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
In the exclusive OR gate (XOR gate) has two inputs The ouput will be 1 when either input is 1 i.e X = 1 and Y = 1, but not when both the inputs are 1. If both the input is zero i.e X = 0 and Y = 0, then the output is 0.
The combined inductance of two coils connected in series L = L1 + L2 + 2M In series adding case L1 + L2 + 2M = 0.6 H ——–(1) In series opposing case L1 + L2 − 2M = 0.1 H ——–(2) Subtracting eqn (2) from eqn (1) we get 4M = 0.5 H M = 0.125 H Let L1 = 0.2 H (since the coil when isolated, its self-inductance is 0.2 H) Putting the value of M & L1 in equation (2) 0.2 + L2 + 2 ×0.125 = 0.06 0.2 + L2 = 0.35 L2 = 0.15 H
EXAMIANS