 # LE260 : Electrical Machines 1

## 1. FUNDAMENTAL ELECTRICITY AND MAGNETISM

### Exercise 1

• In phase means that one phasor is the part of another phasor
• out of phase means two phasors go in the different direction.
• Lead phase can be checked by looking at the magnitude of the phasor.
• Lag phase can be checked by rotate clockwise and in phase with another phasor.
Which one is correct?

### Exercise 2

• Conventional current flow is the direction of current flow in electric power technology.
• You can tell that it is the source by seeing the current flow out of the positive terminal.
• The difference in potential caused by the electron at the negative terminal compare to the positive terminal.
• Electron and electricity flow in the same direction in the electrical circuit.
Which one is wrong?

### Exercise 3

• increase magnetic flux density
• reverse current
• increase resistance
• increase magnetic field intensity in the opposite direction
• reduce the flux density
• increase induced voltage
State 2 ways to eliminate the residual flux

### Exercise 4

#### Hint: there are 2 ways to reduce the eddy current in the iron core.

• Make the conductor become laminates.
• increase the size of the iron core
• mix silicon and steel to the iron core
• reduce the air gap between the core
• increase the cross-sectional area of the core
• increase the volume of the core
State how to reduce eddy current ( There will be more than 1 way to reduce eddy current).

### Exercise 5

• The sum of current in the close loop equal to zero.
• current that pass every resistance in the circuit will be equal.
• Voltage around the close loop equal to zero.
• current in the parallel circuit will equal to the current in series circuit.
What is Kirchhoff's current law ?

### Exercise 6

#### Hint: use the formula of voltage induced in a conductor.

• 10V
• 20V
• 30V
• 40V

A conductor 2m long moves at a speed of 60km/h through a magnetic field having a flux density of 0.6 T. Calculate the induced voltage.

### Exercise 7

• 1.8V
• 1.25V
• 1.35V
• 1.67V

A coil having 200 turns links a flux of 3mWb, produced by a permanent magnet. The magnet is moved,and the flux linking the coil falls to 1.2m Wb in 0.2s. Calculate the average voltage induced.

### Exercise 8

• 600N
• 200N
• 100N
• 300N

A conductor 3 m long carrying a current of 200A is placed in a magnetic field whose density is 0.5T. Calculate the force acting on the conductor if it is perpendicular to the lines of force.

### Exercise 9

• a. 6A b. 200 V c. 1660W d. 3124 W
• a. 12A b. 169.7 V c. 1440 W d. 2880 W
• a. 8A b. 170.65 V c. 2567 W d. 1367 W
• a. 5A b. 143.29V c. 1329 W d. 1236 W

A sinusoidal voltage of 120 V is applied to a resistor 10 ohms. Calculate

• a. the effective current in the resistor
• b. the peak voltage across the resistor
• c. the power dissipated by the resistor
• d. the peak power dissipated by the resistor

### Exercise 10

• I1=3A I2=2A I3=5A
• I1=1.9A I2=-2.7A I3=4.87A
• I1=1.43A I2=4A I3=-5.43A
• I1=-1.25A I2=2.32A I3=-7.87A Calculate the currents  in this circuit.

## 2. MAGNETIC CIRCUIT

### Exercise 1

• conductance
• conductivity
• resistance
• flux density
Permeability in a magnetic circuit corresponds to ........ in an electric circuit

### Exercise 2

• increase the resistivity
• decrease the resistivity
• use a laminated core
• increase the energy density in the core
How to reduce the eddy current loss

### Exercise 3

• magnetising components in phase with flux and symmetrical with respect to e
• energy components in phase with voltage e accounting for the core loss
• electric components in phase with flux and symmetrical with respect to e
• power components in phase with voltage e accounting for the core loss
The exciting current can be split into two components are

### Exercise 4

• reluctivity
• magnetomotive force
• permeance
• reluctance
The property of a material which oppose the creation of magnetic flux in it is know as

### Exercise 5

• increase their cost of manufacture
• lower their power efficiency
• leads to their increased weight
• produces fringing
Point out the wrong statement. Magnetic leakage is undesirable in electric machines because is

### Exercise 6

• 10 V.
• 20 V.
• 30 V.
• 40 v.
A conductor 2 m long moves at a speed of 60 km/h through a magnetic  field having a flux density of 0.6 T. Calculate the induced voltage.

