The equivalent circuit of a 22KVA, 11700/520 V, 60HZ, single-phase loaded transformer when referred to its low-voltage side is given. Find the percent voltage regulation of this transformer under this loading conditions. 4= 40.25 22.5 1.0 + j1.25 e Vs = 4 Ko j6.1 Ko 381/0° Select one: O a. %VR = -6.0 %
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- A single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R1=0.75 Resistance of the 240-V winding: R2=0.0075 Leakage reactance of the 2400-V winding: X1=1.0 Leakage reactance of the 240-V winding: X2=0.01 Exciting admittance on the 240-V side =0.003j0.02S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.Consider an ideal transformer with N1=3000andN2=1000 turns. Let winding 1 be connected to a source whose voltage is e1(t)=100(1| t |)volts for 1t1ande1(t)=0 for | t |1 second. A2- farad capacitor is connected across winding 2. Sketch e1(t),e2(t),i1(t),andi2(t) versus time t.An ideal transformer has no real or reactive power loss. (a) True (b) False
- (a) An ideal single-phase two-winding transformer with turns ratio at=N1/N2 is connected with a series impedance Z2 across winding 2. If one wants to replace Z2, with a series impedance Z1 across winding 1 and keep the terminal behavior of the two circuits to be identical, find Z1 in terms of Z2. (b) Would the above result be true if instead of a series impedance there is a shunt impedance? (c) Can one refer a ladder network on the secondary (2) side to the primary (1) side simply by multiplying every impedance byat2 ?The transformer of Problem 3.16 is supplying a rated load of 50 kVA at a rated secondary voltage of 240 V and at 0.8 posr factor lagging. Neglect the transformer exciting current. (a) Determine the input terminal voltage of the transformer on the high-voltage side. (b) Sketch the corresponding phasor diagram. (c) If the transformer is used as a step-down transformer at the load end of a feeder whose impedance is 0.5+j2.0, find the voltage VS and the power factor at the sending end of the feeder.The direct electrical connection of the windings allows transient over voltages to pass through the auto transfonner more easily, and that is an important disadvantage of the autotransformer. (a) True (b) False
- An infinite bus, which is a constant voltage source, is connected to the primary of the three-winding transformer of Problem 3.53. A 7.5-MVA,13.2-kV synchronous motor with a sub transient reactance of 0.2 per unit is connected to the transformer secondary. A5-MW,2.3-kV three-phase resistive load is connected to the tertiary Choosing a base of 66 kV and 15 MVA in the primary, draw the impedance diagram of the system showing per-unit impedances. Neglect transformer exciting current, phase shifts, and all resistances except the resistive load.Consider a source of voltage v(t)=102sin(2t)V, with an internal resistance of 1800. A transformer that can be considered as ideal is used to couple a 50 resistive load to the source. (a) Determine the transformer primary-to-secondary turns ratio required to ensure maxi mum power transfer by matching the load and source resistances. (b) Find the average power delivered to the load, assenting maximum power transfer.Example: A single-phase, 100-kV A, 1000:100-V, 60-Hz transformer has the following test results: Open-circuit test (HV side open): 100 V, 6 A, 400 W Short-circuit test (LV side shorted): 50 V, 100 A, 1800 W * Draw the equivalent circuit of the transformer referred to the high- voltage side. Label impedances numerically in ohms and in per unit. » Determine the voltage regulation at rated secondary current with 0.6 power factor lagging. Assume the primary is supplied with rated voltage * Determine the efficiency of the transformer when the secondary current is 75% of its rated value and the power factor at the load is 0.8 lagging with a secondary voltage of 98 V across the load 55 EMI Transformers Dr Thamir Hassan ATYIA Tikrit University
- Q/ A 20 KUA , 200/500 voit single phase step-up transformer hes: primary resistance: 0.022 ; secondary resistance = 0.0482; magnetizing branch with Ro - 1001 and Xo = 750.52. Findi magnetizing Current ? The efficiency at rated current unity ef: wa efteciemy at hall rated current with Pt 0.7 leading? The effecieing at 0:8 Vated current Zerol.f._; Maxoff At vnity PF ?A 1000-VA, 230/115-V traformer has been tested to determine its equivalent circuit. The results are shown below: Open-Circuit Test results from the low voltage side: VOC = 115 V, IOC = 0.1 A, POC = 5 W; Short-Circuit Test results from the high voltage side: VSC = 20 V, ISC = 10 A, PSC = 50 W; Find the transformer’s voltage regulation, in percent, at 50% loading conditions and 0.8 PF lagging. Note: (If %1, the you should write as '1', not as '0,01')3) The results of an open circuit test and a short circuit test In a 450 kVA transformer, 42000/2400 V, 60 Hz are the following: Open circuit test (low voltage side) VCA = 2400 V ICA = 18,1 A PCA = 3785 in Short circuit test (High voltage side) Vcc = 1910 V Icc = 11,9 A PCC3850 W a) Find the equivalent circuit of the transformer referred to the bass side voltage. b) Calculate the efficiency of the transformer at full load with a factor of Power of 0.85 in delay.