h−Parameters (Solved Problem)

[@sutejas7333]

Im passing this sem due to NESO hatsoff we r paying fees to college and getting knowledge here

[@cholomsury1548]

Absolutely this Guy keeps things easy

[@newbie8051]

Same bro sameeeeeeeeeeeeeeeeeeeeeee

[@nagarajusonti7773]

👍👍

[@PrasannHalagall-pp1qu]

It's truth of student life

[@bruce_lee_brawlstars2564]

Very true

[@sankhamitra3862]

Great explanation sir with great concept clearing teaching. Keep doing such lectures and will be so beneficial for me if u kindly increase the frequency of lectures uploading on control system. H.W. ans h11=19/5Ω, h12=2/5, h21= -18/5, h22= 1/5 s. Thank you 😊.

[@coderfek]

KZbin is the largest platform to avail anti-premium courses, guided by professionals and this channel is a living proof. Thanks a lot NESO Academy!

[@sasha751]

In the hw ques...if u r not able to solve......the current in top left resistance (1ohm) is i1 and in bottom left 2ohm resistance it will be 5i1+i2 and in top mid 2 ohm resistance it is 4i1+i2 and in the top right 1ohm resistance it is i2 multiply these current value with resistance and you will get 2 eq which are V1=11i1+2i2 V2=18i1+5i2 You will get z parameters frm here just convert them to h

[@maheshsaiprudhvinadhvanumu6333]

tq

[@himanshu010]

Thanks a lot f

[@rohanram7197]

Hunny Desi Gaming 873 very kind of you thank you ☺️

[@GagandeepSingh-mm3br]

Thnks bro

[@piyushjain6909]

tq

[@Shounen12345]

*HOMEWORK* *IN LOOP ONE OPEN* Entering current = I1 Exiting current = I1 *MIDDLE LOOP CLOSED *:- *Upper branch across 2ohm* Current = 4I1 + I2 *Left 2ohm branch* Current= 4I1 + I2 + I1 *Lower without resistance* *branch:-* Current= 4I1 + I2 *LOOP 3 OPEN:-* Entering current = I2 Exiting current = I2 *EQUATION FOR Z* *parameter:-* V1 = 11I1 + 2I2 V2 = 18I1 + 5I2 *Rest simply use the* *relation to convert Z* *parameter To h parameter*

[@sakshichaudhary9891]

Can you pls explain that how equation of V2 came?

[@aniruddhc4613]

@@sakshichaudhary9891 You use the concept of supermesh 1. Write the current through every element while the dependent current source is present 2. Remove the dependent current source 3. Middle and right loop become one single loop 4. Solve normally with KVL

[@nandagopankm9280]

why doesn't the current in the left side resistor have (5I1+I2)

[@mohankishoresuru7228]

H11=3.8,H12=0.4,H21=-3.6,H22=0.2

[@anshul8258]

Answer: h11 = 19/5 h12 = 2/5 h21 = -18/5 h22 = 1/5

[@kalpeshganvir7473]

CAN U TELL HOW IT IS 19/5 FOR H11

[@priyakhurana5274]

@@kalpeshganvir7473 its determine of |z| =19 divided by value of z22 = 5

[@aryandhakad8495]

I m correct

[@anishsharma5097]

thats correct!

[@sarthakvaish2520]

correct

[@YashPalF-]

Sir thanks for giving us homework problem 😊in every lecture

[@archanagowda1285]

Sir...u are my saviour sir...i just thank u a lott sir...😁😁 Coz of u I can solve my NT with confidence...

[@thechor800]

Thank you for the wonderful explanation ! Now , please make vedio on abcd parameters with suitable example.

[@prashantsharma1592]

Can we solve H.W question by using nodal analysis. Using V1 and V2 node thus finding value of Y parameters . Then convert Y parameters to H parameters

[@tus206]

thank you for your effort for teaching us such very informative lesson. I know how to calculate this type of excercise from now

[@vishwasai4325]

Can you please provide the multistage amplifier concepts and problems.we are very much thankful to u sir for providing this online class.we hope you provide that what we have asked for...

[@alphasatari]

Thank you Master Ji.

[@chaithanyasaiparuchuri7104]

Thanks a lot for these videos , they are very helpful ...

[@samreensadiya6638]

These neso videos are amazing, thanks a lot for these videos

[@auroshisray9140]

Thank you Sir

[@dhruvgarg2003]

Why you didn't added 4I1 to I2 current in 2nd loop?

[@varma4545]

sir can u explain one more question on h parameters..

