MGVCL Exam Paper (30-07-2021 Shift 1)
Replace the underlined phrase in braket grammatically and conceptually with the help of the given options. If the given sentence is correct then select the option 'The given sentence is correct'.
The new client had to pay money to initiates a account and activate their service
Current flow through the circuit, I = V/Z = 400/10 = 41.5 A Power factor = 8/10 = 0.8 After connecting a capacitor bank, the reactive power component of current changes but the active power component of current is unchanged. I1*cosφ1 = I2*cosφ2 41.5*0.8 = I2*0.9 I2 = 36.88 A Q1 (before connecting capacitor) = √3*VL*IL*sinφ1 = √3*415*41.5*sin(36.86) = 26.847 kVAR Q2 (after connecting capacitor) = √3*VL*IL*sinφ2 = √3*415*36.88*sin(25.84) = 15.867 kVAR kVAR supplied by capacitor bank = Q1 - Q2 = 26.847 - 15.867 = 10.98 kVAR
For M no. of series configuration of solar cell, Open circuit voltage of PV module= M*Voc = 20*0.7 = 14 V Short circuit current of PV module = M*Isc = 20*4 = 80 A
Intern turn winding fault can be easily detect by differential protection relay. Also detect by buchholz relay (gas operated relay), cause buchholz relay is used to detect the internal fault of transformer.
Bus Type - Known Parameter - Unknown Parameter Load Bus -P, Q - V, phase angle Generator Bus - P, V (magnitude) - Q, Voltage phase angle Slack Bus Voltage - magnitude and phase angle - P, Q
EXAMIANS