SSC JE Electrical 2019 with solution SET-1 At f = _____R – L – C series circuit operates at unity power factor 1/RC 1 ⁄ RLC 1 ⁄ LC 1/2π√LC 1/RC 1 ⁄ RLC 1 ⁄ LC 1/2π√LC ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 An electric heater is connected across 230V and it draws a current of 2A. Then the resistance offered by the heater is? 115 ohm 290 ohm 29 ohm 11.5 ohm 115 ohm 290 ohm 29 ohm 11.5 ohm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP GivenVoltage V = 230 VCurrent I = 2 AResistance R = V/IR = 230/2 = 115 Ω
SSC JE Electrical 2019 with solution SET-1 A synchronous motor runs at 600rpm, which of the following case is true? P = 12, f = 60 Hz P = 8, f = 50 Hz P = 10, f = 60 Hz P= 12, f = 50 Hz P = 12, f = 60 Hz P = 8, f = 50 Hz P = 10, f = 60 Hz P= 12, f = 50 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 In the speed-time curve of a train, speed in and time are plotted km/s; sec km/h; hour km/h; sec km/s; hour km/s; sec km/h; hour km/h; sec km/s; hour 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 V1 (V2 − V1) −(V2 − V1) V2 V1 (V2 − V1) −(V2 − V1) V2 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Net voltage = −V2 + V1Net voltage = −(V2 − V1)
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: 0.4 mWb 4 mWb 0.04 mWb 0.004 mWb 0.4 mWb 4 mWb 0.04 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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