Test: Three-Phase AC Circuits - 2 - Electrical Engineering (EE) MCQ

Given

Speed of Alternator, N = 1200 RPM

Number of Alternator poles, P = 6

Supply frequency, f = PN/120 = 60 Hz

Hence, for number of 8 poles in Induction motor, the synchronous speed or speed of rotating field is: N_s = 120f/P = 900 RPM

Let N_r is actual speed of motor.

Thus, Percentage Slip, S % = ((N_s – N_r)/N_s) x 100

Substituting the values, we get: N_r = N_s (1-S/100) = 873 RPM.

A squirrel cage rotor consists of a laminated steel cylinder with slots in which either Aluminium conductors are die-cast of copper bars are used. These bars are shorted at both ends by heavy end rings of same materials. To reduce the magnetic noise and hence produce a more uniform torque and to prevent possible magnetic locking of the rotor with the stator, the rotor slots are skewed at a certain angle to the rotor shaft.

Slip is the difference between the speed of rotating field and actual speed of rotor. The speed of the rotor must be less than the speed of the rotating field, else there would be no relative motion and the rotary motion would cease to exist. The speed of the rotor should be such that the magnitude of the rotor current is enough to exert the necessary torque. The slip speed denotes the speed of rotor relative to that of the field. Though it is measured in revolutions per minute, more commonly it is denoted in percentage and ranges from 0 to 5%.

The rotating magnetic field produced by the stator winding on being subjected to a three-phase supply, causes an induction of voltage in the stationary rotor windings. As the rotor circuit is complete, current starts flowing due to the induced voltage. This induced current in turn produces its own magnetic field. As the current carrying conductors are placed in the magnetic field, a force is produced, which acts tangentially and develops a torque which causes the rotor conductors to rotate. Thus, operation of the motor depends upon the induced voltage in the rotor conductors and the motor is called Induction Motor.

Let reading of one Wattmeter = W1
Reading of second Wattmeter = W2
Input Power, P = W1+W2 = √3V_LI_Lcosφ = 10000 ________1
Power Factor, cosφ = 0.9
Phase angle, φ = 25.8 degrees
Therefore, W1 = V_LI_Lcos(30-φ) = 0.99V_LI_L = 6350W
W2 = V_LI_Lcos(30+φ) = 0.56V_LI_L = 3650W

Given
Line Voltage, V_L = 400 Volts
Frequency, f = 60 Hz
Line Current, I_L = 25 Amperes
Power per phase, P_ph = 5kW
Phase Voltage, V_ph = V_L/3^1/2 = 230.9 Volts
Phase current, I_ph = 25 Amperes
Hence, power factor, cosφ = P_ph/V_phI_ph = 0.866
Impedance, Z_ph = V_ph/I_ph = 9.236 Ohms
Resistance, R = Z_phcosφ = 8 Ohms

Given values:
Star Connected phase voltage, V_phs = 230 Volts
Phase load resistance, R_phLd = 20 Ohms
Phase load reactance, X_phLd = 40 Ohms

Hence, phase load impedance, = 

Star connected line voltage, V_L =√3V_phs = 398.37 Volts
For the delta connected load, Phase Voltage, V_phLd = V_L = 398.37 Volts
Therefore, current through each phase of load, I_phLd = V_phLd/Z_phLd = 8.9 Amperes
Line current for delta connected load, I_L =  √3I_phLd = 15.41 Amperes
Power Factor, p_fs = R_phLd/Z_phLd = 0.44
Thus, power supplied to load, P_L =√3 V_LI_Lp_fs = 4.7 KW

A delta connected AC circuit is achieved by connecting the start end of a winding to the finish end of another winding such that all three windings form a mesh. Since each end of the windings forms the line connection, voltage across each winding is equal to the potential difference between the corresponding lines taken from that winding. Hence the phase voltage is equal to the line voltage.

A generator having two or more electrical windings which are separated by equal electrical angle generates a polyphase electrical system. The electrical angle or displacement depends upon the number of windings or phases. For example, in a three-phase electrical system, the generated voltages are separated from each other by 120 degrees.

Reading of Wattmeter 1, W₁ = 4000 Watts

Reading of Wattmeter 2, W₂ = 2000 Watts

Phase angle;

Power Factor,  = 0.866