Question

X=30R ㅔ Zoov 60HZ 32,- 4412 XL = 9052 R₂ = 3652. A Series Circuit Connected across a 2001, 60 Hz line Consists of a Capacitor of Capacitance 30%, a non- inductive resistor of 445 and a Coil of inductive Reactance 9052. Determine: @ The Current in the Circuit 6 the p.d across each element © the power factor of the Circuit) @ the power absorbed by the Circuit

Answer 1

S tine giveny, tion X = 301 Let i be the Cuxtent thyough 200valt 3 R = 442 the Circuit frequency= 60 HZ 1 So, w = tf . mooo W = 2 TX60 XL = 902 (@= 1207 dad see R 2=36) Gris connected tn series with Enductive - Jeactance XL) → Doeguency domain quivalent Circuit: - Soequency demandent . -Xe = -130 ) DR - R 200valt 6 R = 442 L = JULY w=lRotballe (6 - - j X c = = 1/2 VAD = Bizor I Lemon i= 200 ZOO 7 367190-120) B = 36 XL=20} RtjXL = (36 +190) By using the otin's law, we get i= 200 Lo (44+ 367j90–139) T80+j605 By replacing demoninaty Part with its polar foom equivalent as 8otibo = 1180 + (fore tanto = 100 / 36.869 we get → i = 200 Loºh o ton Too 136-869° > = 2 436-9099 |Ambrogie Ane Di Potential drop across -ix is given by : Ve= fj%) i Ve= 302–200 xa 436-869° - Ne = 6o L-125-861) Answers Potential drop across the R, is given by; VR = R i VR = 44 X 2 (-36.869° i [VR = 88 (-36-869 Volt Answers 1 (111) Potential cloob across Botjq) & is given by (across the inductive Vz= (36+19x1 V--10932 / 03.198ºX (2 36-8690)

→ z = 193-894 481.png" volt) Answer Hesultant o resultant angle tn ble the current f valtage in the series. R-L-C circuit is called as Power factor angle of Phases of current i warity the voltage is called as power facter angle So, e o V= doovolt 10 = -388897 Pour factor is given by Cos d = Cos (36-967) > PF= cos®z0-8jAns

d) Power absorbed by the Circuit is given by: Voms Isms Casal (Voms = 200 volt 1 Toms = Q Amp P= 200 X 2X 018 foom (2 +36-869 A) so; P = 320 watt Answer