### Exercise 7

• 3840 Ampere turns
• 3920 Ampere turns
• 4100 Ampere turns
• 5350 Ampere turns
We want to produce a flux density of 0.6T in an air gap having a length of 8 mm. Calculate the mmf required.

### Exercise 8

• 65.73 A
• 69.71 A
• 70.71 A
• 71.71 A
A sinusoidal current has an effective value of 50  A. Calculate the peak value of current.

### Exercise 9

• 2400 W, -800 W
• 2600 W, -800 W
• 2700 W, -1000 W
• 2800, -1000 W
The voltage applied to an ac magnet is given by the expression E = 160 sin Φ and the current is I = 20 sin(Φ-60°), all angles begin express in degree. Calculate the peak positive power and the peak negative power in the circuit.

### Exercise 10

• 1.42 ms
• 1.55 ms
• 1.61 ms
• 1.67 ms
The current in a 60 Hz single-phase motor lags 36 degrees behind the voltage. Calculate the time interval between the positive peaks of the voltage and current.

# Y-Y

## 3.6.2 THREE-PHASE TRANSFORMER ON A COMMON MAGNETIC CORE (THREE-PHASE UNIT TRANSFORMER)

A three-phase transformer can be constructed by having three primary and three secondary windings on a common magnetic core. If balanced three-phase sinusoidal volt- ages are applied to the windings, the fluxes Φₐ , Φb,Φc and will also be sinusoidal and balanced.
If the three legs carrying these fluxes are merged, the net flux in the merged leg is zero. Both primary and secondary windings of a phase are placed on the same leg.
a three-phase transformer of this type. Such a transformer weighs less, costs less, and requires less space than a three-phase transformer bank of the same rating. The disadvantage is that if one phase breaks down, the whole transformer must be removed for repair.

## 3.8.1 TRANSFORMER EQUIVALENT CIRCUIT IN PER-UNIT FORM ## 3.8.3 FULL-LOAD COPPER LOSS ### Exercise 1

• maximum resistance on primary side and least resistance on secondary side
• minimum resistance on primary side and maximum resistance on secondary side
• equal resistance on primary and secondary side
• no ohmic resistance on either side
For an ideal transformer the winding should have

### Exercise 2

• primary = 22.7 A, secondary = 602.4 A
• primary = 602 A, secondary = 22.7 A
• primary = 301 A, secondary = 22.7 A
• primary = 11.4 A, secondary = 301 A
A single phase transformer has specifications as 250 KVA, 11000 V/415 V, 50 Hz. What are the values of primary and secondary currents?

### Exercise 3

• maximum flux density
• supply frequency
• both 2 and 3
In a transformer, hysteresis and eddy current losses depend upon

### Exercise 4

• high frequency supply
• direct current
• rated transformer voltage
• rated transformer current
Open circuit in a transformer is prefered with

### Exercise 5

• 3200 W and 2500 W
• 3200 W and 5200 W
• 1600 W and 1250 W
• 1600 W and 5000 W

The full load copper loss and iron loss of transformer are 6400W and 5000W respectively. The copper loss and iron loss at half load will be respectively?

### Exercise 6

• Pi/Pc
• (Pi/Pc)²
• √(Pi/Pc)
• Pc/Pi
If Pi and Pc represent core and full load copper loss respectively, the maximum KVA delivered to load corresponding to maximum efficiency is equal to rated KVA is

### Exercise 7

• ratings
• per unit impedance
• leakage reactance
• efficiency
Two transformers operating in parallel will share the load depending on their

### Exercise 8

• Conduction process only
• Induction process only
• Both 1 and 2
• None of the above
n an auto-transformer, power is transferred through

### Exercise 9

• iron loss
• copper loss
• rotational loss
• all of the above
In a transformer which of the following losses are zero ?

