MGVCL Exam Paper (30-07-2021 Shift 3) Transistors were used in First Generation Computers Fourth Generation Computers Third Generation Computers Second Generation Computers First Generation Computers Fourth Generation Computers Third Generation Computers Second Generation Computers ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In which of the following situations, there is no need to provide directional overcurrent protection. ring main double end fed, single feeder single end fed, parallel feeder single end fed, single feeder ring main double end fed, single feeder single end fed, parallel feeder single end fed, single feeder ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Directional overcurrent relay protection is used for parallel feeder and ring main system.It is not used for the protection of radial system.
MGVCL Exam Paper (30-07-2021 Shift 3) Find the word which is correctly spelt from the given options. Abolishid Obecity Seriuous Syndrome Abolishid Obecity Seriuous Syndrome ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Tiger Woods is associated with which of the following fields? Sports Cinema Politics Literature Sports Cinema Politics Literature ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Which of the following medium is used between CPU and RAM to speed up the processing power of a CPU? Virtual memory Flash memory DRAM Cache memory Virtual memory Flash memory DRAM Cache memory ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In a series RLC circuit, the supply voltage is 230 V at 50 Hz. The resonant current is 2 A at the resonant frequency of 50 Hz. Under the resonant condition, the voltage across the capacitor is measured to be equal to 460 V. What are the values of L and C? 0.73 mH and 13.85 pF 0.53 mH and 15.83 pF 0.73 H and 13.85 pF 0.53 H and 15.83 pF 0.73 mH and 13.85 pF 0.53 mH and 15.83 pF 0.73 H and 13.85 pF 0.53 H and 15.83 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Vc = I*Xc460 = 2*(1/(2*ᴨ*50*C))C = 2/(460*2*50*ᴨ)= 1.385*10⁻⁵ F= 13.85 FV_L = I*X_LL = V/(I*2*ᴨ*f)= 460/(2*ᴨ*50*2)= 0.7345 H
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