EE110_W08_HW 2 Solution

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

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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.

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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 = − = + − + = >
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

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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 - .
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• Winter '08
• Gupta
• Volt, Cos, iL

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