SSC JE Electrical 2019 with solution SET-2
If the connected light load in a house is 3000W and power sub-circuit load 6000W, then what is the total number of sub-circuits required?

3
4
1
6

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SSC JE Electrical 2019 with solution SET-2
IL = Load Current, then the relation between these currents for a DC series generator is

IL ÷ Ise = Ia
IL = Ise + Ia
IL ÷ Ise ÷ Ia
IL = Ise = Ia

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SSC JE Electrical 2019 with solution SET-2
A series magnetic circuit will have

Different magnetic flux
Two or more paths for magnetic flux
Total reluctance of a series circuit = difference of the reluctances in different parts of the circuit
Same magnetic flux

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