Three Phase AC Circuits MCQs With Explanatory Answers

Explanatory Answer:
Total Power in a Three Phase Circuit,
P = 3 x Power per Phase,
P = 3 x V_Ph I_Ph CosФ
P = 3 V_Ph I_Ph CosФ…………(1)

[For a Delta Connection]

[V_Ph = V_L and I_Ph = I_L/√3.]

then putting the values in eq …..(1)
P = 3 x V_L x ( I_L/√3) x CosФ
P = √3 x√3 x V_L x ( I_L/√3) x CosФ …{ 3 = √3x√3 }
P = √3 x V_Lx I_L x CosФ ….Ans.

Also
[For Star Connection]

[V_Ph = V_L/√3 and I_Ph = I_L]
Putting the values again in eq…….(1)
P = 3 x (V_L/√3 ) x IL x CosФ
P = √3 x√3 x (V_L/√3 ) x I_L x CosФ …{ 3 = √3x√3 }
P = √3 x V_L x I_L x CosФ ….Ans.

Explanatory Answer:

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.

Explanatory Answer:

Three phase voltages are generated by having an alternator with three armature windings such that each winding is displaced from the other by 120 degrees. When these windings are placed in a rotating magnetic field or rotated in a stationary magnetic field, electromotive force is generated in each coil, of same magnitude and direction. Consider the below diagram

Figure : 3 Phase AC Waveforms

As seen EMF generated in coil R-R1 is e_R, which is the reference in this case. EMF generated in coil Y-Y1 is e_Y which is 120° degrees ahead of e_R and EMF generated in coil B-B1 is e_B which is 240° degrees ahead of e_R.

Therefore the voltage equations are as given below;

Hence, sum of all three voltages is zero.

Explanatory Answer:

A star connected AC circuit is achieved by connecting each end of the winding to a common point known as neutral point and leaving the other end of each winding free. While voltage across each coil is the phase voltage, potential difference between each free end is the line voltage.

Consider the circuit below;

Now as said above, phase voltages are equal

Hence, V_NR = V_NY = V_NB = V_ph

Therefore, line voltage,

V_RY = √3 V_PH

Since the line conductor is in series with the phase winding, same current will flow through the line conductor as through the phase windings, hence phase current is equal to phase current.

Explanatory Answer:

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.

Explanatory Answer:

Given values are:

Line voltage, V_L = 230 V

Line frequency, f = 60 Hz

Power Factor, cosφ = 0.5

Power consumed = P = 1800 Watts = √3 V_{L x} I_{L x} cosφ

Hence, line current, I_L = 9 Amperes

Since it is a star connection, phase current = line current = 9 Amperes

Phase Voltage, V_ph = V_L/√3 = 132.8 Volts

Phase Impedance, Z_ph = V_ph/I_ph = 14.7 Ohms

Now, Power Factor = Resistance/Impedance

Hence, Resistance of Coil = Impedance X Power Factor = 7.4 Ohms

Substituting values, we get Reactance of coil = 12.7 Ohms

Thus, inductance of coil , L = 0.03H

Explanatory Answer:

Given values:

Star Connected phase voltage, V_PH = 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 = V_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 =  √3 I_PHLd = 15.41 Amperes

Power Factor, p_fs = R_phLd/ Z_PHLd = 0.44

Thus, the power supplied to load then, P_L =  V_L I_L p_fs = 4.7 KW

Explanatory Answer:

Power is measured using a Wattmeter which consists of two coils – Current coil, connected in series with the load, carrying the load current and Voltage coil, connected in parallel to the load.

Explanatory Answer:

The number of Wattmeter required to measure power in a polyphase system is determined using Blondell’s theorem. According to this, the number of Wattmeter required is equal to one less than the number of wires in the circuit. For example, in a three phase, four wire system (Star network), the number of Wattmeter required is three.

Explanatory Answer:

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

Putting the values in the below equation, Reactance, X =  Therefore, Inductance, L = 0.02H

Explanatory Answer:

Reading of Wattmeter 1, W₁ = 4000 Watts

Reading of Wattmeter 2, W₂ = 2000 Watts

Phase angle;

Power Factor,  = 0.866

Explanatory Answer:

Let reading of one Wattmeter = W₁

Reading of second Wattmeter = W₂

Input Power, P = W₁+W₂ = V_L I_L cosφ = 10 kW … (1)

Power Factor, cos φ = 0.9

Phase angle, φ = 25.8 degrees … (i.e. Cos ^-1 = 09 = 25.8°)

Therefore,

W₁ = V_L I_L cos (30 – φ) = 0.99V_L I_L = 6350W

W₂ = V_L I_L cos (30 + φ) = 0.56 V_L I_L = 3650W

Answer:      ( 1 )

Explanatory Answer:

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.

Answer:        ( 2 )

Explanatory Answer:

Three phase voltages are generated by having an alternator with three armature windings such that each winding is displaced from the other by 120 degrees. When these windings are placed in a rotating magnetic field or rotated in a stationary magnetic field, electromotive force is generated in each coil, of same magnitude and direction. Consider the below diagram.

Figure 1: 3 Phase AC Waveforms

As seen EMF generated in coil R-R1 is e_R, which is the reference in this case. EMF generated in coil Y-Y1 is e_Y which is 120 degrees ahead of e_R and EMF generated in coil B-B1 is e_B which is 240 degrees ahead of e_R.

Therefore the voltage equations are as given below

Adding all three equations, we get

Hence, sum of all three voltages is zero.

Answer:      ( 2 )

Explanatory Answer:

A star connected AC circuit is achieved by connecting each end of the winding to a common point known as neutral point and leaving the other end of each winding free. While voltage across each coil is the phase voltage, potential difference between each free endis the line voltage.

