For an air-cored transformer with Lp = 100 mH and Ls = 50 mH what is the mutual inductance (M = ?) if the coefficient of coupling, k = 0.7? a. 49 H b. 49 mH c. No right answer d. 30 mH
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- 7. The no-load current of a transformer is 4A at 0.25 pf when supplied at 25V, 60Hz. Find the magnetizing component of the no-load currentA 40-kVA, 8000/230-V distribution transformer has an impedance referred to the primary of 18 + j80 Ω. The components of the excitation branch referred to the primary side are Rc = 100 kΩ and Xc = 20 kΩ. If the primary voltage is 7920 V and the load impedance is ZL = 1.5 + j*0.7 Ω what is the secondary voltage, in volts?A 480V/240V, 60Hz single transformer has Z1 = 0.1+j0.22 ohms and Z2 = 0.035 + j0.12 ohm. Find the equivalent impedance in ohm referred to the Primary Group of answer choices 0.24+j0.62 0.14+j0.62 0.24+j0.7 0.14+j0.7
- A 2,400/480-V rms step-down ideal transformer delivers 50 kW to a resistive load. Calculate:(a) the turns ratio(b) the primary current(c) the secondary currentA linear transformer couples a load consisting of a 360 Ω resistor inseries with a 0.25 H inductor to a sinusoidal voltage source, as shown.The voltage source has an internal impedance of 184+j0 Ω and amaximum voltage of 245.20 V, and it is operating at 800 rad/s. Thetransformer parameters are R1=100 Ω, L1=0.5 H, R2=40 Ω,L2=0.125 H, and k=0.4. Calculate (a) the reflected impedance; (b) theprimary current; and (c) the secondary currentWhat is the impedance of a transformer coil that has 340' of #18 copper wire and an inductance of 70mH at 60Hz? 1000’ of #18 copper wire is 16 ohms.
- A step-down transformer is rated 10kVA, 7500/250 V and has the following equivalent circuit parameters:Rp = 2.92 Ω Rs = 0.00432 Ω Rm = 51,840 ΩXp = 14.6 Ω Xs = 0.02160 Ω Xm = 12,960 ΩAssume that this transformer is supplying rated load at 250 V and 0.85 PF lagging.(i)Find the equivalent winding impedance referred to the high voltage side.(ii)Determine this transformer’s input voltage.(iii)Determine its voltage regulation.A 5,000-kVA, 3-phase transformer, 13.2/33-kV, A/Y, has a copper loss of 11 kW. The impedance drop at full-load is 7.5%. Calculate the primary voltage when a load of 4000 kW at 0.8 p.f. is delivered at 33 kv.The low voltage side of a particular transformer has 747 turns while its high voltage side has 69471 turns,when connected as a step-up transformer, its secondary impedance is 371907 + J 250821 ohms. What is thenequivalent of this impedance in the primary side of the transformer?
- Given a 115kV/13.2kV distribution transformer rated at 30MVA with an 8% nameplate impedance what is the full Load current in ampsa 500/100 v potential transformer has following constants :primary resistance =47.25ohm, secondary resistance+0.43ohm,primart reactance=33. secondary reactance is negligible,no load primary current=0.1A at 0.6 p.f. calculate the no load angle between primary and reversed secondary voltage.A 20-kVA 8000/277-V distribution transformer has the following resistances and reactance: RP = 32 Ω RS = 0.05Ω X P = 45Ω XS = 0.06Ω RC = 250 kΩ X M = 30 kΩ The excitation branch impedances are given referred to the high-voltage side of the transformer. (a) Find the equivalent circuit of this transformer referred to the high-voltage side. (b) Find the per-unit equivalent circuit of this transformer. (c) Assume that this transformer is supplying rated load at 277 V and 0.8 PF lagging. What is this transformer’s input voltage? What is its voltage regulation? (d) What is the transformer’s efficiency under the conditions of part (c)?