MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 350 kA 25 kA 2500 A 35 kA 350 kA 25 kA 2500 A 35 kA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following statement is TRUE? In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. In the cables, sheaths are used to provide enough strength. The current carrying capacity of cables in DC is less than that in AC. In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. In the cables, sheaths are used to provide enough strength. The current carrying capacity of cables in DC is less than that in AC. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors.
MGVCL Exam Paper (30-07-2021 Shift 2) A delta-connected balanced three-phase load is supplied from a three-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10 kW. Find the power factor. 0.52 0.82 0.62 0.72 0.52 0.82 0.62 0.72 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = 10 kWVL = 400 VIL = 20 A3- phse power, P = √3*VL*IL*cosφcosφ = 10*1000/(√3*400*20)= 0.7216
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase 240 V, 1 kW heater is connected across single phase 240 V, 50 Hz supply through an SCR. For firing angle delay of 90˚, calculate the power absorbed in the heater element. 300 W 225 W 250 W 275 W 300 W 225 W 250 W 275 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Resistance of heater, R = V²/PR = 57.6 ohmRMS value of voltage through controlled SCR = (Vm/2π)*(1 + cosα)= 120 VPower absorb by heater = Vrms²/R= 120²/57.6= 250 W
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 20 to 25 km 10 km to 15 km 75 km to 100 km 30 km to 60 km 20 to 25 km 10 km to 15 km 75 km to 100 km 30 km to 60 km ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A wire pilot is generally economical for distances up to 5 or 10 miles, beyond which a carrier-current pilot usually becomes more economical.
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