Qualification > Sciences
Edexcel PHYSICS Doubts
viktress:
Phew! Thanks, I kept getting 650 N s however much I tried and I just couldn't figure it out. Now I'll be able to sleep tonight :D
wakemeup:
--- Quote from: Hermione Jean Granger on June 20, 2011, 12:31:15 pm ---I am lost in 13b myself too. :(
Ok, Q16:
bi) Capacitor stores charge, as charge flows in during the charging process, pd is also formed.
If the voltage is constant, the capacitor will not discharge. The graph is during the charging process. Hence, pd is constant.
(remember: during discharging process, pd falls.)
ci) Use E= 1/2*C*V^2
(take voltage value from the graph)
cii) Compare the graph with the previous one and see the change in voltage value. (pd decreases)
Capacitor is initially charging from power supply then the capacitor is discharging exponentially.
(remember: the changing position of switches causes the discharge process, instead here. . . we added a power supply)
I hope that's clear. :)
--- End quote ---
Sorry, I'm still pretty lost. How does the capacitor keep the p.d constant? By discharging a constant pd? But the pd varies when the capacitor discharges. And if the capacitor is discharging exponentially why does the graph look like that? Could you please explain cii as well. (You explained bii where you wrote ci)
Romeesa-Chan:
--- Quote from: wakemeup on June 20, 2011, 03:39:46 pm ---Sorry, I'm still pretty lost. How does the capacitor keep the p.d constant? By discharging a constant pd? But the pd varies when the capacitor discharges. And if the capacitor is discharging exponentially why does the graph look like that? Could you please explain cii as well. (You explained bii where you wrote ci)
--- End quote ---
The pd is constant during the CHARGING process. That's cos the pd is equal to the emf of the source.
During discharging, the pd value falls as the charge decreases. (the capacitor decreases exponentially, we must just say that pd falls)
Sorry, if I am not making sense here.^^ :(
cii) T= RC
R= T/C
(value for capacitance is given and T value is taken from the graph)
:)
wakemeup:
--- Quote from: Hermione Jean Granger on June 20, 2011, 03:48:59 pm ---The pd is constant during the CHARGING process. That's cos the pd is equal to the emf of the source.
During discharging, the pd value falls as the charge decreases. (the capacitor decreases exponentially, we must just say that pd falls)
Sorry, if I am not making sense here.^^ :(
cii) T= RC
R= T/C
(value for capacitance is given and T value is taken from the graph)
:)
--- End quote ---
Thanks, I understand a bit better now but as for cii what value of T are we supposed to take? Because the mark scheme has 15-20 ms in it and I don't know how that came about.
Romeesa-Chan:
--- Quote from: wakemeup on June 20, 2011, 04:01:49 pm ---Thanks, I understand a bit better now but as for cii what value of T are we supposed to take? Because the mark scheme has 15-20 ms in it and I don't know how that came about.
--- End quote ---
That's good. :D
Lemme know if you don't.
For value of T, I took 16 ms. :)
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