PATHFINDER SOLUTIONS | JEE ADVANCED | OBJECTIVE-18 RC CIRCUITS | DIELECTRIC REMOVAL | SCHOOL PHYSICS
Вставка
- Опубліковано 15 чер 2024
- DON'T MISS THE IMPORTANT PRACTICE PROBLEM GIVEN AT THE END
A COMPLETE STEP BY STEP EXPLANATION OF AN OFTEN CONFUSED AND MISCONCEIVED CONCEPT OF DIELECTRIC (QUICK) REMOVAL IN RC CIRCUITS
Q : An air filled parallel plate capacitor of capacitance C is connected through a resistance R to an ideal voltage source of electromotive force V. A dielectric plate of dielectric constant k is inserted in the capacitor to occupy whole space between the plates. After a steady state is reached, the plate is quickly pulled out. Which of the following is correct expression for heat generated in the resistance until a steady state is reached again?
(a) (b) (c) (d)
AN EXAM ORIENTED APPROACH SOLUTION PROVIDED FOR OBJECTIVE - 18 OF CHAPTER-13 "ELECTRIC CURRENT" SOLVED FROM THE BOOK "PATHFINDER PHYSICS"👇
HERE'S THE PLAYLISTS LINK FOR OTHER "CURRENT ELECTRICITY" CHALLENGES👇
48_ELECTROSTATICS PART-5(CAPACITORS WITH DIELECTRICS): • 48_ELECTROSTATICS PART...
47_CURRENT ELECTRICITY PART-3 (EXPERIMENTS AND METERS): • 47_CURRENT ELECTRICITY...
46_CURRENT ELECTRICITY PART 2_KIRCHOFFS RULES AND CIRCUIT ANALYSIS: • 46_CURRENT ELECTRICITY...
*HERE'S THE LINK FOR PATHFINDER SOLUTIONS PLAYLIST*👇
• PATHFINDER SOLUTIONS
HERE'S PLAYLIST FOR AITS SELECT SOLUTIONS SERIES👍
• "AITS SELECT" SOLUTION...
HERE'S THE PLAYLIST OF PHYSICSSIRJEE ORIGINALS👇
• PHYSICSSIRJEE ORIGINALS
OTHER PLAYLISTS
HERE'S THE LINK FOR "OLYMPIAD WORKOUT SERIES" BOOKS,SYLLABUS,RESOURCES👇
• ANNOUNCEMENT-2 🧨"OLYMP...
RESOLVED series playlist 👇
• RESOLVED SERIES
HERE'S THE PLAYLIST OF REST OF THE "OLYMPIAD WORKOUT SERIES"👇
• OLYMPIAD WORKOUT SERIES
HERE'S THE JEE ADV SOLUTIONS -THE ALPHA AND OMEGA SERIES👇
• JEE ADV SOLUTIONS - TH...
HERE'S LINK FOR PLAYLIST OF "JEE MAINS DEBATED QUESTIONS" PLAYLIST👇
• JEE MAINS DEBATED QUES...
FOR REST OF THE INTERESTING BRAIN TEASING JEE PHYSICS CHALLENGES AND CONCEPTS , PLEASE SUBSCRIBE TO MY CHANNEL " PHYSICS sir JEE " HERE BELOW
/ physicssirjeejanardhan
/ @physicssirjee
THE CHANNEL IS AN IMPORTANT RESOURCE TO THOSE STUDENTS PREPARING FOR JEE MAINS, JEE ADVANCED EXAMS IN INDIA AND FOR PHYSICS OLYMPIADS LIKE NBPhO,INPhO,IPHO,HKPhO,BPhO ETC GLOBALLY.
