Question

a) A single phase 240V 50Hz supply is supplying a power to a load which consist of 500 resistance and 0.3H inductance through a cable which has an effective impedance of 1+j2 0. The equivalent circuit of the single-phase system is shown in Figure Q1(a). Zcable=1+j2 I Load 5092 240V 50Hz Supply V Load 0.3H Figure Q1(a): A single phase system. i. ii. Determine the load current, low and the load voltage Vood Determine the active and reactive power delivered to the load Determine the active and reactive power supplied by the source. Draw the phasor diagram for current and supply voltage iii. iv. b) Based on a), in order to ensure an efficient power utilization, a minimum power factor of 0.85 lagging is required by the utility company. i. Check the power factor of the single-phase system and if the system doesn't meet the requirement, design a simple power factor correction circuit to improve the power factor to a minimum value. Determine the value of component used in the power factor correction circuit and analyse the change on the load current

please take note the question is extend

Answer 1

(a) Zcable = 1+j2 + 240V ş 502 Боnz load 0.3H (i) So voltage is 240 Lo V. > for I lood supply Total Z = Zload + Zooble - (+j?) + (50 +251403) Z = 51+ 96.24j 240 LO 2.2. L-62.08 I load - v Z 51+j96.24 Now Vload = I land x Zload Veoad. = 235.074 2-0.02° (ii) Now, active & reactive power delivered to load First, we will find apparent power te. S-VI* = 635.074 L-0.02)x (2.20L+62.08) tis due to 517.95 262.05° VA Now read & reactive power are Conjugate )

vollage Read power = 242.73 W Reactive bower = 457.55 VAR (ii) Now, supplied by source the power will be given by supply Again SEVI* - (240L0) (2.20L+67.08) S- 528.8 (+62.08 Now real & reactive power supplied by P=247.59 W 0 = 467. 26 VAR. sowice are iv) here 235.074 Z-0.0pº v Is = 2.20 L-62.080 A X=40_o Ye (Iex Xl Te Re

(6) Now, the power factor of source source must be 0.85 as mentioned by ability company, So power factor is Cos (Cangle between Voltage & Current) je cos 62.08 = p.fi p.f. So, need to improve ove the b.f. that Capacitor connection across the cable must be installed So let's value of capacitive element is 0.468, means series assume new (6 -jX? So Z cable () +j?+ (-; X) Zablo 1 + j(9-x) 1 -j (x-2) Now, tatal load, Z. Zland + Ziable. - 50 siden)-(-;-) Z = 51 + j (309+-x) Now for 0.85 p.j. tan (cos' (0.85) Зоr + 2-х 51

That means reactive part divide by active part is fan of angle be of p.j. So for cos' (0-85) - fan 31.78 = 307+2-X 51 X= (30T +2) – (51x fan 31.78') - (Bor + ?) - (31-606) X = + 64.64 So I Cw 64.64 Now for value of Capacitance C, -C P8x50x 64.64 so C = 49.24 MF