P31_049 - . 609 L = 1 . 609 L R = 1 . 609(6 . 30 10 6 H) 1...

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49. (a) If the battery is switched into the circuit at t = 0, then the current at a later time t is given by i = E R ³ 1 e t/τ L ´ , where τ L = L/R . Our goal is to Fnd the time at which i =0 . 800 E /R . This means 0 . 800 = 1 e t/τ L = e t/τ L =0 . 200 . Taking the natural logarithm of both sides, we obtain ( t/τ L )=ln(0 . 200) = 1 . 609. Thus t =1
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Unformatted text preview: . 609 L = 1 . 609 L R = 1 . 609(6 . 30 10 6 H) 1 . 20 10 3 = 8 . 45 10 9 s . (b) At t = 1 . L the current in the circuit is i = E R ( 1 e 1 . ) = 14 . 0 V 1 . 20 10 3 ( 1 e 1 . ) = 7 . 37 10 3 A ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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