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An ideal transformer with
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Chapter 3 Solutions
Power System Analysis and Design (MindTap Course List)
- Consider Figure 3.25 of the text for a transformer with off-nominal turns ratio. (i) The per-unit equivalent circuit shown in part (c) contains an ideal transformer which cannot be accommodated by some computer programs. (a) True (b) False (ii) In the - circuit representation for real c in part (d), the admittance parameters Y11 and Y12 would be unequal. (a) True (b) False (iii) For complex c, can the admittance parameters be synthesized with a passive RLC circuit? (a) Yes (b) Noarrow_forwardA two-winding single-phase transformer rated 60kVA,240/1200V,60Hz, has an efficiency of 0.96 when operated at rated load, 0.8 power factor lagging. This transformer is to be utilized as a 1440/1200-V step-down autotransformer in a power distribution system. (a) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the ratings as a two-winding transformer. Assume an ideal transformer. (b) Determine the efficiency of the autotransformer with the kVA loading of part (a) and 0.8 power factor leading.arrow_forwardAn ideal transformer has no real or reactive power loss. (a) True (b) Falsearrow_forward
- The parameters of a 1000/250V, 50HZ single phase transformer are: R1=1.3 N, X1=2.8 N, R2=0.11 N, X2=0.2 N, R,=250 N, X,=900 N. If the load voltage is 220V, load power is 10kW, and load power factor is 0.8 lagging. Find by using the exact equivalent circuit: (i) Input current, (ii) The input voltage and power factor, (iii) The magnetizing current, (iv) The copper losses, (v) The voltage regulation.arrow_forwardA Transformer with 10:1 ratio and rated at 50-kVA, 2400/240-V, 50-Hz is used to step down the voltage of a distribution system. The low-tension voltage is to be kept constant, if the transformer has: R = 2.2 0, R2 = 0.014 Q, X, = 4.1 0, X2 = 0.03 0, then, a. What load impedance connected to low-tension side will be loading the transformer fully. b. Calculate the equivalent resistance as referred to primary. c. Calculate the equivalent reactance as referred to primary. d. Calculate the equivalent impedance as referred to primary. e. Calculate the total Cu loss.arrow_forwardA transformer takes a No load current of 0.64 A and power factor of 0.654 lag, when its primary is connected to a 225 V, 50 Hz supply and secondary being open circuited, then the value of iron loss current is,arrow_forward
- A 15-kVA, 2400:240-V, 60-Hz, single-phase transformer hasthe following circuit parameters expressed in ohmsRP = 2.50Ω RS = 0.025Ω XP = 7.00Ω XS = 0.070ΩRC = 32 kΩ Xm = 11.5 kΩIf the transformer is supplying a 10-kW, 0.8 p.f. lagging loadat rated voltage. Assume that the 0.8 p.f. lagging load can be modeled by apassive impedance, but the transformer primary is connected toa 2400-V bus. Determine the resulting (actual) load current andvoltage using approximate circuit of Figure 1.Answer should be 50.7/-37.86° A ...arrow_forwardThe equivalent parameters of a 150kVA, 2400V/240V transformer, are R1=0.2Ω, R2=2mΩ, X1=0.45Ω, X2=4.5mΩ, Rc=10kΩ, and Xm=1.55kΩ. The transformer is operating at rated load and rated voltage with 0.8 lagging power factor. Using the approximate equivalent circuit referred to the primary side, determine: (i) the voltage regulation (ii) the transformer power loss and efficiency Please refer to the answerarrow_forwardA 230-kVA, 2300/230-V, 60-Hz, step-down, two-winding transformer has the following parameters RH = 1.2 Ohm, XH = 3 Ohm, RL = 12 mOhm, XL = 30 mOhm,RcH = 2 kOhm, XmH = 1.8 kOhm. A) If the transformer operates at half load at its rated terminal voltage with a unity power factor, what is its efficiency? B) If the transformer is operating at 80% of its load, at secondary terminal voltage of 216 V and 0.866 pf lagging, what is its efficiency?arrow_forward
- Primary side-reduced equivalent circuit parameters of a 25 KVA, 50 Hz, 4000/400 Volt transformer R1 = 32 Ω, X1 = 45 Ω, Rc = 250000 Ω, R2 '= 66.14 Ω, X2' = 79.34 Ω, Xm It is given as = 30000 Ω. Since the secondary of the transformer is fed at a full load rated voltage with an advanced power factor of 0.9, for these operating conditions; a) Calculate the primary voltage using the T equivalent circuit. (Only the amplitude of the voltage and the phase value will not be entered.) b) Calculate the efficiency of the transformer. c )Calculate the voltage regulation of the transformer.arrow_forwardConsider an ideal transformer with N1 ¼ 3000 and N2 ¼ 1000 turns. Let winding 1 beconnected to a source whose voltage is e1ðtÞ ¼ 100ð1 jtjÞ volts for 1a ta 1 ande1ðtÞ ¼ 0 for jtj > 1 second. A 2-farad capacitor is connected across winding 2. Sketche1ðtÞ, e2ðtÞ, i1ðtÞ, and i2ðtÞ versus time tarrow_forwardA single phase transformer is rated 2400V/240V, 500kVA 50HZ yields the following primary side equivalent parameters R=19ohm, X=0,148ohm, G=4,25x10^-4 S and B=4,14x10^-3 S. By using base values calculate per-unit values for primary side of the transformer (Vbase=2400V, Sbase=500x10^3 VA)arrow_forward
- Power System Analysis and Design (MindTap Course ...Electrical EngineeringISBN:9781305632134Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. SarmaPublisher:Cengage Learning