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Slove this question Find the value of Vo/Vd for the discontinuous mode of conduction for a Buck converter Find the value of Vo/Vd for the discontinuous mode of conduction for a Buck converter
Those are the unique one .. I have to say really.After watching that much of detailed analysis i guess the examiners will lag behind and for us that will be very lucky. Thank you sir
all the lectures of your something new to be listen to me.In this lecture "the voltage across the switch and diode are drawn " . which is very easy to understand but the page is not sufficient for you so i understand it.Now the answer for the question is i.500amps ii.286.57amps iii.62.5amps iv.64.5amps v.180.9amps vi.176.77amps
Outstanding teaching..sir...After listening ur classes..I felt PE is easy and me intersteting one...I really thank you so much sir..It helped me alot...
Sir very good concept it is .....It's really helpful to us.....To day whatever concept in gathered in power electronics because of u and pankaj sukla sir .....
my ans for R load inverter home work 1=1 2=405.68 watt 3=812.18watt 4=500watt 5=1000watt thankuu sir very much i really understood the concept of inverter i have solved homework by myself in very less time i feeling confident thankuuuuuuuu sir
Nice Sir.. 1) IL(peak)=500A 2)Fundamental I(peak)=405.2847A 3)Average Value of thyristor current=62.5A 4)Fundamental average value of thyristor current=64.503 5)RMS VALUE OF THYRISTOR CURRENT=144.337 6)FUNDAMENTAL RMS VALUE OF THYRISTOR CURRENT=143.290
Sir please cover syllabus of power electronics as soon as possible ...my request don"t delay in uploading videos... Bcoz I am doing self study and your students should be able to compete compete coaching students..
Hello sir, thank you so much for such efforts... I just wanted to share a trick, every symmetric waveform ( square wave, triangular, sinusoidal) when divided in half with axis of symmetry, rms value gets divided by √2. Ex. Take full wave rectifier rms V/√2. Its half i.e half wave rms V/2 Its half like the one in video V/(2* √2). This is valid for all waves... Thank you
What a analysis sir ... respect and salute to u sir for #HWR concept to calculate RMS and average.....very keen observation..thanks a lot sir ...12th tak homework nhi kiya but ab wait krta hu ki kab new homework milega...bcz of u sir Homework answer: 1) 500 A 2) 202.64 A 3) 62.5 A 4) 79.577 A 5) 102.062 A 6) 71.644A Sorry for editing bcz i forgot to mention one answer and also in 4th one i put i1(max) instead of i0(max)
Sir in your solution why u are taking Vdc/2 = 100 so 2Vdc= 200V so why all of u putting 2Vdc = 400V kindly check homework 2nd part...plz sir whether i m correct or not??
Got it my mistake...i was mistaking...during fourier simplification u already replaced Vdc=Vdc/2 for square wave of amp. Vdc/2... that's why my answers were wrong....thank u sir...pardon me for silly mistake...
Peak value of Inductor (Load) Current = 500 A Peak value of Fundamental Component of Load Current = 405.285 A Average value of Thyristor Current = 62.5 Amp Average value of Fundamental Component of Thyristor Current = 64.503 A RMS value of Thyristor Current = 144.34 Amp RMS value of Fundamental Component of Thyristor Current = 143.189 Amp
Sir at 18:58 RMS value of thyristor can be easily found sir.. there is a shortcut.. if we want to find the RMS value of any triangular wave then just follow this below method.. Io(rms)^2 = { (1/3)*(amplitude^2)*(duration for which waveform is present) }/Total time period Here it will be like.. Is(rms)^2 = { (1/3)*(Iomax^2)*(T/4) } / T Therefore Is(rms) = Io(max)/(2√3)
Ramesh barnikala.. no man it's not area.. here I've taken square of amplitude as well I have divided the equation by 3 also.. in case of triangular wave we do like this.. we divide it by 3.
I have one major doubt, When we take average of any wave we consider the portion of wave in one time period ,so why in the case of switch and diode you only took the positive wave ,there was negative too in one cycle .. Please help
hello sir thanku for providing such great lectures , i have a douubt while calculating the fundamental power from each source , that is i think whilwe calculating it , it should have been fundamental component of voltage source rather than the average value of the source . please elaborate on this
I think there will be a correction that fundamental component of current of thyristor is less than total current of thyristor as shown in the graphs according to sirs calculations
Sir according to you what will be more beneficial in exam either to mug up the formulas or to derive it there itself....plz answer sir....it will be of great help 🙏🙏🙏
Mug up may fetch you a good rank only if they ask you directly, and after that you need to attend Interview, how could you explain them if they ask u a small doubt why and how? Without knowing actual concept. Take a time and study completely
Sir one doubt in @6:15 sir agar diode on hoga to direction to top to bottom honi chaiye na bcz Anode current(Ia) is greater than the Cathode current(Ik)....and sir aap ye current bottom to top le rahe hai isme to (Ia< Ik) fir to diode off ho jayega na....... ye samj nhi aaya...please reply sir?
