Electromagnetic Induction 2
2011-03-20 8:58 am
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I just know that when the switch is on, the self-induced emf of an inductor of inductance 2H is 12V when the current rises at the rate of 6As^-1.
It seems the statement (1) is unpredictable to me if the value of self-induced emf is not given.
So please could you explain in detail, why the statement (1) is correct.
回答 (1)
2011-03-20 9:30 am
✔ 最佳答案
The rise of current I in the circuit is described by the equation:I = (Io)exp(-Lt/R)
where Io is the final current
L and R are the inductance and resistance in the circuit
t is the time
The rate of rise of current dI/dt is given by
dI/dt = (Io).[(R/L).exp(-Lt/R)]
But Io = 12/6 A = 2 A
Hence, dI/dt = 2[(6/2).exp(-Lt/R) = 6.exp(-Lt/R)
The initial rate of increase of current (at t = 0) = 6 A/s
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110320000051KK00079