P31_076 - rule requires that L 1 di 1/dt which is large and...

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76. For t< 0, no current goes through L 2 ,so i 2 =0and i 1 = E /R . As the switch is opened there will be a very brief sparking across the gap. i 1 drops while i 2 increases, both very quickly. The loop rule can be written as E− i 1 R L 1 di 1 dt i 2 R L 2 di 2 dt =0 , where the initial value of i 1 at t =0isgivenby E /R and that of i 2 at t = 0 is 0. We consider the situation shortly after t = 0. Since the sparking is very brief, we can reasonably assume that both i 1 and i 2 get equalized quickly, before they can change appreciably from their respective initial values. Here, the loop
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Unformatted text preview: rule requires that L 1 ( di 1 /dt ), which is large and negative, must roughly cancel L 2 ( di 2 /dt ), which is large and positive: L 1 di 1 dt ≈ − L 2 di 2 dt . Let the common value reached by i 1 and i 2 be i , then di 1 dt ≈ ∆ i 1 ∆ t = i −E /R ∆ t and di 2 dt ≈ ∆ i 2 ∆ t = i − ∆ t . The equations above yield L 1 µ i − E R ¶ = − L 2 ( i − 0) = ⇒ i = E L 1 L 2 R 1 + L 1 R 2 = L 1 L 1 + L 2 E R ....
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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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