#PHYSICSSIRJEE,#PATHFINDERPHYSICS,#IOQ2021,#JEEADVANCED2021,#PATHFINDERSOLUTIONS,#ELECTRICCURRENT,#RCCIRCUITS,#DIELECTRIC,#CAPACITORS,#HEATINRESISTOR,
Sir i tried that practice problem...i got the answer...i will again check my procedure! And thanks for your efforts 🤩
Awesome explaination 🔥🔥
thank you sir awesome problem 🔥🔥
Very nice explanation
thank you sir
Awesome explanation sir👍
Amazing sir 😁
Thank you sir best teacher of India
You copied my dp from the other channel😁👌. Btw, thank you for the appreciation . I hope , one day your words are close to true🙏
Lovely vdo 😍
Thankyou sir❤️❤️
Thanks sir !
Thank you sir 🙌🙌
Thank you sir 😊❤️
Tnx sir
Sir i considered the voltage constant and performed work energy theorem, thank you for clearing my misconception😅
🔥
Sir,how would things change if there was no resistance present in the circuit ,can we still calculate heat loss?
op problems
Sir can we do pathfinder subjective are they relevant .
Someyimes it is very time consuming sir😔
Where do I find the answer to the practice problem?
Sir please jee mains ke liye crash course start kar do 🙏🙏
8:04 is the final charge key incorrect sir?
Op
Plss take some good questions on alternating current there's lack of questions of this topic over the internet
Do question from pathfinder there are many
jee main ka topic hai wo, utna bhi tough ayega nahi mains me
@@sunritroykarmakar4406 advanced ke syllabus me hai aur kai questions aa chuke hain usse
@@sakshampaliwal8170 all super easy questions compared to normal advanced level
in first question i can't write kvl equation
Please anyone help
Sir,I request you to please explain that why at 7:04, intial potential energy is not K(CV^2)/2😥.🙏I got wrong answer taking that.And,even integration gives what you solved...but I am not able to figure out,why it's not what I thought.Please respond sir🙏🙏
You should first need to understand about the initial point from where we are taking note of the potential energy.
So at the moment when you have removed the dielectric abruptly, then the capacitance has immediately changed to C (KC-->C) but the charge can not change back to CV immediately (Sir has explained about the same)
So the potential energy is not 1/2*(2C)V^2
The very easier way of writing the PE is use (Q^2/2C) where you will put Q =KCV
Again I repeat at the moment when we are writing the initial energy:
Capacitance=C (not KC)
Q=KCV
Voltage across cap= Q/C =KV
@@rishu_Kumar07 Thanks a lot..I had been considering it the potential energy before removing the dielectric which was absolutely wrong.
Sir i have a doubt in this form of Work Energy Theorem.
When kinetic friction do negative work on a block. Why dont we take the Change in Microscopic Kinetic Energy in work energy theorem
Dont know if you’ll see this , as its after 8 months , but here’s a explanation . I am assuming by micorscopic KE you mean kinetic energy of molecules which correlates to the temperature of the body , in mechanics we only deal with the mechanical energy of the system and any negative work done by resistive forces is considered as “lost” energy , in true sense its not lost but just converted to other form of energy such as heat , the reason why we dont include those energy in work energy theorem is simply because we dont care about those in mechanics.
First like
For HW Q , I am getting
a) Imax= AV/(rho(epsilonr +1)d)
b) Initially:
Qcap= epsilono AV/d (epsilonr/epsilonr+1)
Qdie= epsilono AV/d (epsilonr-1/epsilonr+1)
Finally:
Q cap= Qdie=epsilonoAV/d (such that electric field in dielectric becomes zero)
c) Time constant = 2rho epsilono
Sir are my answers correct?
Sir please daily upload kijiyega videos
Possible hoga toh zaroor karlenge🙏
@@PHYSICSSIRJEE ok sir... Thanks a lot ☺️
8:30 Sir i didn't find any mistake in the HW problem. Sir Can u tell where I may be missing out.
Please kindly wait for my Solution, let others give it a try 😀
@@PHYSICSSIRJEE pls make the solution
2rhoepsilon time constant?
my answer coming rho epsilon not twice can u please tell whats wrong please
Why work done by agent in dielectric removal ignored?
