SSC JE Electrical 2019 with solution SET-2
The combined inductance of two coils connected in a series is 0.6H and 0.1H, depending on the relative directions of the currents to the coils. If one of the coils when isolated has a self-inductance of 0.2 H, then calculate the natural inductance?
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
In the PMMC type instrument, the deflection is directly proportional to the current flowing through the instrument, we get a uniform scale for the instrument. It gives a uniform scale of upto 270° or more. Td ∝ I
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
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