SSC JE Electrical 2019 with solution SET-1 The cheap and temporary system of the internal wiring is? Casing-capping Conduit wiring CTS or TRS wiring Cleat wiring Casing-capping Conduit wiring CTS or TRS wiring Cleat wiring ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 Identify the machine shown in the circuit DC long shunt compound generator DC shunt motor DC short shunt compound motor DC short shunt compound generator DC long shunt compound generator DC shunt motor DC short shunt compound motor DC short shunt compound generator ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 Two inductors of 4H and 6H are connected in series. The equivalent inductance of this combination is 4 H 2.4 H 10 H 6 H 4 H 2.4 H 10 H 6 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Total inductance when inductor are connected in seriesL = L1 + L2L = 4 + 6 =10H
SSC JE Electrical 2019 with solution SET-1 The direction of the arrow represents the direction of when the diode is forward biased P-type material N-type material Conventional current flow P-N Junction P-type material N-type material Conventional current flow P-N Junction ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The direction of the arrow represents the direction of the conventional current flow when the diode is forward biased.
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: 4 mWb 0.4 mWb 0.004 mWb 0.04 mWb 4 mWb 0.4 mWb 0.004 mWb 0.04 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 In the two wattmeter method, the readings of the two wattmeters are 500W, 500W respectively. The load power factor in a balanced 3-phase 3-wire circuit is: 0.8 0.5 0.9 1 0.8 0.5 0.9 1 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In two wattmeter method the phase angle is tanφ = √3(W1 − W2)/(W1 + W2) tanφ = √3(500 − 500)/(500 + 500) tanφ = 0° φ = tan−10° = 0° Power factor = cosφ PF = cos0° = 1
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