### Exercise 10

• Coulombs law.
• Amperes law.
• Faradays law of electromagnetic induction.
• Newtons law.
The working principle of transformer depends on

## LINEAR SYSTEM

If the reluctance of the magnetic core path is negligible compared to that of the air gap path, the A-i relation becomes linear. For this idealized system

λ =L(x)i

where L(x) is the inductance of the coil, whose value depends on the air gap length. The field energy is

Wf  =  ∫i dλ  = 1/2 L(x)i²

fₘm  = 1/2(i²) [dL(x)/dx]

For a linear system ,                                               Wf  =  W'f  =  1/2 L(x)i²

if the reluctance of the magnetic core path is neglected,

Ni   =  Hg2g   =  Bg/μ₀ (2g )

Wf  =  (Bg²/2μ₀) x volume of air gab    =  (Bg²/2μ₀) x (Ag2g)

fm  =  (Bg²/2μ₀) x 2Ag

The total cross-sectional area of the air gap is 2Ag . Hence, the force per unit area of air gap, called magnetic pressure Fm , is

Fm  =  Bg²/2μ₀ (2Ag) N/m²

## Consider the following cases.

1. ωr =0, α=0 ,ωm = ωs. The rotor current is a direct current and the machine rotates at the synchronous speed. If the machine is brought up to its synchronous speed (ωm = ωs), it will develop an average unidirectional torque and continuous energy conversion can take place at the synchronous speed. This is the basic principle of operation of a synchronous machine, which normally has dc excitation in the rotor and ac excitation in the stator.

at ω = 0, the machine does not develop an average torque and therefore the machine is not self-starting.    With one winding on the stator, the machine is called a single-phase synchronous machine. Although it develops an average torque, the instantaneous torque is pulsating.

The pulsating torque may result in vibration, speed fluctua- tion, noise, and waste of energy. This may be acceptable in smaller machines but not in larger ones. The pulsating torque can be avoided in a polyphase machine, and all large machines are polyphase machines.

2.  ωm  =  ωs  -  ωr, Both stator and rotor windings carry ac currents at different frequencies and the motor runs at an asynchronous speed This is the basic principle of operation of an induction machine, in which the stator winding is excited by an ac current and ac current is induced in the rotor winding.

Note that the single-phase induction machine is also not self-starting, because at ωm = 0 no average torque is developed.The machine is brought up to the speed  ωm=ωs -ωr so that it can produce an average torque. To eliminate pulsating torque, polyphase induction machines are used for high-power applications.

### Exercise 1

• increase the co-energy at constant flux
• increase the co-energy at constant mmf
• decrease the stored energy at constant mmf
• decrease the stored energy at constant flux

The developed electromagnetic force and/or torque in electromechanical energy conversion systems, acts in a direction that tends to.....

Which of the above statements are correct?

### Exercise 2

• electrical system in order to extract energy from mechanical system
• mechanical system in order to extract energy from mechanical system
• electrical system in order to extract energy from electrical system
• electrical or mechanical system for electro-mechanical energy conversion
A coupling magnetic field must react with

### Exercise 3

• electrical side is associated with emf and current
• electrical side is associated with torque and speed
• mechanical side is associated with emf and current
• mechanical side is associated with torque and speed
In an electro-mechanical energy conversion device, the coupling field on the

### Exercise 4

• Electrical energy to mechanical energy only
• Mechanical energy to electrical energy only
• All of the mentioned
• None of the mentioned
An electro-mechanical energy conversion device is one which converts.......

### Exercise 5

• Magnetic field
• Electric field
• Magnetic field or Electric field
• None of the mentioned
What is the coupling field used between the electrical and mechanical systems in an energy conversion devices?

### Exercise 6

• stored in the magnetic field
• stored in the electric field
• divided equally between the magnetic and electric fields
• zero
A physical system of electromechanical energy conversion, consists of a stationary part creating a magnetic field with electric energy input, and a moving part giving mechanical energy output. If the movable part is kept fixed, the entire electrical energy input will be.....

### Exercise 7

• increase the stored energy at constant mmf
• decrease the stored energy at constant mmf
• decrease the co-energy at constant mmf
• increase the stored energy at constant flux
The developed electromagnetic force and/or torque in electromechanical energy conversion system, acts in such a direction that tends to.....

