prelab ( exp 4).docx

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1-PRELAB: 1. Characteristic equation: S 2 + 2*α*S + W 0 2 =0 With: W 0= LC / 1 and L R 2 / This equation has two solutions S1 and S2: S1= 2 0 2 and S2= 2 0 2 In this case, R=10 kΩ L=0.25H C=0.1 µ F Schematic: a- In this case: W 0= LC / 1 = 6324.55 L R 2 / = 20000
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S1= 2 0 2 = -1026.33 and S2= 2 0 2 = -38973.66 b- We set R=Rc; To get Rc, we get two double roots for the characteristic equation: S 2 + 2*α*S + W 0 2 =0 = 4 α 2 – 4W 0 =0 6 2 6 2 2 10 1 . 0 1 10 1 . 0 25 . 0 4 25 . 0 4 c c c R R LC L R So R c = 3162.27766 Ω In this case: V c (t) = (A 1 + A 2 t) e - αt with A 1 and A 2 two constants to determine L Rc 2 / = 6324.55 = W 0 So V c (t) = (A 1 + A 2 t) e - αt = (A 1 +A 2 t) e -6324.55 t c- Now, we changed the value of R and we fix it to R=150 Ω
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In this case: W 0= LC / 1 = 6324.55 L R 2 / = 300 W 0 > α W d 2 = W 0 2 – α 2 S1= - α + j W d = -300 + 6317.43 j and S2= - α - j W d = -300 – 6317.43 j 2- Case 1: R=10 kΩ
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After we use the transient analysis, we get the following results.
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  • Winter '19
  • jihad daba
  • Following, The Circuit, Characteristic polynomial, Complex number, RC circuit, Pallavolo Modena

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