Consider the circuit shown: R1 100 Ω V1 sine ☑ 60 Hz 10 Vrms 1:1 سنا ли R2 1 Ω Figure 1: Impedance Matching Circuit 1. Calculate the magnitude (in RMS) of the voltage across R2. 2. Calculate the average power being dissipated in R2. 3. Determine the transformer ratio that provides for maximum power dissipation in R2 (and what is that maximum power).
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- 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) NoFor the circuit element of Problem 2.3, calculate (a) the instantaneous power absorbed, (b) the real power (state whether it is delivered or absorbed). (c) the reactive power (state whether delivered or absorbed). (d) the power factor (state whether lagging or leading). [Note: By convention the power factor cos() is positive. If | | is greater than 90, then the reference direction for current may be reversed, resulting in a positive value of cos() ].A 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.
- A single-phase 100-kVA,2400/240-volt,60-Hz distribution transformer is used as a step-down transformer. The load, which is connected to the 240-volt secondary winding, absorbs 60 kVA at 0.8 power factor lagging and is at 230 volts. Assuming an ideal transformer, calculate the following: (a) primary voltage, (b) load impedance, (c) load impedance referred to the primary, and (d) the real and reactive power supplied to the primary winding.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 Figure 3.4. For an ideal phase-shifting transformer, the imda nce is unchanged when it is referred from one side to the other. (a) True (b) False
- 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.A 50 MVA, 11/132 kV three-phase, 50 Hz, generator transformer has guaranteed parameters as, load loss 160 kW at full load at 75°C, no load loss 30 kW and percentage reactance 9% at rated load. Calculate the following:*percentage regulation a full load and unity power factor*percentage regulation at 3/4 load and 0.85 lagging power factor*efficiency at full load and unity power factor*efficiency at 3/4 load and 0.85 lagging power factor(Use 5 decimal places)The following data were obtained from a short-circuit test performed upon a 50 kVA, 2300/115 V, 60 Hz transformer: Esc = 87 V, Isc = 21.75 A, Psc = 590 W. Calculate (a) equivalent resistance, reactance and impedance referred to the primary, and (b) voltage regulation at a power factor of 0.866 lagging. Subject: Electrical Apparatus and Devices
- Which termination would result to a phase change of 180 degrees at the load if Zo = 100 ohm? a. 50 ohm b. short circuit c. 75 ohm d. all of these What is the purpose of impedance matching? a. maximum return loss b. maximum SWR c. maximum load power d. maximum reflection A short piece of transmission line that may be open or shorted and used for impedance matching purposes. a. converter b. decoder c. stub d. transformerData from short-circuit and open-circuit tests of a 25-kVA, 6900-230 V, 60 Hz transformer are given in table below: Sr No Open circuit parameters VOC = 230 IOC = 5.4 A POC = 260 W Short circuit parameters VSC = 513 ISC = 3.6 A PSC = 465 W Determine: (i) The magnetizing reactance referred to the high side (ii) The per unit parameters (iii) Efficiency (iv) Voltage regulation at 0.65 per unit load and 85% power factor leading (v) Low side voltage when the load is removed (vi) Voltage that must be applied to the primary in order to obtain the low side voltage in the part (v)A two-winding transformer rated at 100 kVA, is connected as an autotransformer to provide 10kV from a 15kV supply. Calculate the new rated kVA output. consider the power transformer mode and calculate the related currents Where, Vp = 10000VIp = S/(√3Vp) Is = S/(√3Vs) When find the new currents and voltages proceed to find the primary and secondary powers of the autotransformer. And conclude the rating of the autotransformer.