Concept explainers
A single-phase,
Resistance of the 2400-V winding:
Resistance of the 240-V winding:
Leakage reactance of the 2400-V winding:
Leakage reactance of the 240-V winding:
Exciting admittance on the 240-V side
(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.
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Chapter 3 Solutions
POWER SYSTEM ANALYSIS+DESIGN-EBK >I<
- A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding short-circuited, rated current flows when the voltage applied to the primary is 5.5 kV. The power input is read as 100 kW. Determine Req1andXeq1 in ohms referred to the high-voltage winding.arrow_forwardFor an ideal 2-winding transformer, an impedance Z2 connected across winding 2 (secondary) is referred to winding 1 (primary) by multiplying Z2 by (a) The turns ratio (N1/N2) (b) The square of the turns ratio (N1/N2)2 (c) The cubed turns ratio (N1/N2)3arrow_forwardFor an ideal transformer, the efficiency is (a) 0 (b) 100 (c) 50arrow_forward
- An ideal transformer has no real or reactive power loss. (a) True (b) Falsearrow_forwardThe ratings of a three-phase three-winding transformer are Primary(1): Y connected 66kV,15MVA Secondary (2): Y connected, 13.2kV,10MVA Tertiary (3): A connected, 2.3kV,5MVA Neglecting winding resistances and exciting current, the per-unit leakage reactances are X12=0.08 on a 15-MVA,66-kV base X13=0.10 on a 15-MVA,66-kV base X23=0.09 on a 10-MVA,13.2-kV base (a) Determine the per-unit reactances X1,X2,X3 of the equivalent circuit on a 15-MVA,66-kV base at the primary terminals. (b) Purely resistive loads of 7.5 MW at 13.2 kV and 5 MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per- unit impedance diagram, showing the per-unit impedances on a 15-MVA,66-kV base at the primary terminals.arrow_forwardThe symbol shown is a(n) a. iron core transformer. b. auto transformer. c. current transformer. d. air core transformer.arrow_forward
- (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 ?arrow_forwardConsider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in , place the polarity marks, and label the terminals a, b, and c in accordance with the American standard. (b) Relabel the terminals a, b, and c such that VAN is 90 out of phase with Va for positive sequence.arrow_forwardRework Problem 3.14 if the transformer is delivering rated load at rated secondary voltage and at (a) unity power factor, (b) 0.8 power factor leading. Compare the results with those of Problem 3.14. -arrow_forward
- A 23/230-kV step-up transformer feeds a three-phase transmission line, which in turn supplies a 150-MVA,0.8 lagging power factor load through a step-down 230/23-kV transformer. The impedance of the line and transformers at 230kVis18+j60. Determine the tap setting for each transformer to maintain the voltage at the load at 23 kV.arrow_forwardThe 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) Falsearrow_forwardA single-phase l0-kVA,2300/230-volt,60-Hz two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 to 2530 volts (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of this kVA rating is transformed by magnetic induction? (C) The following data are obtained from tests carried out on the transformer when it is connected as a two-winding transformer: Open-circuit test with the low-voltage terminals excited: Applied voltage =230V, input current =0.45A, input power =70W. Short-circuit test with the high-voltage terminals excited: Applied voltage =120, input current =4.5A, input power =240W. Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransformer than as a two-winding transformer.arrow_forward
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