EE110_W08_HW 2 Solution - Q1 1-stage V1 1 0 DC 20 R1 1 2...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Q1. 1-stage: V1 1 0 DC 20 R1 1 2 10Ohm R2 2 0 10Ohm .OP .PRINT DC I(V1) VOLTAGE SOURCE CURRENTS NAME CURRENT V1 -1.000E+00 Thus, Req = 20/1.0 = 20 Ohm 3-stages: V1 1 0 DC 20 *stage 1 R1 1 2 10Ohm R2 2 0 10Ohm *stage 2 R3 2 3 10Ohm R4 3 0 10Ohm *stage 3 R5 3 4 10Ohm R6 4 0 10Ohm VOLTAGE SOURCE CURRENTS NAME CURRENT V1 -1.231E+00 Thus, Req = 20/1.231 = 16.25 Ohm Repeat similarly for 5-stage and 7-stages. We then obtain the plot: 0 0.5 1 1.5 2 2.5 0 1 2 3 4 5 6 7 8 No. of Stages Req/R Yes, Req/R converges to 1.618, which is (sqrt(5)+1)/2, the value obtained in HW1.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Q2. Parallel RLC circuit: (a) For an underdamped response, we require 0 α ω < , so that 15 18 1 1 1 1 10 or ; 2 2 2 2 10 L R R RC C LC < > > × . Thus, R > 11.18 Ω . (b) For critical damping, 1 11.18 2 L R C = = Ω (c) For overdamped, R < 11.18 Ω . Q3. 100 200 40 30 mA, C 1mF, (0) 0.25V = = = − t t c i e e v (a) (b) (c) 100 200 100 200 100 200 1 ( ) 0.25 (40 30 ) 0.25 C ( ) 0.4( 1) 0.15( 1) 0.25 ( ) 0.4 0.15 V t t t t c o o t t t t v t i dt e e dt v t e e v t e e = = = − + = − + 2 2 2 2 1 2 3 2 2 100 200 R 100 , 200 300 2 , 150 1 1 500 150 ,R 3.333 Also, 2R10 150 200 150 22500 20000 1 100 20000 , L 0.5H LC L i ( ) 0.12 0.045 A R = − = −α + α − ω = − = −α − α − ω ∴− = − α α = + = = Ω = − − ω ∴ω = = = = = = + o o o o t t s s s v t e e 100 200 100 200 ( ) ( ) ( ) (0.12 0.04) ( 0.045 0.03) ( ) 80 15 mA, 0 t t R c t t i t i t i t e e i t e e t = − = + − + = >
Image of page 2
Q4. Referring to Fig. 9.43, L = 1250 mH so Since α > ω o , this circuit is over damped. The capacitor stores 390 J at t = 0 : 2 1 1 2 2 So (0 ) 125 V (0 ) + = = = = c c c c c W C v W v v C The inductor initially stores zero energy, so Thus, 8 2 ( ) = + t t v t Ae Be Using the initial conditions, (0) 125 [1] = = + v A B 3 8 2 3 (0 ) (0 ) (0 ) (0 ) 0 (0 ) 0 2 (0 ) 125 So (0 ) 62.5 V 2 2 50 10 [ 8 2 ] (0 ) 62.5 50 10 (8 2 ) [2] + + + + + + + + + + = + + = = − = − = − = = × = − = − × + L R c c c t t c c v i i i i v i dv i C Ae Be dt i A B Solving Eqs. [1] and [2], A = 150 V B = 25 V Thus, 8 2 ( ) 166.7 41.67 , 0 = > t t v t e e t 1 1 4rad/s 1 5 2 ω = = α = = o LC s RC 2 2 1,2 (0 ) (0 ) 0 5 3 8, 2 + = = = −α ± α − ω = − ± = − L L o i i S
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Q5. (a) (b) Q6. It’s probably easiest to begin by sketching the waveform v x : (a) The source current ( = i L ( t ) ) = 0 at t = 0 - .
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern