Three equal impedances are first connected in delta across a three-phase balanced supply. If the same impedances are connected in star across the same supplya.phase current will be one-thirdb.line current will be one-thirdc.noned.power consumed will be one-third

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.

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 three-phase load is balanced if all the three phases have the samea.impedanceb.power factorc.impedance and power factord.none of the above

A balanced star connected load of (2+j3) Ω per phase and a balanced delta-connected load of(6+j6) Ω per phase are connected in parallel to a three phase, 400V, 50Hz supply. Calculate thephase current in each load, the total current in each supply line, the total power supplied and theoverall power factor.

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. ####