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
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 t = 0 in RC circuit, After t = 0+ condition, capacitor is behave as short circuit. Voltage across the it 0. At final time, Capacitor is behave as open circuit, voltage across the capacitor at final time is equal to 15 V.
Conductor with 75 °C temperature rating must have ampacity from the 60°C column or the 75°C column. It is permissible to use the 75°C rating if the installed conductor is rated at least 75 °C. Because all of the connection points in this example have at least a 75°C rating, the conductor’s ampacity can be based on the 75°C column
The magnitude of earth fault current for a given fault position within a winding demands upon the winding connections and method of neutral grounding. Earth fault protection for an electric motor is provide by instantaneous relay having a setting of approximately 30% of motor rated current in the residual circuits of two CTs and with ground wire.
EXAMIANS