I figured out the 2Idc^2 part. In Idc*I, Idc is a constant so it will come out of the integration. Now write I as Iac+Idc. Integration of Iac will be zero over a full cycle. Hope it helps. This is also why the average value is Idc.
As Idc is constant , Idc(2/2π) integral 0 to 2π I(dwt) To solve :- Integral 0 to 2π I(dwt) 0 to π I(dwt) + π to 2π I(dwt) As for half wave rectifier π to 2π I(dwt) =0 We only left with 0 to π I(dwt) As I=I' sin(wt) Where I' is peak value of output current Such that Integral 0 to π I d(wt) become integral 0 to π I' sinwt d(wt) I' is constant Therefore by integration of sin(wt) we get (-cos wt ) and putting the limits we get value is 2 Thus the value of Integral 0 to 2π I (dwt) = 2I' Thus final value we get is =Idc(2/2π)2I' =Idc(I'/π) As I'/π=Idc ( average value of output dc current) Therefore we get at last is (Idc)^2 🙂🙂
I just want to say your channel is a blessing on UA-cam. Every time I need to quickly revise a topic I always find your videos. Well organized and comprehensive. Thank you so much for all the work.
Integration of I² from 0 to 2π wrt (wt) divided by 2π is (Irms)². "Integration of I from 0 to 2π wrt (wt) divided by 2π is Idc. " :quoted part Written in Millman's electronic devices and circuit page 6.8 Thats explains the third part of the integration. Note:" I " here is output Aslo, Idc=Iav
For everyone who is asking how I×I(dc)=I(dc)^2 Since I & I(dc) both are constants we can take I(dc) out of integration then the remaining integration containing only "I" is the formula for I(dc) . Therefore it gives I(dc)^2
in the integration of the last term , we consider Idc as constant , so we have inside the integral I*Dwt and it give us Iav which is equal to Idc , so the result is 2*Idc^2
You are the only man in whole UA-cam and google who described every step of derivation at best way.... May Allah give you more knowledge and success in your life
How can F.F=I(rms)/I(dc) when in the last lecture Form Factor was defined as F.F=V(rms)/V(av)? Didn't understand how you got I(dc) as 'Average value of output'?
In the Idc*I integration, Idc is constant so its left out..leaving the integration of I.dwt..now that is equal to the average value of I which is also the dc component i.e Idc. Therefore integration of I.dwt=Idc.. giving us Idc^2 after multiplying with the Idc constant..
How you got 121% percent of ac component,I think it can be greatest for AC and it should be 100% for that,so how rectified output has more AC component??
I=I(ac)+I(dc); I(ac)=I(m)sin(wt); If you integrate sin over its period it results in 0. So integration over I(ac) is 0 and we're left with I(dc) component of I
@@abhishekkumaryadav1928 I may not be very clear in explaining, but integrating I(ac) over its period will result in zero. Please refer to the definition of alternating voltage, it periodically varies its polarity and changes its magnitude continuously. So whether you consider sinosudal or rectangular or triangular signal, Its Integration over its period will come out to be zero. Also I(dc) may or may not be a constant wrt time, A pulsating voltage may also be dc, refer to wikipedia if necessary.
@Vinay Kumar Reddy sine waves int. become zero when we consider the int. from 0 to 2pi but in rectified output ac component exist only for 0 to pi range then it becomes dc then again ac component arises from 2pi to 3pi so i am a bit confuse about your explanation
@Vinay Kumar Reddy yes i had Fourier in my last semester but how far I understand is we cancel sine part when we calculate the avg from 0-2npi. But here in output we get a sine wave from 0-pi then a straight line(dc part) from pi-2.pi ....
sir u said RMS value is equivalent to the value of effect created by dc value..but in derivation u have replaced the dc value by AVG value..:( please make me to understand
In the last lecture you told that rms value of ac is equal to the value of dc. So in this lecture how is av o/p current(Iavg) taken as Idc i.e (Iavg=Idc)? According to the last lecture rms o/p current(Irms) should be equal to Idc i.e (Irms=Idc) right? This is proving that Irms and Iavg is equal which is not equal in practice. Please help.
initially even i was confused,then i came to know that average value is equivalent to dc component of current since the opposing peaks of the ac components get cancelled out
in an earlier video you said that 5v rms ac will glow a bulb equivalent to that of 5v dc but here ratio of I(rms) / I(dc) is not 1, I(dc) seems to be equivalent to I(avg), Why so?
I can't understand how can the ripple factor be more than 100%. Ripple factor is the percentage of AC in the output, so if you're taking the entire output to be, say, 100, how can the AC component in that output be 121? I googled it but coudn't find any explanation. Please explain
IN Last lecture you told that I(rms)=I(dc) value but in this lecture you are telling that I(av)=I(dc) i am confused now ...!!!!! Help somebody.....!!!!!!
