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Problem # 6 (20 points): An ac source of 110-V (rms) and angular frequency w=300 rad/s is connected to an LRC series circuit, where R=40 , L =200 mH, and C =600 pF. Find: (a) the inductive reactance, the capacitive reactance, and the impedance of the circuit (b) the peak and rms current, (c) the peak voltage and the rms voltage across each element, (d) the instantaneous voltage across each element.

Answer 1

Answer: R=401 L = 200mH C = 600 MF W = 300 9190/s Vyms = 110V (W) inductive Reactance = XL = WL = 6012 Capacitive Reactance = Xc = I 100 = 5.55_2 ш С. 18 impedence = z = f(x2-xx)? +R2 = 67.56 12 b) we know : Үpesk = 12 Vams = 155.561 Іpesk Xpeak Tyms = Vans Z Z = 155.56 = 8:30 A = 10 = 1.63 A. 67.56 67:56 (5) Across Resis tor: Fehk 7 YREK #13894 ☆ 500 voteskibi Vpeak = Ipeak XR Vyms = Irms XR = 927 = 65.2. Across Inductor: speak Ipeak X XL Xems = Irms XXL = 138 y 2 - (= = 97.87 toia ACMOSS capacitor! Xpeak Vams = Irns x Xc = 12: 765V E = 9.05 Vwo : Geld zu SEE White Ipeak Xc

> ($ = 53.7) VR = 92 Coswt). PO = =Å ale = Vc = 12.765 cos(wt-4) oovito work = 138 Cos (wt + o) A svitiaeg tand = XL-XC 1.36 R Answer: ² = (10 î+ 219 - 62 Å) NIC B = (0.57 +0.7 h +0.43²) UT wege Nowo, 10 i +21 9 - 62k 66.22 B 0.57 +0.7§ +0.43 0:96 direction of Em wave = Êx ^ 1 9 W spory 10 21 -62 = 66.22X0.94 0,5 0:7 0.43 em CO DA È = E 1w 1w 10/10 [ 63.57 = Tî (52.43) - ¡(9-35-3) + (-3.57] 2207ed 0.82î - 0.55 - 0.055€ Foop 69 Power LEI = 8.85 x102x0.5x C100+213 +622) P 1.1.94x108 J/sm2 or Watt/m2