### Exercise 8

• 9.75 V
• 10.46 V
• 11.73 V
• 14.50 V The coil has 400 turns and a resistance of 5 ohms. The reluctance of the magnetic material is negligible. The magnetic core has a square cross section of 5 cm by 5 cm. When the sheet of steel is fitted to the electromagnet ,air gap, each of length = 1 mm, separate them. An average force of 550 newtons is required to lift the sheet of steel.

Determine the dc source voltage.

### Exercise 9

• 78.33°
• 79.33°
• 80.33°
• 81.33° When current flows through the curved solenoid coil, a curved ferromagnetic rod is pulled into the solenoid against the torque of a restraining spring. The inductance of the coil is L = 4.5 + 180 µH, where θ is angle of deflection in radius. The spring constant is 0.65 mN•m/rad.

Determine the deflection in degree for a current of 10 amperes (rms).

### Exercise 10

• 2.18 mm
• 2.63 mm
• 3.05 mm
• 3.14 mm The plunger has a cross-sectional area 0.0016 m². The coil has 2500 turns and a resistance of 10 ohm. A voltage of 15 V (dc) is applied to the coil terminals. Assume that the magnetic material is ideal.

Determine the air gap g in mm for which the flux density in the air gap is 1.5 T. Determine the stored energy for this condition.

## 5. LINEAR MACHINE

### Exercise 1

• Newton's law F=ma
• Law of inertia ∑F = 0
• Faraday's law E = N(∆Φ/∆t)
• F = i(l x B)
• E = l( v x B )
• E = L( ∆i / ∆t )
• Kirchhoff's voltage law ∑v = 0
What are the 4 basic equations for the application of dc machine?

### Exercise 2

• a c d e b
• d e a b c
• c d a e b
• c d e b a

Arrange the steps of the linear dc machine behavior in order

• a. The bar accelerates to the right,producing an induced voltage as it speeds up.
• b. The induced force is thus decreased until eventually F=0. At that point induced voltage = source voltage and current flow = 0, and the bar moves at a constant no-load speed v = source voltage/Bl.
• c. Closing the switch produces a current flow.
• d. The current flow produces a force on the bar.
• e. The induced voltage reduces the current flow.

### Exercise 3

• In a real dc motor, the increased current flow increases its induced torque.
• The induced torque will equal to the load torque of the motor at a new slower speed.
• For the generator, when the induced force equal and opposite to the applied force, the bar will slow down and charging the battery.
• The power convert from electrical form to mechanical form is given by the equation P = Fv. On the other hands, the power converted from electrical to mechanical is given by the equation P=ωτ
Which one is Wrong?

### Exercise 4

• Linear dc machine use 4 basic equation for the application
• We can use Kirchhoff's voltage law for the machine to determine the induced voltage on the bar
• Current along the bar is in proportional to the induced voltage.
• When induced force equal to the load force, the bar will increase its moving speed
Which one is wrong?

### Exercise 5

• Motor changes from mechanical energy to electrical energy and generator changes from electrical energy to mechanical energy.
• For motor, load force applied to the bar in the opposite direction to the motion. For generator, load force applied to the bar in the same direction to the motion.
• Motor has higher F net than the generator
• Motor use higher current acting to the bar than the generator
What is the difference between linear dc machine as motor and linear dc machine as generator?

### Exercise 6

• a. 1100A b.660N c.366.67m/s
• a. 1000A b.650N c.365.50m/s
• a. 1200A b.670N c.370.75m/s
• a. 1500A b.700N c.376.37m/s

A linear machine has a magnetic flux density of 0.6 Wb/m^2 directed into the page, a resistance of 0.2 Ohms, a bar length = 1m, and a battery voltage of 220V, calculate

• a. What is the starting current flow?
• b. What is the initial force on the bar at starting
• c. What is the no-load steady-state speed

### Exercise 7

• a. 337.87m/s b.7555W c. 7333.3 W d. motor
• a. 422.22m/s b.7444W c. 7222.2 W d. motor
• a. 377.78m/s b.7555W c. 7333.3 W d. generator
• a. 222.22m/s b.7444W c. 7222.2 W d. generator

A linear machine has a magnetic flux density of 0.6 Wb/m^2 directed into the page, a resistance of 0.2 Ohms, a bar length = 1m, and a battery voltage of 220V. If a 20 N force pointing to the same direction as no-load steady state is applied to the bar, calculate

• a. What is the steady state speed?
• b. How much power would the bar be producing or consuming ?
• c. How much power would the battery be producing or consuming ?
• d. Is the machine acting as a motor or a generator?