Heyy ,The shortcuts are good but I am wondering at 23:48 , why is that if we know the rms value of half wave rectifier , why does dividibg by 2 to get the answer for the fundemntal rms value of inverter with L load work .If I just divide by 2 the rms value of half wave rectifier , I cant get the answer , yet this works for the average value
sir i have one doubt........... in the formula of Io1(max) [max value of fundamental load current] do we need to take wo(fundamental freq) term in the denominator???? If yes, what is its value as per in the given homework???
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sir , what is the reason behind the voltage constant even for rl and rlc load
Sir please can help me
😭😭
Slove this question
Find the value of Vo/Vd for the discontinuous mode of conduction for a Buck converter
Find the value of Vo/Vd for the discontinuous mode of conduction for a Buck converter
Can help me
Short cut methods explained in this lecture are excellent. Witty,simple & easy to remember. Thank you sir.
I have not listen such a lectures till now. I have got a great source for my GATE preparation. Thank you Sir,.
Ur most welcome...share this channel with ur frnds/juniors so that they can also get benefit from this lecture...
Those are the unique one .. I have to say really.After watching that much of detailed analysis i guess the examiners will lag behind and for us that will be very lucky. Thank you sir
Thank you so much sir and short cut is super
1:500
2:405.29
3:62.5
4:64.5
5:144.33
6:143.29
Awesome analysis bro...great effort and i bet no coaching institute would provide atleast 40% of this crystal clarity explanation.
sir i first time learn this rms method.thank u sir.
all the lectures of your something new to be listen to me.In this lecture "the voltage across the switch and diode are drawn " . which is very easy to understand but the page is not sufficient for you so i understand it.Now the answer for the question is
i.500amps
ii.286.57amps
iii.62.5amps
iv.64.5amps
v.180.9amps
vi.176.77amps
Outstanding teaching..sir...After listening ur classes..I felt PE is easy and me intersteting one...I really thank you so much sir..It helped me alot...
Thankyou sir to give that trick, i really help a lot to remember the formula.
1.500 A
2.405.6A
3.62.5 A
4.64.49A
5.144.43A
6.143.4A
Thank you sir i clearly understood concept
Excellent shortcut method learnt today . Thanks sir for your valuable lecture .👍👍👍
Very good shorcut method sir... That you had explained in this video
Such selfless contribution ...wow
1. 500A
2. 405.28A
3. 62.5A
4. 64.503A
5. 144.33A
6. 143.289A
Thank u so much sir for clearing each topic.....
Extraordinary class.
Sir very good concept it is .....It's really helpful to us.....To day whatever concept in gathered in power electronics because of u and pankaj sukla sir .....
great shortcuts for free of cost very thankyou sir
Most welcome
Sir you are truly amazing, I have been following all your lectures, they are excellent.
my ans for R load inverter home work
1=1
2=405.68 watt
3=812.18watt
4=500watt
5=1000watt
thankuu sir very much i really understood the concept of inverter
i have solved homework by myself in very less time i feeling confident
thankuuuuuuuu sir
tq very much sir for great effort to make this sub quite easy..and hats off sir
Thanks sir .... your really great . .... your my favourite teacher..... shortcut method is outstanding Nd very much clear to understand ...
most welcome..share with ur frnds..that is what i want from ur side..all the best.
questions given at end are very imp as they help to retain concepts .Thanku for such a beautiful explanation
Sir I always hit like your lectures😁😀
Nice Sir..
1) IL(peak)=500A
2)Fundamental I(peak)=405.2847A
3)Average Value of thyristor current=62.5A
4)Fundamental average value of thyristor current=64.503
5)RMS VALUE OF THYRISTOR CURRENT=144.337
6)FUNDAMENTAL RMS VALUE OF THYRISTOR CURRENT=143.290
Thank you sir your videos helped me alot basically I have nil knowledge on PE but now I am able to solve maximum problems after watching your videos
Understood all concepts. Thank you sir.
