Good @ Maths?

[festo]

The instanteous current i amps flowing in a capacitor at a time t secs is given by:
I= I(1-e^-t/CR), where I = 6A, R = 52Kohms, C= 25^-6 F
Determine the time t when the instanteous current I is 4A

[Josh]

Pardon?

And no, you shoul'dnt be making the forums do your maths homework.

[festo]

lol its part of an assignment im behind on!

[Crazy]

hmm




























































a.q h43r:

[TunaFishyMe]

uhhh...
arent you just plugging in the numbers?

4= 6(1-e^(-t/25^-6)(52))
0.6 = 1-e^(-t/25^-6)(52))
0.3=e^(-t/25^-6)(52))
ln 0.3 = (-t/25^-6)(52))
(1.2)(52)(25^-6) = t
t = 2.5 * 10^-7


Btw, are you use the value of C is not 25*(10^-6)?

[TunaFishyMe]

oops i made a mistake....
4= 6(1-e^(-t/25^-6)(52))
0.6 = 1-e^(-t/25^-6)(52))
0.3=e^(-t/25^-6)(52))
ln 0.3 = (-t/25^-6)(52))
(1.2)(25^-6)/(52) = t
t = 9.45*10^-11


Btw, are you use the value of C is not 25*(10^-6)?

[0wnoR]

:wacko:

[_TaLenT_]

bildiğim matematiğide unuttum aq bu ne :wacko:

translate:fuck maths :wub: ^_^

[festo]

yeah its 25*10-6 or 25uF if u want to be technical

[bleaF]

wow... how the fuck can you guys do that lol i cant do 7x7 h43r:

[Crazy]

49 h43r:

[DayDreamer]

2x2 ftw? :unsure:











3yeay!! B)

[Hostage]

Originally posted by TunaFishyMe
uhhh...
arent you just plugging in the numbers?

4= 6(1-e^(-t/25^-6)(52))
0.6 = 1-e^(-t/25^-6)(52))
0.3=e^(-t/25^-6)(52))
ln 0.3 = (-t/25^-6)(52))
(1.2)(52)(25^-6) = t
t = 2.5 * 10^-7


Btw, are you use the value of C is not 25*(10^-6)?

Can you explain the math reasoning behind this? It was the only part I didnt understand..

[ChanTicO]

@ my job I use a crap load of formulas like that. . not really hard, just ALOT of formulas to remember :blink:

[Imhotep]

and I thought my maths coursework was hard