MGVCL Exam Paper (30-07-2021 Shift 1) A new generation of smart meters is known as ____ SMETS1 SMETS4 SMETS2 SMETS3 SMETS1 SMETS4 SMETS2 SMETS3 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The second generation of smart meters are known as SMETS 2.Full form of SMET is Smart Metering Equipment Technical Specifications.It is with the previous generation known as SMETS 1.SMETS 2 smart meters were introduced in 2018.
MGVCL Exam Paper (30-07-2021 Shift 1) જીવનમાં તડકોછાયો એતો એક ઘટમાળ છે.વાક્યમાં ક્યા સમાસનો ઉપયોગ કરવામાં આવ્યો છે? કર્મધારય દ્વન્દ્વ તત્પુરુષ ઉપપદ કર્મધારય દ્વન્દ્વ તત્પુરુષ ઉપપદ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Three resistances 750 Ω, 600 Ω and 200 Ω are connected in parallel. The total current is 1 A. Determine the voltage applied. 200 V 125 V 100 V 150 V 200 V 125 V 100 V 150 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equivalent resistance of the circuit, Req = (750|| 600 || 200)Req = 125 ΩCurrent from source = 1 ASource voltage = 1*125Source voltage = 125 V
MGVCL Exam Paper (30-07-2021 Shift 1) આપેલ વિકલ્પો પૈકી દ્રવ્યવાચક સંજ્ઞાનું વાકય કયું છે? મહારાષ્ટ્રમાં અનેક શહેરો છે. રણવીરને બોલીવુડનો બાહુબલી કહી શકાય. ગુરુ એવો હોવો જોઈએ જેની વાણીમાં મીઠાસ હોય. આજકાલ ઠંડી હવા પ્રસરવા લાગી છે. મહારાષ્ટ્રમાં અનેક શહેરો છે. રણવીરને બોલીવુડનો બાહુબલી કહી શકાય. ગુરુ એવો હોવો જોઈએ જેની વાણીમાં મીઠાસ હોય. આજકાલ ઠંડી હવા પ્રસરવા લાગી છે. ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) While selecting the gas for a circuit breaker, the property of gas that should be considered is its(i) dielectric strength(ii) non-inflammability(iii) non-toxicity (i) only (i) and (ii) only (i), (ii) and (iii) (i) and (iii) only (i) only (i) and (ii) only (i), (ii) and (iii) (i) and (iii) only ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Properties of gas while selecting circuit breaker,1. Dielectric strength.2. Thermal conductivity.3. Arc quenching.4. Electrical properties.5. Chemical properties. Toxicity of arc products.
MGVCL Exam Paper (30-07-2021 Shift 1) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 15 Ω and the resistance of the resistor connected to the sound core was 45 Ω. Calculate the distance of the fault point from the test end. 75 m 450 m 225 m 150 m 75 m 450 m 225 m 150 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (15/60)*600= 150 m
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