC resonance .. series as well as parallel
Sir, at 8:24. The phase is shown as phi=tan-1(1/wCR). Shouldn't it be phi=tan-1(-1/wCR). I mean to say the negative sign is missing. I'll be thankful if you reply fast. I have an exam tomorrow.
Yes, correct.
Sir, please do a phasor analysis of pure L-C circuit, I have difficulty in finding the angle φ of resultant impedence in argon plane.
Excellent as usual, keep up
thq sir ..its so useful
In last example how 6.67
At 5 : 40
Sir why lzl=√r^2+(wL)^2 ?
How u remove j from the equation
Z=R+jwL ?
🤔 Plz
Thank you so much. You’re simply amazing
Well in the last problem how you have explained the z=r+jX as a+jb i didn't grt it Can you please explain it?
It is just representation of the impedance in rectangular form, as impedance we got was in polar form. So, the real part will be the resistance and imaginary part will be the reactance.
I hope it is clear to you now. If you still have any doubt then do let me know here.
You see, there is a 30 of Z degrees, then the sin30=Z/X=1/2 => X=6.67/2=3.33 then you find R=(Z^2-X^2)^(1/2)
why there are no subtitles in this video, I hope u could provide it...
Soon, it will be get added for this video too.
how can you find the angle in the the 1st example for current i.e,36.53
Awesome man, it was such a confusion.keep it up👍👌✌
It should be tan inverse( -1/wCR) correct there should be a minus sign ?
awesome vid, keep it up
IN RL CAN WE TAKE V IN PHASOR DIAGRAM DOWN SIDE 90 DEGREES
Source V=Vm sinwt
When presenting in polar form, why use Vm(=Vmax?) but not Vrms?
in the last problem the answer of R is actually ~5.78Ohms, not 5.57
how did you get this
4:37 why does the current have the same amplitude as the voltage?
No, the amplitude may not be the same. It will depend on the value of R.
Let's say I= 2A and R= 2 ohm, then VR = 4V. In that case, both I and Vr are in phase but the amplitude is not the same.
@@ALLABOUTELECTRONICS I think there is a mistake and you wrote "Vm" instead of "Im" at 4:37
@@charleswolfe6849 No, that's alright. If you see the phasor diagram, the voltage leads the current. Or in another way, I can say that the current lags the voltage by some angle.
That's why I = Im sin (wt - Φ).
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS No it's not about which one leads the other, it's that I think you wrote I = Vm sin(wt - theta)
(4:42) I think current will be Vm/Z not Vm
Thanks.. Sir Aviti and his supplementary exam are going down.. 💪💪
Thanks a lot for your viedeo sir . Your viedeo was easy to understand
Could you please upload so many problems related to rlc circuits and current locus
I have provided the link in the description for the more videos.
Also check the community tab section of the channel, where regularly the quiz related to the different topic is posted.
If you go to the earlier quiz then you will find some quiz related to RLC circuit as well.
kindly sir tell me how to solve and draw diagrams of circuit having Both Inductors and Both Capacitors
Thank you very much
wht bout lc circuit ?
hello how did you find the rectangular form of the impedence from 6.67
If r
Yeah I don't get that part I tried using the solution provided but I had R=5.94 and X=3.03 please help me
Can u do a video on phasor diagram for combination of parallel and series rlc circuits
How did you find out the value of R+jx using A+jB
The impedance will have a real and imaginary part. So, it can be treated as a complex number. And the algebra which is used for the complex number can be applied over here too.
which software sir?? hope that u tell me ...
Sir did you have taken current as reference for phasor diagram
Yes.
Bro tell about how calculate indecture in dc circuit
I think you mean to say the impedance of the inductor right?
The impedance of the inductor or reactance is 2*pi*f*L
At DC, f = 0. So, in steady state condition its impedance will be zero.
why avoid the minus sign for the reactance (X) for last question final ans
In the question, only the magnitude of the reactance is considered.
Thanks you're a genius..
I didn’t understand how you got the rectangular form in the last problem
If r ∠ θ is impedance in polar form, then in the rectangular form it is x + j y, where x = r cos θ and y = r sin θ.
Hai bro how to possible z=5.57-j(3.33) once tell me that one bro
thanks a lot
What is j??
