Discuss the potential transformer with neat schematic diagram.
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Q: 1) How can you prove that Lsec = Lan (RI/ Rs) for matching impedance with the ideal transformer?
A: We are authorized to answer the only single question at a time since you have not mentioned which…
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A: Before differentiate between ideal and practical transformer let us understand the basics of simple…
Q: Along with suitable vector show, how the transformer behaves on load and no load condition
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A: Since both the question are not interrelated as per the guidelines of Bartley we supposed to answer…
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Q: (a) With neat diagram explain potential transformer.
A: a- Potential transformer - Also known voltage step down transformer or voltage step down…
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- For developing per-unit equivalent circuits of single-phase three-winding transformer, a common Sbase is selected for all three windings and voltage bases are selected in proportion to the rated voltage of the windings (a) True (b) FalseThe ideal transformer windings are eliminated from the per-unit equivalent circuit of a transformer. (a) True (b) FalseIn developing per-unit circuits of systems such as the one shown in Figure 3.10. when moving across a transformer, the voltage base is changed in proportion to the transformer voltage ratings. (a) True (b) False
- What should be the focus of transformer preventive maintenance efforts?For an ideal 2-winding transformer, the ampere-turns of the primary winding, N1I1 is equal to the ampere-turns of the secondary winding, N2I2 (a) True (b) FalseA three-phase transformer is rated 1000MVA,220Y/22kV. The Y-equivalent short-circuit impedance, considered equal to the leakage reactance, measured on the low-voltage side is 0.1. Compute the per-unit reactance of the transformer. In a system in which the base on the high-voltage side of the transformer is 100MVA,230kV what value of the per-unit reactance should be used to represent this transformer?
- A single-phase two-winding transformer rated 90MVA,80/120kV is to be connected as an autotransformer rated 80/200kV. Assume that the transformer is ideal. (a) Draw a schematic diagram of the ideal transformer connected as an autotransformer. showing the voltages, currents, and dot notation for polarity. (b) Determine 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 the kA rating is transferred by magnetic induction?The leakage reactance of a three-phase, 300-MVA,230Y/23-kV transformer is 0.06 per unit based on its own ratings. The Y winding has a solidly grounded neutral. Draw the per-unit equivalent circuit. Neglect the exciting admittance and assume the American Standard phase shift.Three single-phase, two-winding transformers, each rated 450MVA,20kV/288.7kV, with leakage reactance Xeq=0.10perunit, are connected to form a three-phase bank. The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit equivalent circuit if the low-voltage windings are connected (a) in with American standard phase shift or (b) in Y with an open neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current are neglected.
- An ideal transformer has no real or reactive power loss. (a) True (b) FalseRework 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. -A 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.