Question 6: (7 Points) For the three phase system shown in Fig. 6 3-phase source seguence:...
Please show how this transformers problem is done. Thanks! Problem 3ph [10 pts]: Y-connected source: A Y-connected balanced three-phase source is feeding a balanced three-phase load (it doesn't matter whether the load is Y or Delta connected). The voltage and current of the source are: van (t) - 340 sin (377t + 0.5236) ia(t) 100sin (377t 0.87266) Calculate the following: (a) The rms phase voltage (Van, magnitude and phase). (b) The rms line-to-line voltage (Vab, magnitude and phase) (c) The...
#3. For the 3-phase power system shown in Fig. 3, the load is 30 kVA at a p.f. of 0.94 lagging. If the phasor load voltage, V. = 480/0 Compute: (i)The phasor line current, I. Ia Fig. 3 (ii) The phasor voltage Vec of the source. A . Contjo. 6). Ba [Zunet (0.1+30.6). 30 kVA hase/p. f. 0.94 las I ho o.1tjo. 652 ů 2
4. (20 points) The three-phase Y-connected power supply has a line-to-neutral voltage KIN of 1.220° KV. A balanced-A load consisting of impedance Z1-20 30 Ω per phase is in parallel with a balanced-Y load having phase impedance of Z2 = 15245° Ω. Identical line impedances of Z1 = 3L15。Ω are in each of the three lines connecting the combined loads to the phase supply (a) Draw the per-phase equivalent circuit; (b) Find the line-to-line voltage Vab (magnitude and angle) at...
QUESTION 4: A 400 V three-phase, 50 Hz system, ABC sequence supplies the following loads: A star-connected load with the following phase impedances: ZA ZZ10410' BN A delta-connected load of equal phase impedances: ZB15435Z=30Z-75Zc=(19+j27 )n 4.1 Use VAB as reference and detemine the following quantities: The magnitude and angle of the currents drawn by the star-connected load (lAN lEN, ICN) 4.1.1 (6) 4.1.2 The magnitude and angle of the phase (IAB; lBc; IcA) and line (IA; lB; Ic) currents drawn...
A balanced 3 phase system is given with the following specifications: - The balanced system has a single delta-connected source and a single Y-connected load side. - Between the source and the load side there is the line_impedance = Rlinr+ jXline. - The average power and reactive power delivered form the source is Pin = 500 Watt and Qin = 500 VAR respectively. - The source side line voltage is Vline_source=100 Volt (rms). - The load side average and reactive...
Question #2 [30 Points) For the three phase system shown in Fig.1, use base of 20 MVA and 2.3 kV of the load. (i) Specify base voltages in all parts of this system. (ii) Find all reactance's in pu on the common base. (iii) Draw the reactance diagram with all pu reactance values marked on the diagram. (iv) Calculate the rated load current in amperes and in pu at the common base. (v) Find the voltage at bus l in...
(15 points) In a balanced three-phase wye-delta system, the source has a positive (abc) phase sequence and a phase voltage of Van 200VrmsZ - 30°. If the line impedance is zero and the line current is LaA -6ArmsZ- 65°, find the load impedance per phase in the delta. AB AB Arms
There is a problem regarding power system. A 190km three phase transmission line is rated at 745kV (Line to line). The receiving bus is at the rated voltage, with the voltage phase angle at 0 degrees. The receiving current is 1,15828.24 A. The ABCD parameters for this line are: B = 72.42281.40 C = 8.856x10-3285.34 A = 0.826520.135 0.826520.135 D = Find the magnitude of the line sending current (Is) in Amperes.
1. A three-phase voltage source is feediing the three phase receiving end voltage through a three phase series impedance. To compensate the lines, a series voltage is injected using the three phase static synchronous series compensator (SSSC). Draw a schematic diagram of this system including the lines, the transformer and the converter. 2. The per unit source voltage magnitude is 1.0 pu with an angle of 10 degrees while per unit phase voltage of the receiving end is 0.95 pu...
In a balanced three-phase Y-A system, the source has an ABC phase sequence and Van=114/_43° V rms. The line and load impedances are Z1 = R1 + jX1 = 2 + j1.82 and Zab = Rab + jXab = 54 + j3922, respective А RijX w IAa a VAN n Lab Ica јХca, Rab VON N VBN Rca jXab C B Roc jXbo с b RjX2 Wm Wm R3 jX3 Find the magnitude of the delta current Iab, Ibc, or...