MGVCL Exam Paper (30-07-2021 Shift 3) Job, in operating system refers to Process Program System software Application Software Process Program System software Application Software ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In a series RLC circuit, the supply voltage is 230 V at 50 Hz. The resonant current is 2 A at the resonant frequency of 50 Hz. Under the resonant condition, the voltage across the capacitor is measured to be equal to 460 V. What are the values of L and C? 0.73 mH and 13.85 pF 0.73 H and 13.85 pF 0.53 mH and 15.83 pF 0.53 H and 15.83 pF 0.73 mH and 13.85 pF 0.73 H and 13.85 pF 0.53 mH and 15.83 pF 0.53 H and 15.83 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Vc = I*Xc460 = 2*(1/(2*ᴨ*50*C))C = 2/(460*2*50*ᴨ)= 1.385*10⁻⁵ F= 13.85 FV_L = I*X_LL = V/(I*2*ᴨ*f)= 460/(2*ᴨ*50*2)= 0.7345 H
MGVCL Exam Paper (30-07-2021 Shift 3) સાબરમતી: અમદાવાદ:: મૂસી: ગોવા લંડન હૈદરાબાદ વેનિસ ગોવા લંડન હૈદરાબાદ વેનિસ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table A = (ii), B = (iii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) A = (ii), B = (iii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Bus Type - Known Parameter - Unknown ParameterLoad Bus -P, Q - V, phase angleGenerator Bus - P, V (magnitude) - Q, Voltage phase angleSlack Bus Voltage - magnitude and phase angle - P, Q
MGVCL Exam Paper (30-07-2021 Shift 3) હું અહી આવી શકું' વાક્યને સ્થળવાચક ક્રીયાવીશેષણમાં ફેરવો. હું ચેન્નાઈ નહિ આવી શકું. હું કદી પણ નહિ આવી શકું. હું આજે નહિ આવી શકું. હું નિયમિત રીતે અહી આવી શકું. હું ચેન્નાઈ નહિ આવી શકું. હું કદી પણ નહિ આવી શકું. હું આજે નહિ આવી શકું. હું નિયમિત રીતે અહી આવી શકું. ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0