Conversion efficiency of half wave rectifier should be ratio of dc power output on secondary side to ac input power on primary side,which comes to approx. 20 per cent......in your video efficiency is ratio of dc power output at load ,to ac input power at load ,which is 40percent,..... please shed some light on definition of efficiency and why this discrepancy
As Idc is constant , Idc(2/2π) integral 0 to 2π I(dwt) To solve :- Integral 0 to 2π I(dwt) 0 to π I(dwt) + π to 2π I(dwt) As for half wave rectifier π to 2π I(dwt) =0 We only left with 0 to π I(dwt) As I=I' sin(wt) Where I' is peak value of output current Such that Integral 0 to π I d(wt) become integral 0 to π I' sinwt d(wt) I' is constant Therefore by integration of sin(wt) we get (-cos wt ) and putting the limits we get value is 2 Thus the value of Integral 0 to 2π I (dwt) = 2I' Thus final value we get is =Idc(2/2π)2I' =Idc(I'/π) As I'/π=Idc ( average value of output dc current) Therefore we get at last is (Idc)^2 🙂🙂
Average value represents - DC content present in the AC (That's why av. value is zero for pure AC), RMS value represents - DC equivalent of AC content. am I correct?
Can somebody please tell me how to derive the voltage regulation of half wave rectifier, full wave center tap and full wave bridge rectifier? Plz plz plz
I figured out the 2Idc^2 part. In Idc*I, Idc is a constant so it will come out of the integration. Now write I as Iac+Idc. Integration of Iac will be zero over a full cycle. Hope it helps. This is also why the average value is Idc.
thanks....
but this case is for half wave rectifier right?
so the integration of Iac in full cycle isnt zero.
Yashodhan Manerikar thanks
iac would not be 0 because we take the mode of integration it will add the mag. of +ve cycle and -ve cycle
@@dwinovianto1250 it isn't exactly 0,but it's small when compared with other terms.
As Idc is constant ,
Idc(2/2π) integral 0 to 2π I(dwt)
To solve :-
Integral 0 to 2π I(dwt)
0 to π I(dwt) + π to 2π I(dwt)
As for half wave rectifier
π to 2π I(dwt) =0
We only left with
0 to π I(dwt)
As I=I' sin(wt)
Where I' is peak value of output current
Such that
Integral 0 to π I d(wt) become
integral 0 to π I' sinwt d(wt)
I' is constant
Therefore by integration of sin(wt) we get (-cos wt ) and putting the limits
we get value is 2
Thus the value of
Integral 0 to 2π I (dwt) = 2I'
Thus final value we get is
=Idc(2/2π)2I'
=Idc(I'/π)
As I'/π=Idc ( average value of output dc current)
Therefore we get at last is (Idc)^2 🙂🙂
Thank you 😊
thank you
thank u
thanks a lot
Thanks a lot dude
I just want to say your channel is a blessing on UA-cam. Every time I need to quickly revise a topic I always find your videos. Well organized and comprehensive. Thank you so much for all the work.
Integration of I² from 0 to 2π wrt (wt) divided by 2π is (Irms)².
"Integration of I from 0 to 2π wrt (wt) divided by 2π is Idc. " :quoted part Written in Millman's electronic devices and circuit page 6.8
Thats explains the third part of the integration.
Note:" I " here is output
Aslo, Idc=Iav
Sir, you are a saviour. Thanks a lot for making these videos. We owe you a lot.
Doing great job guys,taking all points short and crisp factor.
For everyone who is asking how I×I(dc)=I(dc)^2
Since I & I(dc) both are constants we can take I(dc) out of integration then the remaining integration containing only "I" is the formula for I(dc) . Therefore it gives I(dc)^2
in the integration of the last term , we consider Idc as constant , so we have inside the integral I*Dwt and it give us Iav which is equal to Idc , so the result is 2*Idc^2
You are the only man in whole UA-cam and google who described every step of derivation at best way....
May Allah give you more knowledge and success in your life
Integrating I.Idc, how you got I^2dc?
he assume that Iac=0
so I=Idc+Iac=Idc+0
thereby , I=Idc
that is what i know
so we can apply that in 1st integration to.
i don't think you are correct
How can F.F=I(rms)/I(dc) when in the last lecture Form Factor was defined as F.F=V(rms)/V(av)?
Didn't understand how you got I(dc) as 'Average value of output'?
