HPSSSB JE Electrical Mock Test
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A) The level of voltage
Exp-: In a transformer, the energy transfer takes place with a change of voltage supply frequency of output remains the same as that of the input. Power factor level are almost the same on both sides of the transformer.
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D) 95%
Exp-:Transformer is a static machine owing to absence of rotating parts are no there are no friction or windage losses. Further the other losses are also relatively low, so that the efficiency of transformer. is high
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B) by magnetic flux
Exp-: The transformer work on the principle of faraday law of electro-magnetic induction which state that when the rate of change of flux linking change there is emf induce therefore the flux is created by the primary also link with secondary and the flux of secondary interact with primary then there is a emf induce and power is transfer form primary to secondary through magnetic flux
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D) Both primary and secondary windings
Exp-: Because the flux created by the primary also link with secondary and the rate of change of flux linking is change and emf is induce in both the winding bcoz of interaction of primary and secondary flux
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B) Primary supply voltage
Exp-: The Working component or core less component or active component is in phase with the applied voltage. It supplies the iron losses and a small primary copper loss.
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D) .35
Exp-: The exciting or no-load current IO is made up of a relatively large magnetizing component Im and a comparatively small in-phase or energy componennt Ie so the power factor of a transformer on no load is very small and it is due to magnetising reactance of the transformer
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A) Is strictly constant with load changes
Exp-: core flux is directly proportional to supply voltage and inversly proportional to supply frequency so it independent of load
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C) Remain constant irrespective of load
Exp-: Since the secondary flux φ2 produced by seconndary mmf N2I2 is neutralized by the flux φ'1 produced by mmf N1I' set up by conterbalancing primary current I '1 , the flux in the transformer core remains constant from no load to full load
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C) Copper loss are equal to iron losses
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A) Reduce the weight per KVA
flux density=φ/A Therefore flux density is inversely proportional to area, weight will reduces for higher flux density.
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