35 Replies
@Gyro Gearloose
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@Opt
Rip phy doubt solvers ðŸ˜
Yar mera bhi tere vala expression aaraha ha
Sir se puchunga kal
ðŸ˜
Idts u can combine both the capacitors into 1 and give that 4qo charge
Not even the potential drop across that is same
Holy shit your answers are correct tho
Yeah it didnt work in other questions for me
by ur method
Is vale mein galti kya ha?
@integralofe^2x
Solved???
Its cuz maybe i think u took the initial limit of q1 from 4q but this isnt true as the charges will get distributed as evident by setting t=0 in the expression for q2 u get initial charge on q2 as 12/5q anf on q1 on solving will just be 8/5q not 4q
apparently my dq/qt stuff is strange
@Opt @SirLancelotDuLac any inputs on that part ?
@integralofe^2x what is qi , q1 i and q2 i
Those are initial charges and mb they arnt 0 and 4 they are 8/5 and 12/5
q1= 4qo and q2 = 0 hona chahiye tho
t=0 pe switch band kia
usi time battery se current aayega
charges resdistribute honge
hmm makes sense
Yea it seems like that firstly but then firstly the formula dosent support ur statement and moreover the charges on capacitor will distribute just when u connect them alone b4 connecting the battery, tbh im not too sure of this logic
how do u get 8/5 and 12/5 tho
Use the charge conservation
s1 s2 s3 are closed simultaneously tho
C1V1 + C2V2 / C1 + C2
the same time charges are being redistributed amongst capacitors , battery is also supplying current
Yea thats what id understand cuz the circuit woul be incomplete so charges cant redistribute b4 the switches are closed
yeah...
but then t= 0 pe q2 = 12qo/5 bhi ho gya
Yea acc to formula, id see a way for them to redist.
Ill have a closer look aftet class
alr
Okay, I think I'm tripping but the initial charge of 4q_o distributes across the both the capacitors first to make the potential drop across both the same and hence the lower bound of q seems incorrect.
I might be wrong though.
.
ill still try with diff limits tho
Both things are happening infinitely fast, but the charge distributes first, otherwise the potential on the same wire would be different at 2 points.
holy shit man
big brain
this does work
why do we need that to happen tho?
i agree the capacitors are in parallel but while doing classical solving we dont utilise the fact that both are in parallel right?
Both things happen infinitely fast, but the capacitor distribution occurs first because a point on the wire would otherwise have 2 potentials which would be paradoxical irrespective of whether battery is connected or not. The battery only creates a potential difference which starts acting once charge is distributed. (Some infinites are smaller than other infinites kinda thing ig)
To my understanding (Sorry for the late reply, but I wasn't sure how to phrase it)
yep asked my sir and he confirmed that this logic is correct
this is the way to go
so my entire process was correct only limits were wrong
+solved @SirLancelotDuLac @integralofe^2x
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