The sequence components of the current in an unbalanced system, with phasesequence A-B-C, is given as IA0 = 10300 A, IA1 = 5000 A and IA2 = 10–300 A, and thecorresponding sequence components of voltage are given as VA0 = 50600 V, VA1 =20000 V and VA2 = 100300 V.a) Calculate the current in phase B.b) Calculate the total active power and reactive power of the system.

Power in a balanced three-phase system circuit isa.b.c.d.

A 415-V, 50-Hz supply is applied to a three-phase star-connected balanced load. Each [10]phase has an impedance of R+jX ohms, where R=15 n and X = 13.4 n. Find the(i) phase voltage, f'2(ii) phase current,(iii) line current,(iv) power factor,(v) active power,(vi) reactive power,S,vii) total VA

A single phase voltage source Vs =240∠0° V RMS supplies two loads in parallel with thefollowing parameters (all values RMS),1) 10 kVA at a 0.6 lagging power factor2) 2 kW at a 0.90 leading power factora) Draw the power triangle for each of the two loads and complete Table 1. [4]b) Calculate the power triangle of the combined loads and complete Table 1. Draw thecombined power triangle. [2]c) Apply S=VsI s* to the combined power to calculate the phasor current I s.[1]d) Calculate the required kVARs of capacitance in parallel with the loads to correct the powerfactor to 0.9 lagging. Complete Table 1 as required. [2]e) Calculate the new phasor current I sn for the new power factor corrected system.

a) To determine the magnitude and phase of each line current supplied by the source, we use Ohm's law (I = V/Z). Given that the phase voltage V = 240∠30° V and the impedance Z = 50∠25° Ω, we can calculate the current as follows: I = V/Z = 240∠30° / 50∠25° = 4.8∠5° A This is the phase current for phase A. Since it's a balanced star connection, the line current equals the phase current. Therefore, the line current for each phase is 4.8∠5° A. b) To determine the active power and the reactive power supplied by the source, we use the power formula P = √3 * VL * IL * cos(θ) for active power and Q = √3 * VL * IL * sin(θ) for reactive power. Given that the line voltage VL = √3 * phase voltage = √3 * 240 V = 415.69 V, the line current IL = 4.8 A, and the power factor cos(θ) = cos(25°) = 0.9063, we can calculate: P = √3 * 415.69 V * 4.8 A * 0.9063 = 2721.6 W And for reactive power, sin(θ) = sin(25°) = 0.4226, so: Q = √3 * 415.69 V * 4.8 A * 0.4226 = 1341.6 VAR c) To determine the reactive power requirement of each of the three capacitances to improve the power factor to 0.95 lag, we first need to calculate the new reactive power Q_new using the formula Q_new = P * tan(θ_new), where θ_new = cos^(-1)(0.95). Q_new = 2721.6 W * tan(cos^(-1)(0.95)) = 859.5 VAR The reactive power provided by the capacitors should be the difference between the original reactive power and the new reactive power, which is Q - Q_new = 1341.6 VAR - 859.5 VAR = 482.1 VAR. Since the capacitors are connected in star, the reactive power for each capacitor is 482.1 VAR / 3 = 160.7 VAR. ####

A balanced three-phase star-connected supply of phase sequence ABC feeds a balancedstar connected load of impedance 5025°Ω in each phase. Phase voltage of Phase A ismeasured to be 24030° V.a) Determine the magnitude and phase of each line current supplied by the source.b) Determine the active power and the reactive power supplied by the source.c) Determine the reactive power requirement of each of the three capacitances,connected in star, to improve the power factor to 0.95 lag.