Capacitor
Initial Charge = 2V
after switch switched
2uf Capacitor Charge = 2V/5 Net Charge will be same and charge will divide depending on ratio of Capacitence
8uf Capacitor Charge = 8V/5
Now Q^2/2C - q^2/2C
Q = 2V
q = 2v/5
but am not getting ans
78 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.
Vc common potential iska formula hota Hai q1±q2/c1+c2
8 wale pe shuru me 0 h
Energy ka 1/2cv2 wala use Kiya h
what does Vc signify
and whats wrong with what i did
.
ye i dont understanc it properly
This is just gonna be q²(1/Ceff - 1/2) ja?
Minus of that rather
why
Total energy in initial circuit - total energy in final circuit
Charge is conserved
why the total?
isnt it ONLY Talking abt the 2uf capacitor
They've asked energy dissipated na?
That's energy loss
and the charge distribution i did is it right?
No, they mean energy loss in that process
FROM the 2uf?
wha
From circuit
pehle wale case me toh 8uf pe charge hi nhi hai
ye
we are checking
how much
of its stored
energy
is lost
dusre case me jab join karenge tabhi toh aayega
so shouldnt we compare THE ENERGY OF 2uf in the initial AND Final
to compare
Iska formula bhi toh hota haii
Direct
Mod me v1-v2 ka square something karke
am asking
what i did
is wrong?
also from the language it says how much if its stored energy it lost
so shouldnt we compare
the 2uf with 2uf
why u taking total @Opt
You can't say its from just the 2uF one??
look hear me out
Energy loss happens from a circuit, not from one capacitor
at first there is energy in 2uf
then something something happens
If it's going to another capacitor, it isn't getting dissipated....
Dissipated implies loss
then if we check energy of 2uf in the final
and subtract it
shouldnt we get the total loss in energy
of 2uf
ye
thats why
It's not of 2uF that's asked.
Read the question
It's just asking for the energy dissipated
"the percent of ITS energy dissapated"
and it only talked abt 2uf capacitor
Its stored energy. The energy that was stored in it.
yes
we want to see ITS Loss
so shouldnt we COMPARE ITS
Also, if it goes to another capacitor, it isn't getting dissipated
inital and final
It's getting transferred
ye exactly tahts why
Yeah, so you don't consider just 2uF
You consider the entire circuit
Ok, solve it with my logic, and if you get the right answer, I'm correct?
ok , but tell if my charge distribution logic is correct atleast
Charge distribution is correct, but it's not really necessary
what
You can do charge distribution and consider energy of both, yeah, sure.
u are saying to do
intial in 2uf - final (2uf and 8uf)
Yup
u talking abt this right?
With same charge
Total
ye makes no sense to me but still i wll try
2V^2 - 4v^2/50c - 64V^2/50c
2v^2 - 34V^2/25C
= 16V^2/25C
and on taking %
its coming out to be 16
@Opt
Are you sure you're fine today? Why are you doing this so wrong?
Like, I'm not trying to offend you, it's just that you don't generally do these kinds of things.
i have been off for like weeks now
oh wait so u just took the eq capacitnece and found its energy?
damn
Yes
i rechecked the calc i dont see anything wrong
wahts wrong with taking them seprately?
i checked again its coming out to be 16%
wait am sending a pic
You get the same answer
@Gamertug
@Opt what am i doing wrong 😭
Denominator mein kuch gadbad ho raha hai
1/2 - 1/10 / 1/2
5/10 - 1/10 / 1/2
4/10 / 1/2
ooh
its 80
what exactly i cant find anything wrong
Yup
Try doing it again, but writing down the formulae and substituting one by one.
You might have made an error there
i did it twice
i got 16%
wait a second
i why u got 1/8 by 8uf capacitor energy @Opt
and why extra 1/2
Q²/2C is the formula
i did apply that
for 2uf = 2v/5
Why are you doing this with charges squared??
for 8uf = 8v/5
That's the charge....
Not the energy
ik
am using Q^2/2c
Charge
i mean energy
Ok, doesn't matter since you divide
ye
But still, I'm not seeing that C in your working
and wdym by charges squared
There won't be a 64
No chance
8 squared is 64
wdym there wont be
64V^2/50C
But you have to divide by eight for C na??
oh fuk 😭
+solved @Opt
Post locked and archived successfully!
Archived by
<@1035556259417571408> (1035556259417571408)
Time
<t:1748655060:R>
Solved by
<@763645886500175892> (763645886500175892)