SSC JE Electrical 2019 with solution SET-1 With the current direction marked in the circuit shown, the net voltage applied is (V2 − V1) V1 −(V2 − V1) V2 (V2 − V1) V1 −(V2 − V1) V2 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Net voltage = −V2 + V1Net voltage = −(V2 − V1)
SSC JE Electrical 2019 with solution SET-1 If an alternator is operating at unity power factor, then its terminal voltage is Less than the induced EMF, with positive voltage regulation Greater than the induced EMF, with negative voltage regulation Less than the induced EMF, with negative regulation Equal to the induced EMF, with zero voltage regulation Less than the induced EMF, with positive voltage regulation Greater than the induced EMF, with negative voltage regulation Less than the induced EMF, with negative regulation Equal to the induced EMF, with zero voltage regulation ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 ______starting method CANNOT be used for starting a 3-phase squirrel cage induction motor Direct-on-line Rotor-resistance Auto-transformer Star-delta Direct-on-line Rotor-resistance Auto-transformer Star-delta ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 Find the frequency of rotor induced EMF of a 3-phase, 440V, 50 Hz induction motor has a slip of 10% 25Hz 5Hz 50Hz 2.5Hz 25Hz 5Hz 50Hz 2.5Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Given DataVoltage = 440 VFrequency = 50 HzSlip = 10% =0.1Now,Rotor frequency = slip × frequency= 0.1 × 50 = 5 Hz
SSC JE Electrical 2019 with solution SET-1 A 200 V d.c machine as Ra = 0.5 Ω and its full-load Ia= 20A. Determine the induced e.m.f when the machine acts as a motor 190 V 210 V 200 V 215 V 190 V 210 V 200 V 215 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Given DataVoltage Va = 200 vArmature Resistance Ra = 0.5ΩArmature Current Ia = 20 AInduced EMF = Ea = ?The Induced EMF of a DC machine working as a Motor isEa = Va − IaRaEa = 200 − 20 × 0.5Ea = 190 V
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 45° Voltage leads current by 45° Voltage lags current by 30° Voltage and current are in phase with each other Voltage lags current by 45° Voltage leads current by 45° Voltage lags current by 30° 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°
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