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GR0177 #40
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Alternate Solutions |
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Comments |
Imperate
2008-10-06 07:15:15 | One can remember what the time constant is for an R-L circuit by dimensional analysis. Since an R-L circuit contains only these two elements you immediately know the time const mut be a function of R and L. Next V=LdI/dt => [L]=[V][s]/[A]. Secondly the fundamental units of resistance can be obtained from Ohms law [Ohm]=[V]/[A]. As the name suggests the time constant must have units of time! Thus the function of L and R must be rnL/R= [V][s]/[A] *[A]/[V]=[s].rnPlugging this in  . |  | scottopoly
2006-11-02 15:19:06 | Well the site seems to be a bit broken and I can't edit my comment, so here goes:
I misread "200s" as "200ms" at first, and so was confused by the "decreases rather quickly".
But really once you realize that these values for the components in this problem are rather standard values that you find in lab, a little experience will tell you that indeed, there is hardly a circuit that moves in 200s time.
sorry about that. | | scottopoly
2006-10-29 22:47:24 | Furthemore, I might add that "decreases rather quickly" is, ah, crap. The numers are arbitrary, there is no basis to say anything is "fast". It decrases the rate  , as Void said, which is neither fast nor slow.
Richard
2007-10-25 15:55:55 |
nice... （笑）
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wangjj0120
2008-10-27 02:21:55 |
HAHA~
It's funny to see a chinese word here.
這個網站很棒 謝謝你~Yosun
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| | Void
2005-11-11 08:58:21 | To see why (D) is right, recall that the time constant for an RL circuit is given by  . Inputting the values makes (D) the "obvious" choice. | |
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