SSC JE Electrical 2019 with solution SET-2 If ‘V’ is the voltage phasor and ‘I’ is the current phasor, then VI represents Apparent power Total power Active power Reactive power Apparent power Total power Active power Reactive power ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Apparent Power (S): It is defined as the product of r.m.s value of voltage (V) and current (1). It is denoted by S.S = V/I Volt Ampere
SSC JE Electrical 2019 with solution SET-2 The residential distribution system employs_____ Three-phase, three-wire Two-phase, four-wire Single-phase, two-wire Three-phase, four-wire Three-phase, three-wire Two-phase, four-wire Single-phase, two-wire Three-phase, four-wire ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-2 Observe the given figure. Find Thevenin’s resistance as seen from open-circuited terminals? 4 Ω 8 Ω 32 Ω 16 Ω 4 Ω 8 Ω 32 Ω 16 Ω ANSWER EXPLANATION DOWNLOAD EXAMIANS APP To find the Thevenin’s resistance the current source is open-circuited and the voltage source is short-circuited.
SSC JE Electrical 2019 with solution SET-2 The water hammer effect is expected in a Turbine casing Surge tank Draft tube Penstock Turbine casing Surge tank Draft tube Penstock ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-2 A circuit consists of two parallel resistors, having a resistance of 20Ω and 30Ω respectively connected in series with 15Ω. If the current through the 15Ω resistor is 3A, then find the current through 20Ω and 30Ω resistors respectively? 1A, 2A 1.2A, 1.8A 2A, 1 A 1.8A, 1.2A 1A, 2A 1.2A, 1.8A 2A, 1 A 1.8A, 1.2A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-2 The candle power of a lamp placed normal to a working plane is 60 CP. Find the distance if the illumination is 15 lux? 2 meters 2.5 meters 4 meters 1.5 meters 2 meters 2.5 meters 4 meters 1.5 meters ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Candle Power of lamp = 60 CPIllumination of lamp = 15 luxDistance(D) = ? metersIllumination of lamp = (candle power)/(distance)2(D)2 = candle power/Illumination of lamp(D)2 = 60/15 =4D = √4D = 2
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