SSC JE Electrical 2019 with solution SET-1 Prevention of interference with neighboring telephone lines can be done by: Reducing skin effect Using bundled conductors Transposing transmission lines Reducing corona Reducing skin effect Using bundled conductors Transposing transmission lines Reducing corona ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 A synchronous motor runs at 600rpm, which of the following case is true? P= 12, f = 50 Hz P = 8, f = 50 Hz P = 10, f = 60 Hz P = 12, f = 60 Hz P= 12, f = 50 Hz P = 8, f = 50 Hz P = 10, f = 60 Hz P = 12, f = 60 Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Synchronous speed NS = 120f/PWhenP = 12 & F = 60 then Ns = ?NS = 120f/P = (120 × 60)/12 = 600 RPM
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 10 H 2.4 H 6 H 4 H 10 H 2.4 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 The Thevenin’s resistance as seen through the terminals A and B is: 8 Ω 6 Ω 5 Ω 4 Ω 8 Ω 6 Ω 5 Ω 4 Ω ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 If an alternator is operating at unity power factor, then its terminal voltage is Equal to the induced EMF, with zero voltage regulation Less than the induced EMF, with negative regulation Less than the induced EMF, with positive voltage regulation Greater than the induced EMF, with negative voltage regulation Equal to the induced EMF, with zero voltage regulation Less than the induced EMF, with negative regulation Less than the induced EMF, with positive voltage regulation Greater than the induced EMF, with negative voltage regulation ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 With the current direction marked in the circuit shown, the net voltage applied is −(V2 − V1) V2 (V2 − V1) V1 −(V2 − V1) V2 (V2 − V1) V1 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Net voltage = −V2 + V1Net voltage = −(V2 − V1)
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