SSC JE Electrical 2019 with solution SET-1
A magnetic circuit is applied with a current that changes at a rate of 5 A/sec. The circuit has an inductance of 2H, then the self-induced EMF is
Given Inductance L = 2 H Rate of change of current di/dt = 5 A/sec Self induced EMF = − (Rate of change of current × Inductance) = −L(di/dt) = −(5 × 2) = −10V
In electrical engineering, the power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit ” R/Z”.
The self-inductance is given as L = μN2A/I L ∝ N2 where N is the number of turns of the solenoid A is the area of each turn of the coil l is the length of the solenoid and μ is the permeability constant L1/L2 = N21/N22 0.5/0.5 = N21/N22 N1/N2 = 1
During the positive half cycle of the supply, diodes D1 and D2 conduct are forward biased and conduct current while diodes D3 and D4 are reverse biased and they act as an open circuit, the current flows through the load.
EXAMIANS