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 emitter current (IE) of a transistor has two components such as base current (IB) and collector current (IC). The base current (IB) is about 2% of the emitter current (IE), but the collector current (Ia) is about 98% of the emitter current (IE). ∴ IE > IC > IB IE = IB + IC
The strength of a magnetic field is specified by the magnetic induction B or by the magnetic field strength H. A current I through a long, straight wire produces a magnetic field with strength H = I/2πr at a distance r from the wire. Given H = 1 A/m I = 10 A H = I/2πr 1 = 10/2πr r = 10/2π
While estimating the overhead lines, the first and last pole is always earth connected and after every 3rd electrical pole, the fourth pole is earthed. Therefore, the approximate number of required ground set = 4
EXAMIANS
