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. 35 kA 25 kA 350 kA 2500 A 35 kA 25 kA 350 kA 2500 A 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) The operating time of an inverse definite minimum time (IDMT) overcurrent relay is ____ proportional to the plug-setting multiplier (PSM) and____proportional to the time-multiplier setting (TMS). inversely, directly directly, inversely inversely, inversely directly, directly inversely, directly directly, inversely inversely, inversely directly, directly ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Encrypted security payload extension header is new in IPv4 IPv5 IPv6 IP IPv4 IPv5 IPv6 IP ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Find the word which is correctly spelt from the given options. Maximum Wrinkel Sentimant Absurrd Maximum Wrinkel Sentimant Absurrd ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In circuit breaker, the term MCCB stands for miniature compact circuit breaker modern closed circuit breaker miniature case circuit breaker moulded case circuit breaker miniature compact circuit breaker modern closed circuit breaker miniature case circuit breaker moulded case circuit breaker ANSWER EXPLANATION DOWNLOAD EXAMIANS APP MCCB stands for Moulded Case Circuit Breaker.MCB stands for Miniature Circuit Breaker.ELCB stands for Earth Leakage Circuit BreakerRCCB stands for Residual Current Circuit Breaker.
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 86.32 µF 66.32 µF 76.32 µF 96.32 µF 86.32 µF 66.32 µF 76.32 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
EXAMIANS