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ECE2260F07_HW1p3soln

ECE2260F07_HW1p3soln - 2260 F 07 HOMEWORK#1 prob 3 solution...

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2260 F 07 HOMEWORK #1 prob 3 solution E X : + + 2.3 V t = 0 1 k 1 μH 100 m v ( t ) 5 μF After being closed for a long time, the switch opens at t = 0. Find v ( t ) for t > 0. S OL ' N : We may perform the following initial steps in any order: 1) Find characteristic roots for the parallel RLC circuit (for t > 0) 2) Find the final value of v ( t ) as t -> , which is the A 3 (constant) term in the solution. 3) Find the initial values of energy variables: i L (0 + ) = i L (0 ) and v C (0 + ) = v C (0 ) We will perform them in the order listed. First, we find the characteristic roots. s 1,2 = −α ± α 2 − ω o 2 For a parallel RLC circuit, the value of α is one-half the inverse RC time constant: α = 1 2 RC = 1 2 100m Ω⋅ 5 μ F = 1 M/s For both parallel and series RLC circuits, the resonant frequency, ω o , is the inverse of the square root of the product of L and C: ω o 2 = 1 LC = 1 1 μ H 5 μ F = 0.2 M 2 r/s ( ) 2 Substituting into the equation for the roots yields the following:
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s 1,2 = 1 Mr/s ± 1 0.2 Mr/s = 1 ± 2 5 Mr/s
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