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
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I= I(1-e^-t/CR), where ...
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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
Pardon?
And no, you shoul'dnt be making the forums do your maths homework.
lol its part of an assignment im behind on!
hmm
a.q h43r:
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)?
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)?
:wacko:
bildiğim matematiğide unuttum aq bu ne :wacko:
translate:fuck maths :wub: ^_^
yeah its 25*10-6 or 25uF if u want to be technical
wow... how the fuck can you guys do that lol i cant do 7x7 h43r:
49 h43r:
2x2 ftw? :unsure:
3yeay!! B)
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..
@ my job I use a crap load of formulas like that. . not really hard, just ALOT of formulas to remember :blink:
and I thought my maths coursework was hard
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