Thanks sir...
1)500 A
2)405.28
3) 62.5
4) 64.5
5)
6)143.28
getting problem in 5th question sir..
I understood the short cut and the whole lecture also..tq for the free lectures u r providing for us sir..👍
1. 500
2. 405.28
3. 62.5
4. 64.5
5. 144.337
6. 143.288
Thanks a lot sir !!
excellent videos,understanding
nice lecture sir
1.500A
2.405.28A
3.62.5A
4.64.50A
5.144.337A
6.143.2897A
really sir not in any coaching notes.. thanks u sir. you r best,,,, i studied bhimbra book but not understand well.
most welcome..all the best for gate..
Fill this form for next subject : goo.gl/8TLLR2
Sir please cover syllabus of power electronics as soon as possible ...my request don"t delay in uploading videos... Bcoz I am doing self study and your students should be able to compete compete coaching students..
Outstanding VEDIOS 🙏👍
Excellent shortcuts
Thanks for short cut sir...i understood really easily...👍👌👌👌
Hello sir, thank you so much for such efforts...
I just wanted to share a trick, every symmetric waveform ( square wave, triangular, sinusoidal) when divided in half with axis of symmetry, rms value gets divided by √2.
Ex. Take full wave rectifier rms V/√2.
Its half i.e half wave rms V/2
Its half like the one in video V/(2* √2).
This is valid for all waves...
Thank you
Similarly for square wave
For duty cycle 1 its V
For half duty cycle its V/√2
Like that
correct
Very helpful video sir.
What a analysis sir ...
respect and salute to u sir for #HWR concept to calculate RMS and average.....very keen observation..thanks a lot sir ...12th tak homework nhi kiya but ab wait krta hu ki kab new homework milega...bcz of u sir
Homework answer:
1) 500 A
2) 202.64 A
3) 62.5 A
4) 79.577 A
5) 102.062 A
6) 71.644A
Sorry for editing bcz i forgot to mention one answer and also in 4th one i put i1(max) instead of i0(max)
wait for the next lecture..i will give u the correct solution
Sir in your solution why u are taking Vdc/2 = 100 so 2Vdc= 200V so why all of u putting 2Vdc = 400V kindly check homework 2nd part...plz sir whether i m correct or not??
Got it my mistake...i was mistaking...during fourier simplification u already replaced Vdc=Vdc/2 for square wave of amp. Vdc/2... that's why my answers were wrong....thank u sir...pardon me for silly mistake...
All your explanation really easy to understand. I am preparing GATE by self study. Thanks for your help sir
Sir salute to your efforts. But pls pehle ki tarah daily signal ka ek video pls daaliye na. Scoring subject hai jaldi khatam karna hai usko.
Thank you sir I understood all the concept..
Understood the concept clearly
Thanku sir🙏
thank u so much sir u really taught in an excellent manner thank u so much
Thank you sir. Your videous helping me a lot.
Peak value of Inductor (Load) Current = 500 A
Peak value of Fundamental Component of Load Current = 405.285 A
Average value of Thyristor Current = 62.5 Amp
Average value of Fundamental Component of Thyristor Current = 64.503 A
RMS value of Thyristor Current = 144.34 Amp
RMS value of Fundamental Component of Thyristor Current = 143.189 Amp
How u solved 5th one plss help I m not getting tht thanku in advance
at 18.50 if you shift the waveform to zero then it will be 0 to pi/4 ..?
Excellent short cut method.😀
This trick was really amazing 👍👍👍👍
Really good concept
1.500A, 2.405.69A, 3.62.5A, 4.64.6A, 5.62.59 KA,6.143.43A
1.imax=500
2.i01max=286.58
3.itavg=62.5
4.fundamental avg thy i=79.57
5.rms it=144.34
6.rms funda=176.77
Sir your' classes are better then ace academy classes
Sir I like the concept 😄😄
bohot hard sir,,,,................
Sir at 18:58 RMS value of thyristor can be easily found sir.. there is a shortcut.. if we want to find the RMS value of any triangular wave then just follow this below method..
Io(rms)^2 = { (1/3)*(amplitude^2)*(duration for which waveform is present) }/Total time period
Here it will be like..
Is(rms)^2 = { (1/3)*(Iomax^2)*(T/4) } / T
Therefore
Is(rms) = Io(max)/(2√3)
Ramesh barnikala.. no man it's not area.. here I've taken square of amplitude as well I have divided the equation by 3 also.. in case of triangular wave we do like this.. we divide it by 3.
