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, inversely inversely, directly directly, inversely directly, directly inversely, inversely inversely, directly directly, inversely directly, directly ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The necessary equation to be solved during the load flow analysis using Fast-Decoupled method is given below:where B' and B'' are formed using the imaginary part of the bus admittance matrix, Ybus. If the system consists of 6 buses of which one of the bus is slack bus and 2 bues are voltage regulated buses, then then the matrix B' is of order ____and the matrix B'' is of order____, respectivity.∆P/|Vi| = - B'∆δ∆Q/|Vi| = - B''∆|V| 3 and 5 3 and 6 5 and 3 6 and 3 3 and 5 3 and 6 5 and 3 6 and 3 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In Fast-decoupled method,Order of B' = No. of PV buses = 5Order of B'' = No. of PQ buses = 3
MGVCL Exam Paper (30-07-2021 Shift 2) The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents. Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Calculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)= (546.41 + j*156.47) AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)= (-146.41 - j*56.47) AIao = 1/3*(Ia + Ib + Ic)= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)= (100 + j*50) A
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 35 kA 25 kA 2500 A 350 kA 35 kA 25 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) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (ii) and (iii) only (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a synchronous motor with field under excited, the power factor will be zero leading lagging unity zero leading lagging unity ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For synchronous motor,Under excitation - Lagging power factorOver excitation - Leading power factorFor synchronous generatorUnder excitation - Leading power factorOver excitation - lagging power factor
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