HW_2

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Stanford University *

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210

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Electrical Engineering

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Oct 30, 2023

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HW2 P Haoming Mei ② a) E parallel ① 1 In Fi " ② series 25 25 5 =- us 1 2 . 5 3 . 5 T
③ Combining Parallel Resistors : 5, I 5/1 - - - 12 i) + I ② b . ) Ren ② ↓ ① series Parallel - 6 . 6 36 = 3 6 :
origi ③ series : in ji = 2n ⑭ ↓ parallel & series - + 3 + 2 I + + ⑦ Parallel : nison
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I 55 -- - Reg = W 1 . 4 = 4V - ① , I 48 + ② Voltage ⑧ · . Difference -8V 4 ⑯ & I = 4 ⊥ e I = I - 1A
& B KVL Verification : 4 - (4)( - 1) - 7 = 0 4 + 4 - 8 = 0 y - 8 = 0 ⑭ al t - I - I I - - ① KUL : (1) Is = Iz 12 (2) Vs - 31 , -21 , - 12) = 0 (3) 2(I1 - 12) - 312 = 0
- Is - 12 + Iz = 0 W 51 , -212 - Vs = 0 25 - 512 = 0 0 3 4 ② ① b) 01 A A I references V - Vs = 0 (1) V , = Vs Node & : Vi = +V k - V Node : 5 + Is= Y
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Y Vs = 0 2V , -2k = 3k + 2k - 2V3 V2 - V + 31 = 31 V - 4Vy + 3 Is = 0 V - V = 0 2V , -7 + 2V3 = 0 W Y - 4z + 31s = 0
④ Y = k , Vs + kz Is I ① short Is - · Vi
④ d) Us = /oV I = /A . - Is - 12 + Iz = 0 - I2+Iz = 1 51 , -212 - V = 0 - 51 , 212 = 10 25-512 = 0 21 , -512 = 0 A Used Python - HW2 In Iz I3 A = - 0 - I I I L 5 - 20 I =] 2 - 50 b = (i) x = To
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I 1 = 2 . 381A W I2 = 0 . 952 A I3 = 1 . 952A V - V = 0 ④ Use Python : 2V , -7 + 2V3 = 0 - V , Vz V> A = 2 - x2 Y - 4z + 31s = 0 I I 00 I 01-4 -> - b - 10 10 L 0 7 = [s . 303] - 3 : V2 = 3 . 3032 V3 = 1 5772
- # - t - I x) Vs , -51 , -21 , -12) - 3Ix = 0 (2) 3 Ix + 2)1 , - 12) - 51x - Vx = 0 Plug Ex = Iz V si - 71 , + 21 - 3(I2) = 0 VI-7I , -12 = 0
3(2) + 21 , -2 I-3Iz -Vsz = 0 - 212+21 , -Vs = 0 Vs , -71 , -12 = 0 E - Vs2 + 211-212 = 0 ⑤ b) t - t n ⑪ t - I - Ohm's Law ① V = Vs ③ Ix Y ④ Plug : ② +Vs2 = 0 Vy - 31x = 0
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⑮ ⑮ - ) = 0 V -V + V = 0 Node 8 : VR= V + 6V , -6 = 15-15V3 +10-loVx 6 V , -31V2 + 15V s HoVp = 0 Equations : V , = Vs I V = - Vs2 W - v2+3+ = 0 6V , -31U2+15Vz Hoy = 0
④ ] t - t t - Vz = k , Vs , + > Usa ② ① Short Vs , ⑧ ② ⑮ 1 Ve NodeD 3 2 E I - e - -vs - = 5 3Ix OV / t I V - ①
③ ) Python : Vs , -71 , -12 = 0 - I A = -E I - Vs2 + 211-212 = 0 2 - 2 = [] x = I ] V , = Vs I ⑤ e) V = - Vs2 - v2+3+ = 0 6V , -31U2+15Vz Hoy = 0 - I 00 0 I V , V Vs My b To I A = 0 ⑧ 0 I - 4 0 - I I I 6 - 31 15 10
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V , = -loV V2 = - 5V * ] ⊥ 3 = - PV 4 = 4V

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