1. 500 Amps
2.405.28 Amps
3.62.5 Amps
4.64.5 Amps
5.144.33 Amps
6.143.288 Amps
I have one major doubt,
When we take average of any wave we consider the portion of wave in one time period ,so why in the case of switch and diode you only took the positive wave ,there was negative too in one cycle ..
Please help
@yashwant yadav
Do you got the ans for thos
great explanation
Most welcome..
1)500 A
2)405.28 A
3)62.5 A
4)64.5 A
5)144.33 A
6)143.28 A
I am understand sir Ji
Good morning ,sir ji
Excellenttttttttttttttttttt Sir!!!!!!!!!!!!!!!!!!!1
well explained sir
concepts clear now...
Nice short cut
Thank you Sir
excellent sir.....
Thank u sooooooo much sir..
1.500 w
2.405.28 w
3.62.5 w
4.64.5 w
5.144.33 w
6.143.28 w
great sir
1 -500A
2- 406.09
3-62.5
4-64.66
5-144.33
6-143.59
hello sir thanku for providing such great lectures , i have a douubt while calculating the fundamental power from each source , that is i think whilwe calculating it , it should have been fundamental component of voltage source rather than the average value of the source . please elaborate on this
understood sir !
Thank you sir..!
I think there will be a correction that fundamental component of current of thyristor is less than total current of thyristor as shown in the graphs according to sirs calculations
Thank you very much sir
solve the homework as well..it will benefit u..all the best.
Sir very effective short cut thank you so much
most welcome
Sir in half wave rectifier, you said Io1(max) / pi
And in rms shortcut, you wrote Io1(max) / 2
Pi and 2 are contradicting????
Anyone reply
Much helpful....
Nice making video
1)500
2)405.28
3)62.5
4)64.5
5)72.16
6)143.28
Thank you so much sir ... it will help me alot...pls sir make video on induction motor also like transformers lectures....
solve the homework first
Yes sir
Nc short trick sir
Sir according to you what will be more beneficial in exam either to mug up the formulas or to derive it there itself....plz answer sir....it will be of great help 🙏🙏🙏
Mug up may fetch you a good rank only if they ask you directly, and after that you need to attend Interview, how could you explain them if they ask u a small doubt why and how? Without knowing actual concept.
Take a time and study completely
Thank u sir......
thank u sir...
Sir one doubt in @6:15 sir agar diode on hoga to direction to top to bottom honi chaiye na bcz Anode current(Ia) is greater than the Cathode current(Ik)....and sir aap ye current bottom to top le rahe hai isme to (Ia< Ik) fir to diode off ho jayega na....... ye samj nhi aaya...please reply sir?
1.500
2.405.285
3.1.25
4.79.577
5.144.33
6.176.77
1.500A
2.405.28A
3.62.5A
4.64.5A
5.144.3A
6.143.28A
Sirji ,when to take rms when to take avg not clearly understand pls clarify my doubt
Tq for ur help
For calculating power loss in resistor take RMS current, and for power delivered/consumed by a battery take average current
thanks sirji
Thyristor rms current ke liye ye bhi concept laga sakte he agr waveform half hua to value reduce by root2 times
Heyy ,The shortcuts are good but I am wondering at 23:48 , why is that if we know the rms value of half wave rectifier , why does dividibg by 2 to get the answer for the fundemntal rms value of inverter with L load work .If I just divide by 2 the rms value of half wave rectifier , I cant get the answer , yet this works for the average value
1. 500 A 2. 405.28 A 3. 62.5 A 4. 64.5 A 5. 144.337 A 6. 143.228 A
1.) 500
2.) 405.49
3.) 62.5
4.) 64.53
5.) 102.06
6.) 143.36
Sir...
How the diodes are turning on in the circuit?
How the inductor is charging and discharging in one cycle?
Sir will the rms value remain same if we shift it by 60deg instead of 90deg?...plz ans me..
1.500
2.1273.23
3.62.5
4.202.64
5.72.16
6.450.15
sir i have one doubt...........
in the formula of Io1(max) [max value of fundamental load current] do we need to take wo(fundamental freq) term in the denominator???? If yes, what is its value as per in the given homework???
guru dutt u have to take each term fundamental only..here Wo=2pif
Iomax=500;Io1max=405.28, Itavg=62.5,It01avg=64.502, It01rms=143.288