[@srikanthmedaboina7184]

Please make video on thevenins theorem

[@ankshitmaji2399]

Thanks sir for helping us pass our sems..we will be always greatful

[@priyashrivastava2230]

Sir u r really awesome Your lectures help me alot

[@sunilprajapati5183]

h11=19/5 h12=2/5 h21=-18/5 h22=1/5

[@kishanchaurasiya9352]

Kya ye lec+ practice is enough for gate exam? Please reply sir

[@victorba-murielprincewill7006]

I may be wrong but in the loop 2, I solved myself and it was in form of V2 + 2I2 + 4I1 -10(I1+I2)

[@Zoneofk2]

Homework questions answers:-h¹¹=19/5 ,h¹²=2/5 ,h²¹=-18/5 ,h²²=1/5 ... Please reply

[@IndiLores709]

You are right 👍 I have verified using both methods

[@prakashsharma5218]

Anybody please send the solution stepwise

[@ritwikdasmohapatra9476]

Why is 4I1 current not added with I1...and instead only I2 is added...it shld me 5I1 + I2...please clear this.

[@BECSAQUIB]

B/c it is voltage source not current source..

[@safiyamp8305]

Good class

[@ruchitprajapati]

Amazing video. Thank you sir.

[@yashroy7935]

Sir aap bhagwaan ho bhagwaan

[@sarangkumarsingh7901]

Awesome 👌...

[@anirudhyaguchait5842]

Ans: h11 = 19/5 ohm, h12 = 2/5, h21 = -18/5, h22 = 1/5 mho ...😁😁🤗🤗..

[@ogundaresundayomotayo2718]

Your teaching skill is great sir I always enjoy your lecture but I will like to tell you that if you are replying to people comment it would have been more great thanks sir

[@vishalsharma-fd2qm]

Sir plzz complete 2 port netwrk ASAP.

[@deepthimouthkur3494]

Tqsm sir.....its very helpful to us😊

[@saikatkundu9214]

Sir , why from the value of V1 and V2 comparing with Z parameter you didnt make the answer ??

[@knowledgesea6798]

but 4 I1 should also go through the 10 ohm resistance

[@ankurkhandelwal9706]

That is a voltage source not a current source.

[@GaneshKumar-xz6gl]

lectures are great now i can crack placement exams easily

[@saadshahab1403]

hw ans = 19/5, 2/5, -18/5, 1/5 Thanks @nesoacademy

[@nitishkumarprajapati6096]

Thanks Sir 😃😃

[@hubbanadnan4718]

why are we not using the dependant source?

[@RahulKumar-kx8wv]

h11=3.8 ohm h12=0.4 h21= -3.6 h22=0.2 mho

[@KotanaPhaniSriRahulAP]

h11 should be equal to z11 na. why iws it not equal?

[@veerayaragani1776]

Wt about theorems and remaing sir

[@basavarajumeghana2101]

sir give some more problems to solve

[@angelojunior9507]

Good afternoon, what program do you use to write the equations on the screen?

[@chizyniggaosadebe4405]

h11 = 3.8, h12 = 0.4, h21 = -3.6, h22 = 0.2

[@bilalberkdemir9745]

you are excellent

[@akashsinghchauhan1570]

sir can you give me a clue how to do it i'm little bit confused

[@dharmeshparmani8473]

Apply the output KVl taking the 1 ohm , 2 ohm and 2 ohm resistance path into account . Neglect the current source path . You will get V2 = I2 +[ (4I1+I2)*2 ] + [(5I1+I2)*2] ... V2 = 18i1+5i2 ... V1 = 11(i1)+2(i2)

[@anishgarg6193]

@@dharmeshparmani8473 why we neglected the current source (4i1) path......

[@subhrajitmahanti3758]

@@anishgarg6193 coz we are finding out z parameters and we dont know the voltage drop across the current source

[@srujanreroots8412]

What u took in place of Z for getting 8 ans

[@techinfosupport8314]

Answer:- h11=3.8,h12=0.4,h21=3.6,h22=0.2

[@mohanaks5554]

In equation 2 How it come 6i1 instead of 10i1

[@Ceyhus]

4:30 I wonder, why i2 is not: "... 10 * ( i1 + i2 + 4 * i1) ... " but it is: "...10 * ( i1 + i2)..." while in V2 it is "... 2* i2+ 4 * i1

[@shajahan1064]

the 4 i1 is voltage source,not current source

[@chennamsettysumanth6732]

What about 4i1 above

[@WhiteElephantTv]

Sir I'm a student from Usman Institute of technology your videos and uploads are helpling entire nations . I request you to cover more courses on electrical engineering . MAY GOD BLESS YOU SIR . I REQUEST YOU SIR TO FULLY COVER THE COURSE LIKE ELECTRICAL NETWORK ANALYSIS ONLY THEORY AND Mechanical engineering to please sir

[@sauravsuman6316]

Sir Plz give me hint of given question Anyone who solve plz give solution

[@ogundaresundayomotayo2718]

For the homework Am getting my H11=7 H12= 2/5 H21= -2 H22= 1/5

[@pandeyraj2171]

h21 =-18/5 no unit h11=19/5 ohm h22= 1/5 mho h12= 2/5 no unit Have a nice day to all of you... ----Best of luck for your -------exam For those who are seeing this vdo at last moment.😊

