for those who do physics. (1 Viewer)

[Affinity]

1)
the charge (q) held by a capacitor proportional to the capacitance (C) of a capacitor and the voltage across it (V). (q=CV)


a charged capacitor is connected in series to a resistor of resistance R ohms.

a) explain why dq/dt = V/R

b) show that dq/dt = q/RC

c) initially the capacitor holds a charge of A columbs, express the charge in the capacitor as a function of time.

d) determine a formula that can be used to calculate the time required for the charge on a capacitor to decrease to 1/10 it's orginal value, from given values of R, C and A

 

[Dangar]

hmm you've got me there. I would have absolutely no idea, so please tell me this is nothing to do with the HSC?

 

[Constip8edSkunk]

hmmm
a) current = rate of charge over time, ie dq/dt
V=IR ->I=V/R
.'.V/R = dq/dt
b)but q=CV, V=q/C
from a) dq/dt =V/R = q/RC
c)t=0, q=A
capacitor loses charge over time, so q/RC=-dq/dt
1/q dq=-dt/RC
I(A->q) dq/q = -1/CR I (0->t) dt
ln|q/A| = -t/RC
q/A = e^(-t/RC)
q=Ae^(-t/RC)
d)q=A/10=Ae^(-t/RC)
-ln 10 = -t/RC
t=RCln10
so the time is independant of the original charge?

is this right?

 

[Newbie]

lol
if they ask this
they might as well as business studies questions

 

[underthesun]

Find the integral need to write an essay, from 0 to 40 minutes. get the mock paper from the review forum now!

back to the question, i would have absolutely no idea as well, as these days physics' rate of change of essay-type question is increasing. It's more like english than maths now..

 

[freaking_out]

Originally posted by underthesun
Find the integral need to write an essay, from 0 to 40 minutes. get the mock paper from the review forum now!

back to the question, i would have absolutely no idea as well, as these days physics' rate of change of essay-type question is increasing. It's more like english than maths now..

yeah, we only get exposed to these sorta stuff is in 4u, and not physics.