It is used to define the imaginary term of the impedance. (Similar to the complex number)
i think when converting a sinusoidal form to polar form, we must put the rms value in the polar form. 283/rt(2)
yes
When we take v as referrence what is phasor diagram of rc rl and rlc
Please explain
Generally, in Series RL, RC and RLC circuits current is taken as reference because current that is flowing through all the components is same, but the voltage is different. And we don't know about the resultant voltage. In the given circuit suppose if you know the resultant voltage amplitude and phase, then you can take it as a reference and current will lead or lag accordingly.
For parallel RC, RL and RLC circuits, as the voltage is same across all the components, so voltage is taken as reference and resultant current is found.
So, in short, suppose if you take resultant voltage V as a reference then current will lead or lag accordingly. For example, in series RC circuit if you take voltage V as reference, then current will lead the voltage by some angle Phi.
I hope it will clear your doubt. If you still have any confussion do let me know here.
In pure R circuit we take, V as reference.. What's the reason?
At 3:56 , you have taken voltage as reference..but here you are saying that we take reference the value known to us... Plz clarify..
How you calculated the value of 4+j3?
4+ j3 is the total impedance (R + jwL)
In the given example w= 300 and L = 0.01. S0, wL= 3.
Hence, total impedance Z= 4+j3.
I hope it will clear your doubt.
I was asking about "5
In the calculator, convert the A+jB into r
Ok.thanks..
@@ALLABOUTELECTRONICS thank you so much
Hello Sir.....can u provide solution of 2nd problem of practice problem
The link for the solution has been already updated.
I have already pinned that comment.
Here is the link:
drive.google.com/open?id=1bMn2TbCP9fF0EceQqxHpi8mYr_l53ZPw
thanks man, but I hoped you draw the phasor diagrams for those examples also
Last problem not clearly explained u put the direct values without explaining the process 🙁
explained well
Why don't we dived v by √2...?
v=√2Vsin(ωt) na?
5 multiplied by 36.86 how it came sir please explain
Its rectangular form to polar form conversion. If a+jb is a complex number then it can be represented in the polar form as r< theta.
Where r= sqrt (a^2 + b^2) and theta = tan-1(b/a)
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS what is 'a' and 'b'
How to find an angle
how to represent in rectangular form
???
Let's say 1V of phasor is making an angle of 45 degrees with the reference axis. Then in rectangular form it will be 1*Cos(45) + j 1*sin(45) = 0.707 + j 0.707
Thanks
Thanks unlimited sir 🙏🙏🙏🙏🙏🙏🙏🙏
Super antother level
4:34 wrong value of I
2:01 current voltage se bda kese hua ??
The graph is just for the representation of phase difference between voltage and current.
Sir please how 4+j3=5
Its rectangular to polar form conversion.
Let's say you have A+jB in rectangular form. Then in polar form, it is r
Thanks
I should've payed my school tuition here
I didn't know how to convert polarform and polar angle
Three phase ac ckts loo vides lavuuu chayaraaa plz
Video is extremely good but that j is not understanding
It is used to represent the imaginary part, very similar to complex number A+ jB, where B represents the imaginary part of the number.
sir please some practice questions
I have already given in the description along with the solution. Please check the description.
I have collected it... I want some more both questions nd answer.. pls help sir.... nd pls suggest text buk
If you want to solve more problems then I would suggest this book.
Network Analysis and synthesis by Ravish R Singh
www.amazon.in/Network-Analysis-Synthesis-Ravish-Singh/dp/1259062953/ref=sr_1_1?ie=UTF8&qid=1513183137&sr=8-1&keywords=ravish+singh+network+analysis
thanks sir... if u have some material pls put it
Sir phasor diagrams in practical and ideal transformer,, 1st year
As soon as I get the time then I will definitely make a video, or at least I will update the note on it in the description.
Thanks :)
V(t)=220 sin314t
impedance of an AC circuit is a..........
A) phasor B) vector. C) scalar
Ka ka Kachi daze
Ratio of two phasors = V/I
Thank you
How Vl-Vc came
If XL > Xc, then VL>Vc. Because they are connected in series and the same current is flowing through XL and XC. So, in the phasor diagram, the net voltage due to Xc and XL would be VL-Vc.
In last example
Plz replay
R-L-L
R-C-C
Heard the intro, then heard a guy's voice
(still great vid tho lol)
nice thanks alot
Thank you so much sir....🙏🏻🙏🏻🙏🏻🙏🏻
Sir please how 4+j3=5
for magnitude we take underroot of( 4square + 3square)
for angle we take tan inverse of (3/4)
What is j??
It's used to represent the imaginary part. Similar to A+jB in the complex number. Where jB represents the imaginary part.
@@ALLABOUTELECTRONICS thanks