SSC JE Electrical 2019 with solution SET-1 The cheap and temporary system of the internal wiring is? Conduit wiring CTS or TRS wiring Cleat wiring Casing-capping Conduit wiring CTS or TRS wiring Cleat wiring Casing-capping ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 Which of the following losses is together called iron losses? Eddy current loss and frictional loss Hysteresis loss and frictional loss Eddy current loss and hysteresis loss Hysteresis loss and copper loss Eddy current loss and frictional loss Hysteresis loss and frictional loss Eddy current loss and hysteresis loss Hysteresis loss and copper loss ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 Prevention of interference with neighboring telephone lines can be done by: Transposing transmission lines Using bundled conductors Reducing skin effect Reducing corona Transposing transmission lines Using bundled conductors Reducing skin effect Reducing corona ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 The trade name of the Nickel-Copper alloy, that is used as a heating element is: Nichrome Steel Kanthal Eureka Nichrome Steel Kanthal Eureka ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Constantin‘, also known as ‘Eureka wire‘, is the trade-name for a copper-nickel alloy (approx. 60:40 ratio) formulated in the late 1800s by Edward Weston.
SSC JE Electrical 2019 with solution SET-1 If N = linkage flux, then the linkage flux per unit current is defined as: Magnetomotive force Leakage coefficient Inductive reactance Inductance Magnetomotive force Leakage coefficient Inductive reactance Inductance ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The inductance L of this coil is defined as the flux linkage per unit current and its unit is Henry.L =Λ/IWhereΛ = Flux linkageI = current
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)
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