Question

Impedance, Power, Phasor of RC Circuit a. Given a circuit below, calculate and draw the following quantities: Z, Ez, Iz, P, Q, S and draw their Phasors Iz 11 Es 100 V E1 Ez R 171 22 Xc 2002

Answer 1

ja) Given círcuét Iz Iz + I GT mm Xc Ez 2002 Es 100 V 111-2 Impedance! Here Rand xc. are in pasllal so IR) (ix) R_3Xc Z - => (170) (-1200) 2 - 171-1200 2= 129.97/139.4699% (ii) Ezt Here Ez T8 source voltage itself because Ammeter "t" has no internal Resistance. Ez = Es = 100 V (Rms) so I I Es Iq- Z. Too Iz 11 129.97/139.469 Iq = 0.769 L 139.46 A covy P, B and ! S = Es I 2 * S = (100) (0.769 [139.46) S = (100) (0.769/139.46)

S: 76.94VĄ S: (58.471+350.009) VA. The above Cuation Creation room load side end seen b seen from source and side S: (58.471- 550.009)VA. P = 58.471w Q = 50 VAR. at means source delivering Active power (P) and obserbing Reachive power(Q). Power factor! Tz = 0.769 [139.46 A. P-F = cos (139146) PoFs -0.7599 power factor is leading Nature. in

phasor diagram! Iz 12 SOVAR 76.94VA 외 58.47W P

ㄴ R А Source © А mm R V source where "A means Ammeter for current i measuremea tvy means voltmeter for voltage measurement @ from the Graph:- At bottom they already mentioned values. Voms 99.30V Tome = 23.07 A Varg -0.23 V Laug - -0.03 (ORY For voltage! Upeak Ell division Peek to peek -22 division But per division Howomany volts Not Given. 6 Uspeck? Vorms= Vmbro Up Na Vp = Vomst = 99.3x2 Vp= 140.431V

111) Ispeak? I peak = 1Rms 2 23.0712 Peak = 32.625 A Vilis Vs (Pecak to Peak) 2 Vpeak Vs - 2x140.431 (P-P) NS = 280.862 V (P-P) Irme = 23.074A را ) | Vams = 99.30 v (vi (Vit) Ve (Aug) -0.23 VNOV Is = -0.03 NOA (Ang) (vity Phase Angle blo vs & De at both zero crossing point difference is 3 divisions. .so time gap blw them equal to (3) x 5 = ismsec Given System Has frequency of .50H3. time period = 20msec phase of is worot Vs = 270 lag loro gol lead- (*) phase shift is qo leading (X compared to "Vs' Is is lagging, AS