Can you please post timestamp of your doubt??
Sir there is great problem on nlm in physics galaxy book in olympiad section of chapter 2 questions no. 50 no one including my teachers are able to solve it . Can you plz provide the solution. I have tried it multiple times but in vein. Plz sir its a request 😊😊
I don't have the book with me . I will try to look into it
@@PHYSICSSIRJEE sir you can look for pdf but plz give the solution its very crucial for me
@@PHYSICSSIRJEE I uploaded the photo of the question he asked
@@metablaze3523 where
@@ashwin2263 for some reason UA-cam is constantly deleting the drive link I’m posting
Why k-1 is taken in Wbat Sir?
👍🏻
Difference in charge on capacitor flows through the battery
@@PHYSICSSIRJEE ok Sir
Sir if you are reading this comment. Can you please help me in solving by Kvl can you please give some hints
You can DM me in the app
If it was only capacitor and Battery and no resistor , then too answer would be same right ?
it would be different i think becoz then there would be restriction for current to flow
No
@@mysticshadow4561 still final charge will be CV and initial KCV ... And work done also same by battery so same ans
If there is no resistor wer ll the heat be dissipated so resistor shld be der
Why, and then what will be the ans
But during so why wont the capacitor also generate heat and by that we will get net heat ie by resistor and capacitor and not only resistor 6:42
Capacitors don't 'generate' heat . Heat is dissipated in wires connected due to resistance
@@PHYSICSSIRJEE yes I understood thank you
Sir I'm scoring 285 in jee mains 24 Feb shift 1.... Can i expect under 150 rank??
Yes
First world problems 🤣🤣
Super bro you will get top 100 I guess
I don't think bro... For 50k rank 290 cutoff is there, for 150 rank you need to grab full 300 marks .....
@@amiyancandol4499 ok bro my shift is damn tough
26 last one 250 for me itself is good score😂
Janardhan sir I have a doubt sir when we use the usual energy analysis in circuit containing only capacitor and battery we see that only half of the work done by the battery is stored as field energy in the capacitor and we tend to explain the loss of energy as heat in the connecting wires despite the fact that there was no resistance and when a circuit contains resistor as in this problem we still do the energy analysis and find out the heat generated but my doubt is why should all the heat we calculated be generated in the resistor only I mean why cant we now say a fraction of heat is lost as heat in the connecting wires and a fraction in the resistor ,I hope you understood my doubt sir
R is total resistance of the wire plus externally connected resistance in the Analysis . In an ideal case , R tends to zero and Not R=0. When you perform integration of i^2R dt , we realise as R tends to zero , time tends to zero , but i tends to infinity to keep heat loss same ... Which is "the famous half" taught in books
@@PHYSICSSIRJEE thank you so very much for responding ,completely understood it now sir
In all such problems of energy balance in circuits having capacitor and cell, there is dissipation of energy as heat and electromagnetic energy due to acceleration of charge. So even if we consider an ideal situation with circuit having no resistance then there would be almost instant charging of capacitor and steady state attained. The energy supplied by cell is not all stored up by capacitor. The radiation of electromagnetic energy is there. So it is better to reason dissipation of energy as heat plus em energy.
@@trilokv2444 well said 👍🙂
@@trilokv2444 👍👍
sir can you know the name of the guy who disliked this video
UA-cam doesn't , for a good reason , allow us to know . It helps us keep improving 👍😉
Sir I guess that is you 🤣
Nobody can dislike your videos sir
@@ashwin2263 😂😂
@@PHYSICSSIRJEE sir waiting for premiere 💥
200 op sir
Rohit sharma fan sir😂😎
Another method:
Initial charge - KCV
Final charge - CV
Heat = delta Q²/2C = ((K-1)CV)²/2C
= (K-1)²CV²/2
Is it correct sir?
Wrong , please check the concept correctly . ∆U is negative