SSC JE Electrical 2019 with solution SET-1 The Thevenin’s resistance as seen through the terminals A and B is: 4 Ω 8 Ω 5 Ω 6 Ω 4 Ω 8 Ω 5 Ω 6 Ω 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 2.4 H 10 H 4 H 6 H 2.4 H 10 H 4 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 Prevention of interference with neighboring telephone lines can be done by: Reducing corona Transposing transmission lines Reducing skin effect Using bundled conductors Reducing corona Transposing transmission lines Reducing skin effect Using bundled conductors ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 Which of the following lamps employs a bimetallic strip? Incandescent lamp Fluorescent lamp Sodium vapor lamp Mercury vapor lamp Incandescent lamp Fluorescent lamp Sodium vapor lamp Mercury vapor lamp ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 In an AC network, the load connected is (10 + j10). The phase relation between the voltage applied and the current through the load is: Voltage lags current by 30° Voltage lags current by 45° Voltage leads current by 45° Voltage and current are in phase with each other Voltage lags current by 30° Voltage lags current by 45° Voltage leads current by 45° Voltage and current are in phase with each other ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L0ad impedance Z = R + jXz = 10 + 10jPhase angle θ= tan−1(IL/IR)θ= tan−1(10/10)tanθ = 1tanθ =45°
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 4 mWb 0.4 mWb 0.004 mWb 0.04 mWb 4 mWb 0.4 mWb 0.004 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
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