MGVCL Exam Paper (30-07-2021 Shift 2) A single phase motor connected to 415 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging. 96.32 µF 66.32 µF 76.32 µF 86.32 µF 96.32 µF 66.32 µF 76.32 µF 86.32 µF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = VIcosφ = 8715 wattQ = VAR required to improve PF = P(tanφ₁ - tanφ₂) = 4670 VARCapacitive Reactance, Xc = V²/Q = 36.8790 ohm.C = (2πfXc)⁻¹ = 86.314 μF
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 2500 A 25 kA 35 kA 350 kA 2500 A 25 kA 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) Find the word which is correctly spelt from the given options. Sentimant Wrinkel Absurrd Maximum Sentimant Wrinkel Absurrd Maximum ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is an application layer service? Mail service Network virtual terminal Mail service File transfer, access and management Mail service Network virtual terminal Mail service File transfer, access and management ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following statement is TRUE? In the cables, sheaths are used to provide enough strength. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. The current carrying capacity of cables in DC is less than that in AC. In the cables, sheaths are used to provide enough strength. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. 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 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 930 kVA 2480 kVA 1240 kVA 620 kVA 930 kVA 2480 kVA 1240 kVA 620 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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