Question

112 mann Is → I 2652 250 20° V(rms) -j100 12 j392 Quiz 12b The load L is the series resistor and inductor in parallel with the capacitor. a) Calculate the rms phasors Vi and IL | Vel= Vrms. Phase angle Vų = ° (Degrees) (smallest negative angle) | IL 1 = Arms Phase angle IL = (Degrees) (smallest negative angle) b) Calculate the average power and magnetizing reactive power absorbed by the (26+j 39) 22 (Ohm) elements. Pavg = w Q= VAR c) Calculate the power factor and the power factor angle seen by the voltage source. pf = pf angle = ° (Degrees)

Answer 1

34 oon VL M Is + 1 262 AC 250L VL joor Voms j 394 ov (reference node) as shown V be let voltage no de Nodal Analys (node vel V VL + (VL - 25000) (1+34) (1100) (26+339) a so + (troo) 26+; 39 1+14 1+34 jo.24 - 0.071 -( 243) 14.706 - 158.823 14.706 - j58.823 = 239.51 (-2.25ºv 0071- 10.243 Now, 1L = VL 26+ j39 239.54 (-2.259 (26+ 339) 5.11 (-58.56 I

llula 239.51 Voms 1<" --2-257 27-58.56] [17. 7. Son Arms O P Q PAVO 2 678.91 absorbed by (26 +339) 1913 (26) (5:11) (26) .] TIL 1 (39) (50 (39) 137 JUAR 10/8.37 (1 + i +) acco)-(234-814-223") (+ 34) 3.45 (-34.59° . Is = Arms L's (Is 0-(-34.599) = 34.59 Pf = COS Cos 3459 0.823