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# Help! A level Capacitance qu. watch

1. Can’t do qu11 plz help !!! Thanks 🙏🏼
Don’t know how to go about it!
2. What have you already tried? show your working so I can guide you better.
A hint: the total charge between the two capacitors is constant so try use that with Q=CV
What have you already tried? show your working so I can guide you better.
A hint: the total charge between the two capacitors is constant so try use that with Q=CV
Didn’t even know where to begin all I did was write E=1/2CV^2
Didn’t even know where to begin all I did was write E=1/2CV^2
Okay thats good but that formula gives the energy stored in the capacitor but you re asked bout the energy lost in the resistor so first you need the total energy and then you need the power dissipated in that resistor
Okay thats good but that formula gives the energy stored in the capacitor but you re asked bout the energy lost in the resistor so first you need the total energy and then you need the power dissipated in that resistor
But how do I do that
6. Do so you know that because they are in series you can re write them as one capacitor with a total capacitance of 3C and you know that the total charge, which is the intial charge of the charged capacitor is constant okay?. So mathematically, that means that
1/2Q^2/C (initial charge) is equal to 1/2Q^2/3C ( final charge when the switch is closed) dividing the two you get that the final energy is 1/3 of the intial energy meaning that 2/3 of the energy was lost in that resistor. And thats it.
Do so you know that because they are in series you can re write them as one capacitor with a total capacitance of 3C and you know that the total charge, which is the intial charge of the charged capacitor is constant okay?. So mathematically, that means that
1/2Q^2/C (initial charge) is equal to 1/2Q^2/3C ( final charge when the switch is closed) dividing the two you get that the final energy is 1/3 of the intial energy meaning that 2/3 of the energy was lost in that resistor. And thats it.
The capacitors are effectively in parallel for the steady state condition. Only the current flow path is treated as series for the duration during which charge equilibrium is achieved.
8. (Original post by uberteknik)
The capacitors are effectively in parallel for the steady state condition. Only the current flow path is treated as series for the duration during which charge equilibrium is achieved.
Yup youre sorry bout that meant to right parrallel.
Yup youre sorry bout that meant to right parrallel.
I think I understand what you just wrote! lol.
10. (Original post by uberteknik)
I think I understand what you just wrote! lol.
Lol sorry again, am on my phone and like to type quick so i think of most of what I wanna write instead of writing it.

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