Question

You have an idea that maybe the time rate of discharge on a capacitor should be proportional to the actual charge on a capacitor. To test this idea, you grab your multi-meter and RC-circuit. So, you power up the capacitor and then switch your circuit off. At that instant (at t=0), you find that the capacitor has 1 coulomb of charge. Ten minutes later, you come back to the circuit and measure 0.89 coulombs (or 89% of the initial charge). Wanting to see how accurate your mathematical model is, you decide that you will predict just how long it will take for the capacitor to reach 1% of its i

Answer 1

of its what?

Answer 2

dC/dt = k C(t)

Answer 3

and what is the question? it is incomplete

Answer 4

1% of its initial charge. How many minutes does your model predict this will take?

whoops

Answer 5

so solve dC/dt = k C(t), you get dC/C = k dt, integrate both sides
ln C = kt + constant
C = Ae^(kt)

Answer 6

when t = 0 , C(t) = .1, so
C = e^(kt)

Answer 7

1*

Answer 8

\[\frac{dq(t)}{dt}=-k \times q(t)\]it is -k because its discharging
\[\frac{dq}{q}=-k dt\]

Answer 9

ok q(t) = -e^(kt)

Answer 10

now when t = 10, we have q(10) = .89

Answer 11

.89 = -e^(k*10) ,
...

Answer 12

wait, that doesnt work.

Answer 13

i messed up

Answer 14

q(t) = A e^(-kt)

Answer 15

you dont have to assume - , since it will come out negative later
q(t) = e^(-.011653 t )

Answer 16

now solve .1 = e^(-.011653 t)

Answer 17

it will take almost 200 minutes ,
i get 197.595 minutes

Answer 18

integrating u get
\[\ln(\frac{q}{q_0})=-kt\] q0 is the chaarge when t=0
wer given that when t=10\[\frac{q}{q_0}=0.89\]
solve for k then find t when \[\frac{q}{q_0}=0.01\]

Answer 19

197.59! Thank you! :)

Answer 20

i got t=395 min:S

Answer 21

wait,

Answer 22

i used .1 not .01

Answer 23

should be 395.18

Answer 24

yeh thats what i got:D

gdjob:D

Answer 25

lalay i dont get why you wrote q / qo = .89
its ln (q/q0) = .89

Answer 26

89% of the initial charge
so ln(0.89)

Answer 27

errr, ln ( .89/ q0) = -k (10)

Answer 28

:)