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 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.
Io = V/R = 20/100 Io = 0.2 A Since the given diode is an ideal diode therefore there is no voltage drop across it. Vo = Io × RD Vo = 0.2 × 0 Vo = 0 Hence Io = 0.2 A & Vo = 0
EXAMIANS