Y−Parameters (Solved Problem 1)

[@AbrahamAddisu-o4o]

Underrated Channel, Amazing and clear explanation. Go ahead Sir.

[@kapilumrania]

You're god sent sir!!

[@100courses7]

Sir, U make the problem-solving look so easy. Amazing job sir. Honestly, your channel is underrated.

[@divyarao6280]

After redrawing the circuit. We can use a short cut sir as it was a T network.... name the resistors as Ź a, zb and Zc...now Z parameters will be Za+zb Zb Zb Zb+Zc

[@meerashende9470]

Whats about -ve sign

[@meerashende9470]

How he gave -z12 n 21 we don't take like this whil calculating adj.

[@srisaiteja9375]

Love u sir 💜 from vizag

[@soar_hub]

Thanks so much for your lessons!

[@avethakur9198]

We can also solve this problem by applying KVL in delta loop in circuit to obtain value of i' in form of i1 and can put it in our z parameters equations

[@basarrimon3301]

Yess

[@VAISHNADEVIGECEUG-]

NICE AND DETAILED EXPLANATION SIR THANK YOU SIR

[@bruce_wayne8383]

The video is really informative .. Change the playback speed to 1.25x and thank me later

[@Priyajit_Ghosh]

bruh, I am in 2x speed

[@sirjantudu7553]

thank you so much for clearing doubts

[@Vijayvro777]

Super Teaching sir ......Amazing ✨

[@rupayansahaakapglucifer598]

The method shown is good but very lengthy. If you can convert this star network to delta or pi network and then directly using the short circuit admittance therom it will very easy and short

[@nitishkumarprajapati6096]

Thanks Sir 😃😃

[@venkattelugumusic472]

Thank you so much sir for your explanation mind blowing my doubt was cleared in jntua question paper the Z paramaeters are given but find y- Parameters like addmittence my mind is not working to see the question,after seen this video it's easly iam solved.thank you bro for these video 👌🙏🎇🎆😍😍

[@knoxxcruise6939]

Can we use I1=Y11V1+Y12V2 and I2=Y21V1+Y22V2 I

[@nani2543]

Directly we can solve through nodal analysis right ?

[@venkatramansampath]

Thank you Sir for the video

[@avijitdeyelectrical1551]

please sir upload more numerical problem on z , y parameter, h hybrids

[@sharmachavali]

sir for the above problem if we have delta connection of capacitor how to solve it sir ?

[@securitynandhmoni1553]

Which reference do you used🤔 please ,give a name

[@user-st9rw2vi3s]

I am getting Y11 = -Y21 = 31/260 and Y21 = - Y22 = 33/260

[@syedameenali7361]

Can we convert the inner star to delta and apply kvl?

[@starkendeavours7072]

yes you can, both are convenient though.

[@syedameenali7361]

@@starkendeavours7072 thanks 👍

[@gradient695]

this is also for aktu?

[@abhinavkarn5090]

After changing from Delta to star......what happened to those 10ohm and 5 ohm resistors

[@AcchuRocky]

All 3 resistors in delta is converted into 3 resistors in star form.. hence 20, 10, 5 ohms are disappeared to those 3 resistors..

[@sazidabintaislam2676]

In Z12 and Z21, how did get - sign?

[@levinraj2404]

its matrics rule

[@-MEET--mv1eb]

Sir why whole current i2 will flow through 5 ohm only I think i2 will be divided into two branches initially through 5 ohm and 20 ohm

[@rishabhprasad8333]

Same doubt

[@EshetuKase]

GOOD LICTUR CONTIUE

[@ganeshrajjoshi7559]

how can i solve if there is capacitor instead of 10ohm and 5ohm resistor

[@rachitmittal448]

Sir can we solve the problem using KCL Equation? If we assume lower terminal as ground. Then V1 and V2 will be voltages as upper terminals. 20 ohm resistor between v1 and v2. T Network of other three in v1mand v2. We can assume A as intersection. Write KCL at 1,2,A.

[@theaerogirl]

I got the same idea. Can`t we do that?

[@biswadeepchakraborty2460]

can we solve it directly using eq of Y parameters?

[@JaspreetSingh-er2qr]

same doubt i have

[@tanyashankar9966]

can anyone please explain why can't we calculate simply y parameters simply from the circuit and not through the inverse process?

[@happysingh6586]

i think we can do that also by using kirchoffs current law and write current equations

[@Evansdus]

Me I calculated by using circuit for addmittance without using impedance and I get the same answer. as follows: Y11=0.12,Y12=-0.11,Y21=-0.11,Y22 =0.12

[@karthikvenkata6484]

I am getting different answers as Y11=0.025,Y12=-0.015,Y21=-0.025,Y22 =0.035

[@Evansdus]

Real answer are: Y11=0.12,Y12=-0.11,Y21=-0.11,Y22 =0.12

[@hansrajbagaria794]

Sir why can't we take current(I2") through the 20 ohm resistor as : I2"= (V1-V2)/20. ??? Problem becomes very easy using the value of this current(I2") in terms of V1 and V2. I solved and got almost same answers not exactly the same. If anybody can tell my method is right or wrong? Thanks

[@yashmahalle5675]

Forget everything and use mesh analysis

[@yashmahalle5675]

Don't get complicated, use mesh analysis to calaculate current.

[@Evansdus]

Real answer are: Y11=0.12,Y12=-0.11,Y21=-0.11,Y22 =0.12

[@akaneeii]

Sir pls provide transient topic of network which is most imp

[@kunnu_4]

13:32 / 13:53

[@anshumanmishra4172]

Guys the best method is not this it is to use nodal analysis and directly obtain the Y parameters

[@nesoacademy]

This lecture is not about the best method but the conventional method. This solution is important to set the tone for the upcoming lectures.

[@SarveshGupta-bu5ho]

Sir how to solve problem for capacitor and inductor

[@sumitsinghietlucknow7104]

Sir please make video on MATLAB

[@rickjljr11]

>> z1=10 z1 = 10 >> z2=20 z2 = 20 >> z3=5 z3 = 5 >> Ya=1/z1 Ya = 0.10000 >> Yb=1/z2 Yb = 0.050000 >> Yc=1/z3 Yc = 0.20000 >> Y=[Ya+Yb, -Yb; -Yb, Yb+Yc] Y = 0.150000 -0.050000 -0.050000 0.250000 >> Zs=inverse(Y)+40*ones(2) Zs = 47.143 41.429 41.429 44.286 🙂

[@syedismail1849]

🔥🔥🔥

[@abinokselvin]

Wheree did I²' & I²'' go

[@s.m.senses7199]

🙏❣

[@ranjithk1877]

Ok

[@prakharpandey7062]

For delta to wye connection, refer to this video: kzbin.info/www/bejne/bZ_TYpdrmt9kntU

[@fyesilyurt]

Bu çıkmaz

[@kishoreperam5162]

Sir,after getting Z values, can't we get Y parameters just by reciprocal of Z values