In the Idc*I integration, Idc is constant so its left out..leaving the integration of I.dwt..now that is equal to the average value of I which is also the dc component i.e Idc. Therefore integration of I.dwt=Idc.. giving us Idc^2 after multiplying with the Idc constant..
thanks bro
Thank you!!
thanks bro
How you got 121% percent of ac component,I think it can be greatest for AC and it should be 100% for that,so how rectified output has more AC component??
@6:02 integration you are directly writing , I am not getting it ! how Idc square comes ?
Sir in the last lecture you didn't told about form factor =I RMS/Idc.in you told FF=V RMS/V avg
Please anyone help me
Same doubt
how could you get I(dc)^2 from just multiplying I times I(dc) -in the third part of integration - ?
I=I(ac)+I(dc);
I(ac)=I(m)sin(wt);
If you integrate sin over its period it results in 0. So integration over I(ac) is 0 and we're left with I(dc) component of I
@aayush
Thank you very much
@@ayushkumar-xk7kp dont try to fool them.. his doubt is correct.. it cant be 0 because.. in positive half cycle it doesn't comes out to be zero
@@abhishekkumaryadav1928 I may not be very clear in explaining, but integrating I(ac) over its period will result in zero.
Please refer to the definition of alternating voltage, it periodically varies its polarity and changes its magnitude continuously. So whether you consider sinosudal or rectangular or triangular signal, Its Integration over its period will come out to be zero.
Also I(dc) may or may not be a constant wrt time, A pulsating voltage may also be dc, refer to wikipedia if necessary.
@@balaji2.095 first of all the integral will be -cos(wt)/w. (Though it doesn't matter)
And after you put limits the integral will vanish(1-1=0) the
Doubt at 6:01 goes did you get it
And how will the percentage be 121 generally percentage men's out of 100
Sir,
How come avg. Value of output is Idc?
@Vinay Kumar Reddy ❌
@Vinay Kumar Reddy Thankyou, I understood 85% of what u said..
@Vinay Kumar Reddy
Which year student U r???
U have good knowledge..
Fourier series is currently going on in our class.
@Vinay Kumar Reddy sine waves int. become zero when we consider the int. from 0 to 2pi but in rectified output ac component exist only for 0 to pi range then it becomes dc then again ac component arises from 2pi to 3pi so i am a bit confuse about your explanation
@Vinay Kumar Reddy yes i had Fourier in my last semester but how far I understand is we cancel sine part when we calculate the avg from 0-2npi. But here in output we get a sine wave from 0-pi then a straight line(dc part) from pi-2.pi ....
Sir ur videos are great!!!!
can we find the Ripple factor also with Voltage?
sir how avarage value of the output will be Idc ??
sir u said RMS value is equivalent to the value of effect created by dc value..but in derivation u have replaced the dc value by AVG value..:( please make me to understand
I can't understand the same point ...how can you replace the avg value with dc value ? Sir please explain it...
Heat generated in time T is heat generated for T/2 plus zero hence it comes out to be v/2
In the last lecture you told that rms value of ac is equal to the value of dc.
So in this lecture how is av o/p current(Iavg) taken as Idc i.e (Iavg=Idc)?
According to the last lecture rms o/p current(Irms) should be equal to Idc i.e (Irms=Idc) right?
This is proving that Irms and Iavg is equal which is not equal in practice.
Please help.
No he told rms value is the value of supply so that the supply would give power or brightness to lamp connected to it
initially even i was confused,then i came to know that average value is equivalent to dc component of current since the opposing peaks of the ac components get cancelled out
@@aditipandey6964 how Iav= Idc?
well explained sir!! 😍
How we get - 2 I^2(dc )??
in an earlier video you said that 5v rms ac will glow a bulb equivalent to that of 5v dc but here ratio of I(rms) / I(dc) is not 1, I(dc) seems to be equivalent to I(avg), Why so?
There's a mistake in the final calculation , it's 1.1 not 1.21 , you probably forgot a bracket while using the calculator
But i think that it comes 0.57 instead of 1.21 or 1.1
@@shivamkashyap1543 it's 1.1 dude, I think it's written in R Murugesan's modern physics, not sure, but the final value is 1.1.
Thank You Sir
thanks for ripple factor theory
I don't understand ripple factor
Thanks Bro!
Very good explaination thank you bro
Thanks a lot sir
I can't understand how can the ripple factor be more than 100%. Ripple factor is the percentage of AC in the output, so if you're taking the entire output to be, say, 100, how can the AC component in that output be 121? I googled it but coudn't find any explanation. Please explain
ripple factor is the ratio of % of AC comp in output to the DC comp. AC comp is 121% of the DC value not the entire value. Hope it helps.
shubham maniyar yes I'd figured that out before my exam. Thanks anyway :)
Oh. Great thanks guys for solving my doubt !
shouldn't it be rms value of ac component
I didn't get the 3rd part of the integration which you skipped. Please explain that.
