SSC JE Electrical 2019 with solution SET-2 ____comes under the category of high-frequency heating? Resistance heating Eddy current heating Arc heating Infrared heating Resistance heating Eddy current heating Arc heating Infrared heating ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-2 Observe the given table. The truth table represents gate? XOR NAND OR AND XOR NAND OR AND ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In the exclusive OR gate (XOR gate) has two inputsThe ouput will be 1 when either input is 1 i.e X = 1 and Y = 1, but not when both the inputs are 1.If both the input is zero i.e X = 0 and Y = 0, then the output is 0.
SSC JE Electrical 2019 with solution SET-2 If a zero-centered voltmeter has a scale from 5V to − 5V, then the span of it is 5 V −5 V 0 V 10V 5 V −5 V 0 V 10V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Voltage Scale Span of instrument = Vmax − VminGivenVmax = 5 VVmin = −5VSpan = 5 −(−5) = 10 V
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? 2A, 1 A 1A, 2A 1.8A, 1.2A 1.2A, 1.8A 2A, 1 A 1A, 2A 1.8A, 1.2A 1.2A, 1.8A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-2 An ideal voltage source is one that____? Supplies constant power Has zero internal resistance Has infinite internal resistance Has medium internal resistance Supplies constant power Has zero internal resistance Has infinite internal resistance Has medium internal resistance ANSWER 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 1.5 meters 2.5 meters 4 meters 2 meters 1.5 meters 2.5 meters 4 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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