[@-Lauren-gs8mm]

Can you please help me with the answer I'm unable to get can you send it to my mail the complete answer step by step please....my mail ([e-mail])

[@kunnu_4]

9:43 / 10:22

[@aditimishra1718]

h11- 3.8 h12- 0.4 h21- 3.6 h22- 0.2

[@nishitmeena4791]

right

[@bhaskaralakshmiganisetti3779]

@@nishitmeena4791 no that's wrong h21= -3.6

[@nishitmeena4791]

Yaa h21=-3.6

[@ganeshmula4508]

H. W ans h11 = 19/5 h12 = 2/5 h 21 = - 18/5 h22 = 1/5

[@shikhonahmed8186]

Alhumdullia solved

[@avunooritagore6089]

6ohm=h11 2/5no units=h12 -2no units=h21 1/5mho=h22

[@Softwareengineeringdev]

good ♥♥♥

[@debasmitaadak8880]

Why is it +4I1.....why not -4I1..?? 3:21 3:21

[@itik.6549]

Going anti-clockwise in loop 2, you get out of the +ve terminal , that's why+4I1

[@JEDhanraj]

(19/5,2/5,-18/5,1/5)

[@kvmsaikumar5865]

19/5,2/5,-18/5,1/5

[@anishsharma5097]

we want neso institute of technology so that i can do m.tech

[@muhammadsamiullah6614]

0:20

[@022_firdousnabi5]

h21=-18/5 h22=1/5 h12=2/5 h11=19/5 This is the answer

[@shishirallu3439]

H11 = 19/5. H12=2/5. H21=-18/5. H22= 1/19.

[@riyazveersingh8330]

h parameters {3.8 0.4} {-3.6 0.2}

[@tusharrohilla7154]

find z parameter and then h

[@syedismail1849]

❤️❤️❤️

[@sudheerkulal3784]

*h11=1* *h12=0.25* *h21=2* *h22=0.5* **Mic drop**

[@s.m.senses7199]

🙏❣

[@nagamahendrareddybogala6415]

Thank you sir

[@storm_breaker1312]

h11 =19/5 h12 =2/5 h21= -18/5 h22= 1/5

[@nitishkumarprajapati6096]

Thanks Sir 😃😃

[@renishsurani]

Answer: h11 = 19/5 h12 = 2/5 h21 = -18/5 h22 = 1/5

[@alphasatari]

h11=3.8 ohm h12=0.4 h21=-3.6 h22=0.2 mho

[@dharmeshparmani8473]

Can you explain how you got the answer

[@manojknowmore]

19/5,2/5,-18/5,1/5

[@nagamahendrareddybogala6415]

h11=19/5,h12=2/5,h21=-18/5,h22=1/5

[@prashantranjan2282]

can you please tell me the solution

[@eldrago4321]

@@prashantranjan2282 at 1R =i1 ,, at horizontal 2R=i2+4i1 ,, at vertical 2R=5i1+i2 ,,, using these apply KVL in 2 loops and obtain h parameters

[@santanubiswas8200]

h11=3.8 h12=0.4 h21=-3.6 h22=0.2

[@dharmeshparmani8473]

Can you please explain it in detail

[@shouryaprasad2088]

Yup mate!!!! Same here

[@shouryaprasad2088]

Just compare the impedance equation to the h-parameter equation. Any further hint will make it very easy

[@nstcg7271]

Same here also..

[@venkateshmamidi8407]

h11= 19/5 h12= 2/5 h21= -18/5 h22= 1/5

[@priyakumarisinha1106]

h11=19/5 h12=2/5 h21=-18/5 h22=1/5

[@عبداللهالوحداني-ش6ذ]

Can you show me your solving

[@rishabhsisodiya877]

h11=19/5 h12=2/5 h21=-18/5 h22=1/5

[@manavmalhotra8513]

is this is correct?

[@nishantvedwal6244]

Can you please post the solution

[@sunshineacademy]

h11=19/5 h12=2/5 h21=-18/5 h22=1/5

[@afjalhossain1497]

h11= 19/5 h12= 2/5 h 21= -18/5 h 22= 1/5

[@reficeismail6520]

h11=19/5 h12=2/5 h21=-18/5 h22=1/5

[@adari.kumarswamykumaradari7423]

Can u show me the answer in a paper

[@zahidhasankhoka1649]

h11=19/5 ohm h12=2/5 h21= -18/5 h22=1/5 mho

[@Vrikodara9]

h11=3.6 h12=0.4 h21= -3.6 h22=0.2

[@nishitmeena4791]

wrong

[@Vrikodara9]

@@nishitmeena4791 please give me answer

[@nishitmeena4791]

Only h11=3.8

[@cypher_ayan]

h11=3.8 h12=-3.6 h21=0.4 h22=0.2