I is a sum of iac and idc for 0 to 2pai interval Av of iac is zero but integration of IDc over dwt in intervel o to 2pai is 2pai into idc.
IN Last lecture you told that I(rms)=I(dc) value but in this lecture you are telling that I(av)=I(dc)
i am confused now ...!!!!!
Help somebody.....!!!!!!
@Vinay Kumar Reddy what do you mean by DC component of the wave?
@Vinay Kumar Reddy I don't understand yet...can u please elaborate that Fourier series part
Can you have any videos about full wave rectifier??
Hi sir,thanks for such a helpful lecture....bt i m little bit confused .....how the average current is equal to Idc(output has also ac current)......
Conversion efficiency of half wave rectifier should be ratio of dc power output on secondary side to ac input power on primary side,which comes to approx. 20 per cent......in your video efficiency is ratio of dc power output at load ,to ac input power at load ,which is 40percent,..... please shed some light on definition of efficiency and why this discrepancy
well expained😍
You never answer to the comment section, then what is point of asking us to comment here?
this video was posted 8 years ago mate . he isnt gonna answer .its common sense
sir...i did not understand how u integrate I Idc d(wt).....3 term
As Idc is constant ,
Idc(2/2π) integral 0 to 2π I(dwt)
To solve :-
Integral 0 to 2π I(dwt)
0 to π I(dwt) + π to 2π I(dwt)
As for half wave rectifier
π to 2π I(dwt) =0
We only left with
0 to π I(dwt)
As I=I' sin(wt)
Where I' is peak value of output current
Such that
Integral 0 to π I d(wt) become
integral 0 to π I' sinwt d(wt)
I' is constant
Therefore by integration of sin(wt) we get (-cos wt ) and putting the limits
we get value is 2
Thus the value of
Integral 0 to 2π I (dwt) = 2I'
Thus final value we get is
=Idc(2/2π)2I'
=Idc(I'/π)
As I'/π=Idc ( average value of output dc current)
Therefore we get at last is (Idc)^2 🙂🙂
@@harjitkaur6729 finally we have Idc(2I'/π). How Idc(I'/π) ???
please some one explain me the clear difference between the RMS value and AVG value
Hii sir i am preparing for ese.. in ESE exam have they asked derivation of any of these ????
sir can you just increase your voice a little bit?
its amazing to learn from you
I didn't quite get how to do the integration of the part you skipped
The part with 2/pi integral from zero to 2pi of ( I*Idc *d(wt)) at the minute of 6. 02
i didn't get it too
integration from 0 to 2pi of ( I * dwt ) / 2pi = Im / pi = Iav = Idc , as for mixed signal Iav represent dc current
what does the mixed signal means
it's the average value of the ripple current I witch is equal to Idc
Can anyone explain what is meant by ac and dc component of output current?
Ripple factor is equal to Vp-p\Vavg
Thanks sir
L type LC filter? ?
Best sir...
Can we take Iac=Idc sin(wt)
I dnt understand how u integrated 2*(I total)*(I dc)
Please explain 6:05 further. the third expression inside the square root. thanks
This is so perfect !!! 😍
Average value represents - DC content present in the AC (That's why av. value is zero for pure AC),
RMS value represents - DC equivalent of AC content.
am I correct?
sir how the does output current have both ac and dc current?
I too have the same doubt
Im sorry sir, in 6:06
How did u get 2(Idc^2) cz depend on my note should be 2I(Idc)
thanks anyway.
exactly... , have the same ques too
we'll take Idc outside as a constant
I = I av + I dc
but integration of I av through complete circle is zero
then ...
Thanks :)
This video not properly understanding.sir please make one more video on this topic
what is differance between RMS and AVERAGE value....???????
Can somebody please tell me how to derive the voltage regulation of half wave rectifier, full wave center tap and full wave bridge rectifier? Plz plz plz
Can anyone tell me how I.Idc is equal to 2Idc sqr ?
Anybody understood the 3rd part of the integration!
I is sum of iac +idc av of iac for 0 to 2pai is zero.
Bro, wonderful. Got it!
Studying just for my mdcat exam held by pmc😂🤣👍
I is sum of iac and idc and av of iac is zero but average of idc is idc
av of Idc is idc because idc is constant but in rectifer av of iac is zero because we want only constant supply that is over requirement.
121℅ of 200℅ or what I'm confused121℅ of 200℅ or what I'm confused how it can be more than 100℅ how
121 or 12.1?
Gyama
F.f what is this