AC MACHINES-1 (66761) Theory

8.6 Prob.

1. A 5kVA, 200 V/ 100 V, 50 Hz, single phase ideal two winding transformer is to used to step up a voltage of 200 V to 300 V by connecting it like an auto transformer. Show the connection diagram to achieve this. Calculate the maximum kVA that can be handled by the autotransformer (without over loading any of the HV and LV coil). How much of this kVA is transferred magnetically and how much is transferred by electrical conduction.

Solution

Two connect a two winding transformer as an auto transformer, it is essential to know the dot markings on the two coils. The coils are to be now series connected appropriately so as to identify clearly between which two terminals to give supply and between which two to connect the load. Since the input voltage here is 200 V, supply must be connected across the HV terminals. The induced voltage in the LV side in turn gets fixed to 100 V. But we require 300 V as output, so LV coil is to be connected in additive series with the HV coil. This is what has been shown in figure 28.10.

Solution Two connect a two winding transformer as an auto transformer, it is essential to know the dot markings on the two coils. The coils are to be now series connected appropriately so as to identify clearly between which two terminals to give supply and between which two to connect the load. Since the input voltage here is 200 V, supply must be connected across the HV terminals. The induced voltage in the LV side in turn gets fixed to 100 V. But we require 300 V as output, so LV coil is to be connected in additive series with the HV coil. This is what has been shown in figure 28.10. 50 A Figure 28.10: Two winding transformer as an autotransformer. 300 V Load 100 V 25 75 A 200 V supply P A Here the idea is not to exceed the voltage and current rating of HV and LV coils of the two winding transformer. Now for the transformer having rating 5 kVA, 200 V/ 100 V, 50 Hz we have: Rated voltage of HV coil is = 200 V Rated voltage of LV coil is = 100 V Phase turns ratio is a = 200/100 = 2 Rated current of each HV coil is = 5000/200 = 25 A Rated current of each LV coil is = 5000/100 = 50 A Since the load is in series with LV coil, so load current is same as the current flowing through the LV coil. Thus a maximum of 50 A can be drawn by the load otherwise overloading of the coils take place. Output kVA = 300 × 50 VA = 15 kVA input kVA = Output kVA = 15 kVA ∵ transformer is ideal Version 2