SSC JE Electrical 2019 with solution SET-1 The Thevenin’s resistance as seen through the terminals A and B is: 8 Ω 5 Ω 6 Ω 4 Ω 8 Ω 5 Ω 6 Ω 4 Ω ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 A magnetic circuit is applied with a current that changes at a rate of 5 A/sec. The circuit has an inductance of 2H, then the self-induced EMF is 0.4 V −10 V −2.5 V −4 V 0.4 V −10 V −2.5 V −4 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP GivenInductance L = 2 HRate of change of current di/dt = 5 A/secSelf induced EMF = − (Rate of change of current × Inductance) = −L(di/dt)= −(5 × 2) = −10V
SSC JE Electrical 2019 with solution SET-1 The ratio of average energy demand to maximum demand during a specific period in Utilization factor Power factor Form factor Load factor Utilization factor Power factor Form factor Load factor ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The ratio of average energy demand to maximum demand during a specific period is called the load factor.Load Factor = Average Load/Maximum Demand
SSC JE Electrical 2019 with solution SET-1 The function of condenser in a Thermal power plant is: Purify water Condense used steam into water Condense water Purify steam Purify water Condense used steam into water Condense water Purify steam ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 The direction of the arrow represents the direction of when the diode is forward biased N-type material P-type material Conventional current flow P-N Junction N-type material P-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 Find the frequency of rotor induced EMF of a 3-phase, 440V, 50 Hz induction motor has a slip of 10% 5Hz 2.5Hz 50Hz 25Hz 5Hz 2.5Hz 50Hz 25Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Given DataVoltage = 440 VFrequency = 50 HzSlip = 10% =0.1Now,Rotor frequency = slip × frequency= 0.1 × 50 = 5 Hz
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