SSC JE Electrical 2019 with solution SET-1 Two decimal coils A and B of 1000 turns each lies in the parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is: 0.04 mWb 0.4 mWb 0.004 mWb 4 mWb 0.04 mWb 0.4 mWb 0.004 mWb 4 mWb ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φANo. of turns NA = NB = 1000Flux linkage in a coil with B = Flux linkage in coil A × 80/100= 0.8 × 5 × 10−5= 4 × 10−5 wb =0.04 mwb
SSC JE Electrical 2019 with solution SET-1 Identify the machine shown in the circuit DC short shunt compound motor DC shunt motor DC short shunt compound generator DC long shunt compound generator DC short shunt compound motor DC shunt motor DC short shunt compound generator DC long shunt compound generator ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 If N = linkage flux, then the linkage flux per unit current is defined as: Inductive reactance Leakage coefficient Magnetomotive force Inductance Inductive reactance Leakage coefficient Magnetomotive force Inductance ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The inductance L of this coil is defined as the flux linkage per unit current and its unit is Henry.L =Λ/IWhereΛ = Flux linkageI = current
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