Heat loss
why would there be heat loss in this , also capactior and inductors dont do heat loss? only resistors do
37 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.sure in dc but Xc=1/wc and w=0 in dc so Xc is max and thus ig all energy stored in capcitor will be in heat loss ?
what's the ans
B
what
why will energy stored be in heat loss
its just energy stored
no idea
heat loss would mean the energy left the circuit no?
but here energy is still in capacitor
yeah im not sure about that too
Heat loss is equal to energy supplied by cell - energy stored in capacitor
makes no sense , heat loss should be loss in energy of the circuit
No, the logic is, all losses must be due to heat and entropy.
Though the resistance isn't explicitly mentioned, it's taken into consideration
So energy supplied - energy stored gives you loss in energy
the new energy supplied is stored so how will it get lost
Not all of it is stored.
Q²/C energy is supplied, but only Q²/2C is stored, remember?
Half of it is lost
ooh wait i do remember something vaguely like this
Energy supplied is QV = Q(Q/C)
But energy stored is 0.5CV² or Q²/2C
can u remind me the reason i forgot
of why half is lost
Heat loss. The maths behind it is integration
Isnt Capacitor and Inductor and no loss component tho?
Heat loss + induction effort actually, but the latter is considered negligible compared to former
like thats what i read
only resistance gives heat loss
Circuit has some "implicit resistance"
so for ideal no loss?
but we take energy Q^2/2c for all capacitors ideal and non ideal alike
It's actually always gonna be there, but we don't want to complicate stuff so we say heat loss
There's also polarisation loss, induction loss, dissipation of fields is always going to happen, magnetic field generation loss
Lot of stuff
ic
So just bundle it up and say 'entropy hogaya'
ye
oh
i had also fogotten about it
wow i am cooked
us
Still don't get ans
Energy diff is 3/2cv2 heat should be same 3/2
energy supplied is QV =CV^2 and energy stored is CV^2/2
and thus Us-Uc=CV^2/2
at first Energy = 1/2CV^2
After 2V = Energy = 1/2C(2V)^2 = 2CV^2
energy diffrence = 3/2CV^2
this much energy is stored so this will also be the heat that is lost as both are half half
mb i said 3/4 it should come 3/2 but still its 1/2
energy supplied by battery is CV^2 since U=QV and max charge on capacitor is 2CV and it has CV already so U=CV^2
and del U capacitor is Q^2/2C and capacitor is charged by CV by the battery
now energy stored - supplied will be CV^2/2
U get wrong ans with this my teacher said this only applies for uncharged
Oh. Then how does this question work?
Wdym
Oh wait I read smth else
He did work done by batter = change in C energy + heat
Oh, makes sense
wohi toh bola hai
Ye noice
+solved @727 @Opt
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