Consider the below circuit

Now as said above, phase voltages are equal

Hence, V_NR = V_NY = V_NB = V_ph

Now,

Therefore, line voltage, V_{Ry =} V_ph √3

Since the line conductor is in series with the phase winding, same current will flow through the line conductor as through the phase windings, hence phase current is equal to phase current.

Answer:      ( 2 )

Explanatory Answer:

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.

Answer:      ( 3 )

Explanatory Answer:

Given values are:

Line voltage, V_L = 230 V

Line frequency, f = 60 Hz

Power Factor, cosφ = 0.5

Power consumed = P = 1800 Watts = √3V_LI_Lcosφ

Hence, line current, I_L = 9 Amperes

Since it is a star connection, phase current = line current = 9 Amperes

Phase Voltage, V_ph = V_L/3^1/2 = 132.8 Volts

Phase Impedance, Z_ph = V_ph/I_ph = 14.7 Ohms

Now, Power Factor = Resistance/Impedance

Hence, Resistance of Coil = Impedance X Power Factor = 7.4 Ohms

Reactance of Coil, 

Substituting values, we get Reactance

e of coil = 12.7 Ohms

Thus, inductance of coil , L = 0.03H

Answer:     ( 2 )

Explanatory Answer:

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

Answer:       ( 1 )

Explanatory Answer:

Power is measured using a Wattmeter which consists of two coils – Current coil, connected in series with the load, carrying the load current and Voltage coil, connected in parallel to the load.

Explanatory Answer:

The number of Wattmeter required to measure power in a polyphase system is determined using Blondell’s theorem. According to this, the number of Wattmeter required is equal to one less than the number of wires in the circuit. For example, in a three phase, four wire system (Star network), the number of Wattmeter required is three.

Explanatory Answer:

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

Reactance, 

Therefore, Inductance, L = 0.02H

Answer:       ( 3 )

Explanatory Answer:

Reading of Wattmeter 1, W1 = 4000 Watts

Reading of Wattmeter 2, W2 = 2000 Watts

Phase angle,  = 30 degrees

Power Factor,  = 0.866

Answer:       ( 2 )

Explanatory Answer:

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

Answer:      ( 1 )

Explanatory Answer:

                 Figure 1: 3 Phase Induction Motor

A three-phase induction motor consists of two parts – Stator (the stationary part) and Rotor (the rotatory part) with the latter being separated from the former by a small air gap. Three phase voltage supplied to the stator windings produces a rotating magnetic field. As the magnetic flux cuts the rotor windings through the air gaps, EMF is induced in the winding which in turn induces current. As the induced current interacts with the stator field, forces are produced which causes the rotor windings to rotate.

Answer:      ( 3 )

Explanatory Answer:

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.

Answer:      ( 1 )

Explanatory Answer:

As the rotation of the rotor conductors are initiated normally due to the force produced because of interaction between induced current and rotor magnetic field, no external supply is required and the Induction motor is self-starting.

Answer:      ( 3 )

Explanatory Answer:

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%.

Answer:     ( 2 )

Explanatory Answer:

An induction motor is likea transformer in the fact that both comprise of transfer of energy from a primary winding to secondary winding. However though in a transformer, the frequency of primary voltage is equal to frequency of secondary voltage, in an induction motor, the primary frequency, i.e. the frequency of stator currents is different from that of the rotor current.

Frequency of stator current with N_s synchronous speed and P number of poles is –

 f= PNs/120 … (1)

As the rotor starts revolving, frequency is variable and depends upon the slip or difference between the synchronous speed and rotor speed N_r.

Thus, rotor frequency,  … (2)

Now, slip,  … (3)

Thus, combining equations 1, 2 and 3, we get rotor frequency, f_r = Sf.

Answer:     ( 2 )

Explanatory Answer:

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.

Answer:      ( 1 )

Explanatory Answer:

A slip ring rotor consists of star connected rotor windings wherein the open ends of the star circuit are connected to three slip rings, which are mounted on the shaft with brushed resting on them. These slip rings and the brushes provide a means for connecting external resistors. These resistors serve the purpose of increasing the starting torque, thereby decreasing the starting current and controlling the speed of the motor.

Answer:      ( 2 )

Explanatory Answer:

The flux cut per second by the stator or rotor conductors when the rotor is at stand-still is given as, PφN/60

Now, RMS value of induced EMF per conductor, 

Where, f is the stator frequency

Hence, stator induced EMF for 2T₁ number of conductors is (considering pitch factor and distribution factor):

E₁ = 4.44K_c K_D fφT₁

Thus, rotor induced EMF at standstill for 2T₂ number of conductors is –

E₂₀ = 4.44K_c K_D fφT₂

As rotor starts rotating, the induced EMF is

E₂ = 4.44K_c K_D SfφT₂

Thus, the induced EMF is maximum at the start and varies due to the variation of slip. As the value of slip under normal conditions is about 5%, the value of induced EMF is 20% of its maximum value.

Answer:      ( 3 )

Explanatory Answer:

Under running conditions, the motor torque is given as:

Where, R₂ is rotor resistance and X₂ is rotor reactance and S is the slip

Below the curve between the torque and the slip

Figure 2: Torque-Slip Characteristic Curve

At S = 0, T = 0. Hence the curve starts at the origin. At normal speed, S is small, hence SX₂ is negligible small. Thus Torque, T ∝ S  and Torque-Slip characteristics is a straight line from zero slip to slip at full load.

As the slip increases further beyond full load, the torque increases and becomes maximum at S = R₂/X₂. As the slip increases beyond the value of slip at maximum torque, the value SX₂ is much larger compared to R₂ and thus . T ∝ 1/S

Thus, for smaller values of slip, torque is directly proportional and for larger values of slip, it is inversely proportional.

Explanatory Answer:

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.