Product Of Sums - POS | Boolean Algebra & Logic Gates

[@SimpleSnippets]

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[@chyavanphadke4813]

The final answer was : (A+B) . (A+B') . (A+B) can't we remove the common terms as we did in SSOP ? So the answer will be : (A+B) . (A+B')

[@ishaanbhola2428]

Same doubt , whats the answer ?

[@madsoul9304]

Ur answer is right bro

[@harekrishna7063]

Using idempotent law

[@techguru6851]

Bro u teach in very short duration and clear our concept... Keep it up... Helpful.

[@SimpleSnippets]

Thank you so much bro, I have a request, please do share the videos and our channel with your friends as well. That will be the biggest help 😇

[@techguru6851]

Already did bro.. And will share with more frnds

[@SimpleSnippets]

Thank you so much 🤟😁

[@colonelmuska8216]

Hi, At 2:13, isn't the expression you provided already in SOP form? If it's not in SOP form, why not? If it is in SOP form, why would someone want to convert it to POS form? Thanks so much for these great videos!

[@yinkak3921]

I don't know if ur still active but I have a question. Why did (A+B) . (A+BB') evaluate to (A+B) . (A+B).(A+B'). shouldn't it be (A+B).((A+B).(A+B')) because we evaluated A = (A+B).(A+B') and the "." operation was performed on A so shouldn't the "." operation be performed on ((A+B).(A+B')) as a whole?

[@anirudhsengupta9638]

Bro thats a product take an example x.(y.z) = x.y.z

[@LIFE_OF_COPYWRITER]

Thanx it really helpful nd clear my all doubt. Tomorrow is my internal. Let c.....✌

[@SimpleSnippets]

Hi Nikita, I'm happy that you've cleared your doubts 😇✌️ All the best for your exams 😇✌️

[@LIFE_OF_COPYWRITER]

Simple Snippets thankuh tanmay😊

[@SimpleSnippets]

Do share it with your classmates 😉 Maybe even they'll find it helpful 😋✌️

[@vivekprasad_official_music2267]

Can you help me in this 3 topic 1.Other logic operating 2.Digital logic operation 3.integrated circuits

[@aksharsalunke6176]

Please Sir at for next playlist use white board and thanks for this all knowledge to gave us.

[@WinifredKokosu]

Thanks a lot I want to know if the question given to you will state that you should leave your answer in the standard product of sum form And also should the expression be simplified to the lowest level

[@Funland-v5d]

Very nice and easy lecture

[@raunitjaiswal3629]

hello bhaiya your channel content is just too good .

[@vishalm2766]

i think you can still modify that answer as [(A+B).(A+B')]. because A.A=A

[@SimpleSnippets]

correct ✌

[@ekeshwarlalsahu5747]

(A+B) is not equal to (A+B')

[@geemochi9938]

@Lifes Lessons what

[@SAMRATsofficial]

In last Ans why don't you use Idempotent Law for (A+B)(A+B)

[@vpgsalty]

hey dude. what would I need to do if more than one of the variables is missing from a term and I want to convert the expression to standard form? I understood how to do that if one is missing. like (A+B) where only C is missing. but what if B or A were missing too?

[@SimpleSnippets]

You introduce that variable the same way you introduce the first missing variable ✌️

[@yashkadostt]

Concept cleared bro✌️😎

[@SimpleSnippets]

Thank you so much Yash. Really happy to hear this from you 😇✌️

[@himanshu7103]

thanks sir , beautiful

[@AsmaaDiis]

thank u❤

[@lubienaali2763]

Thanks sir it's really helpful for me

[@SimpleSnippets]

Most welcome my friend. Please do share the video with your friends too. That'll be the biggest help and support ✌️☺️

[@chinmayeegouda151]

your video helped me thx

[@SimpleSnippets]

Most welcome Chinmayee 😊 glad to know this. Please do subscribe on this channel and share the videos with your friends too. That's the biggest help and support ✌️

[@beastwasil2337]

5:46 Can't we to using 3 variables ?

[@siddumair2302]

Thank you 🙏💕

[@SimpleSnippets]

Most welcome Umair 😇 please share the video with your friends too ✌️

[@wilfredalmeida6109]

How did A+(B×Bbar) become (A+B)×(A+Bbar)???

[@SimpleSnippets]

That's the distributive law ✌️

[@rinnivatsa4569]

Hey i have doubt

[@kathanshah3305]

can't we dominace law here.?

[@chandaoswal5396]

Why we really need POS and SOP? And why we really need to convert Sum to Product & Product to Sum as already we have AND gate & OR gate?

[@seeni2005]

dude i feel its easier than logic gates i think algebra screwed you

[@LegitGamer2345]

to get marks , its simple .

[@chandaoswal5396]

Can POS contain single complementary variables. Like for ex: A'+A.B+B'+A'.B?

[@SimpleSnippets]

+Chanda Oswal yes but then A'+A ultimately is 1 right? So i in your above mentioned example the final answer after simplifying this will be 1.1.B

[@chandaoswal5396]

+Simple Snippets :-In our classes it was really confusing me.Thank you so much sir for such a good explanation.Now i think its one of the easiest topics ever. Its so easy to catch, Your videos should be viraled all over so that everyone can get this topic in 5 mins.

[@mirandoo300]

✌

[@SimpleSnippets]

Thank you 😇 please do share the videos with your friends and contacts 😇

[@akshayawasthy5989]

A+B . A+B . A+B' isn't this exp suppose to be like this A+B . A+B' as A.A=A

[@SimpleSnippets]

A+B . A+B' here the 2 sums ie (A+B) & (A+B') are different because in one term we have B and in other we have B' so we cant apply the law that you're mentioning.

[@haythama-n-m2071]

No, he is talking about (A+B)(A+B) (A+B`)=(A+B)(A+B`)

[@kasimkarachiwala1391]

??

[@Amitkrdas17]

apply idempotent law dude..

[@SimpleSnippets]

These laws have been discussed in other tutorials.Here we simply explain what POS means, why and how to use them.

[@justastudent4965]

how does this come about? At kzbin.info/www/bejne/noTKlqyCiNKfick Because i dont see how you jumped from A+(B.B')

[@vickysamraj8980]

peace

[@SimpleSnippets]

✌

[@typicalplayer7151]

Pta ni kya pda ra h