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?
Indian Electricity Rules specify that the maximum load on a light/fan sub-circuit should not exceed 800 watts and the number of points should be limited to 10. Hence for light load number of sub-circuit = 3000/800 = 3.75 = 4 Indian Electricity Rules specify that the maximum load on a power sub-circuit should not exceed 3000 watts and the number of outlets should be limited to two. For the power sub-circuit load number of sub-circuit = 6000/3000 = 2 Total subcircuit = 4 + 2 = 6
The factor by which, the induced E.M.F gets reduced due to short pitching is called pitch factor or coil span factor denoted by Kc. It is given as Kc = cosα/2 given α = 60° Kc = cos60/2 Kc = cos30° Kc = √3/2
EXAMIANS