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Part III
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Electromechanics
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If a load current i2 is now required by the connection of a load to the output circuit (by closing the switch in the gure), the corresponding mmf is F2 = N2 i2 This mmf, generated by the load current i2 , would cause the ux in the core to change; however, this is not possible, since a change in would cause a corresponding change in the voltage induced across the input coil But this voltage is determined ( xed) by the source v1 (and is therefore d /dt), so that the input coil is forced to generate a counter mmf to oppose the mmf of the output coil; this is accomplished as the input coil draws a current i1 from the source v1 such that i1 N1 = i2 N2 or N1 i2 = = i1 N2 (1637) (1636)
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where is the ratio of primary to secondary turns (the transformer ratio) and N1 and N2 are the primary and secondary turns, respectively If there were any net difference between the input and output mmf, ux balance required by the input voltage source would not be satis ed Thus, the two mmf s must be equal As you can easily verify, these results are the same as in 7; in particular, the ideal transformer does not dissipate any power, since v1 i1 = v2 i2 (1638)
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Note the distinction we have made between the induced voltages (emf s), e, and the terminal voltages, v In general, these are not the same The results obtained for the ideal case do not completely represent the physical nature of transformers A number of loss mechanisms need to be included in a practical transformer model, to account for the effects of leakage ux, for various magnetic core losses (eg, hysteresis), and for the unavoidable resistance of the wires that form the coils Commercial transformer ratings are usually given on the so-called nameplate, which indicates the normal operating conditions The nameplate includes the following parameters:
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Primary-to-secondary voltage ratio Design frequency of operation (Apparent) rated output power For example, a typical nameplate might read 480:240 V, 60 Hz, 2 kVA The voltage ratio can be used to determine the turns ratio, while the rated output power represents the continuous power level that can be sustained without overheating It is important that this power be rated as the apparent power in kVA, rather than real power in kW, since a load with low power factor would still draw current and therefore operate near rated power Another important performance characteristic of a transformer is its power ef ciency, de ned by: Power ef ciency = = Output power Input power (1639)
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The following examples illustrate the use of the nameplate ratings and the calculation of ef ciency in a practical transformer, in addition to demonstrating the application of the circuit models
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Principles of Electromechanics
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EXAMPLE 167 Transformer Nameplate
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Problem
Determine the turns ratio and the rated currents of a transformer from nameplate data
Solution
Known Quantities: Nameplate data Find: = N1 /N2 ; I1 ; I2 Schematics, Diagrams, Circuits, and Given Data: Nameplate data: 120 V/480 V;
48 kVA; 60 Hz
Assumptions: Assume an ideal transformer Analysis: The rst element in the nameplate data is a pair of voltages, indicating the
primary and secondary voltages for which the transformer is rated The ratio, , is found as follows: N1 480 = =4 = N2 120 To nd the primary and secondary currents, we use the kVA rating (apparent power) of the transformer: |S| 48 kVA |S| 48 kVA = 100 A I2 = = 400 A = = I1 = V1 480 V V2 120 V
Comments: In computing the rated currents, we have assumed that no losses take place
in the transformer; in fact, there will be losses due to coil resistance and magnetic core effects These losses result in heating of the transformer, and limit its rated performance
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