### Exercise 8

• a. 345.88 m/s b. 69.97% c. generator
• a. 315.56 m/s b. 75.77% c. generator
• a. 345.56 m/s b. 29.85% c. motor
• a. 355.56 m/s b. 96.97% c. motor

From the last question, if the bar is loaded down with a force of 20N opposite the direction of motion.

• a. What is the new steady-state speed?
• b. What is the efficiency of the machine under this circumstance
• c. Is this machine acting as a motor or generator?

### Exercise 9

• a. 120 m/s b. 111m/s
• a. 100 m/s b. 109m/s
• a. 115m/s b. 171m/s
• a. 140 m/s b. 123m/s

A linear machine has a magnetic flux density of 0.1T directed into the page, a resistance of 0.3 Ohms, a bar length equal to 10 m,and a battery voltage of 120 V.

• a. What is the no-load steady state speed of the bar?
• b. If 30 N force pointing to the left were applied to the bar. What would be the new steady state speed be? Is this machine a motor or a generator?

### Exercise 10

• 160 m/s
• 170 m/s
• 140 m/s
• 150 m/s

A linear machine has a magnetic flux density of 0.1T directed into the page, a resistance of 0.3 Ohms, a bar length equal to 10 m,and a battery voltage of 120 V. Assume that the bar is unloaded and that is suddenly runs into a region where the magnetic field is weakened to 0.08 T. How fast will the bar go now?

## 6. DC GENERATOR

### Exercise 1

• Abrasion from dust
• Excessive spring pressure
• Rough commutator bars
• Double the number of poles
The rapid wear of brushes takes place due to

### Exercise 2

• Brushes should be of proper grade and size
• Making the magnetic losses of forces parallel
• Brushes should smoothly run in the holders
• Smooth, concentric commutator properly undercut
Satisfactory commutation of D.C. machines requires

### Exercise 3

• Separately excited generator
• Shunt generator
• Series generator
• Compound generator
An exciter for a turbo generator is a....

### Exercise 4

• 100% regulation
• Infinite regulation
• 50% regulation
• 1% regulation
With a D.C. generator which of the following regulation is preferred?

### Exercise 5

• Copper lugs
• Resistance wires
• Brazing
The commutator segments are connected to the armature conductors by means of

### Exercise 6

• Imperfect brush contact
• Field resistance less than the critical resistance
• No residual magnetism in the generator
• Faulty shunt connections tending to reduce the residual magnetism
If a D.C. generator fails to build up the probable cause could not be

### Exercise 7

• 103 V / 288 W / 10.39 N·m
• 103 V / 278 W / 10.12 N·m
• 106 V / 288 W / 10.12 N·m
• 106 V / 278 W / 10.43 N·m

A separate excited dc generator turning at 1400 r/min produces an induced voltage of 127 V. The armature resistance is 2 ohm and the machine delivers a current of 12 A.

Calculate

a. The terminal voltage

b. The heat dissipated in the armature

c.The breaking torque exert by the armature

### Exercise 8

• 2600 A / 5000 A
• 4500 A / 6400 A
• 5000 A / 7800 A
• 6000 A / 8200 A

Each pole of a 100 kW, 250 V flat-compound generator has a shunt field of 7 turns. If the total shunt-field resistance is 100 ohm. Calculate the mmf when the machine operates rated voltage

### Exercise 9

• 275.78
• 276.78
• 277.78
• 278.78

The generator of Fig. 4.38 revolves at 960 r/min and the flux per pole is 20 mWb. Calculate the no-load armature voltage if each armature voltage if each armature coil has 6 turns ### Exercise 10

• To supply traction load
• To supply industrial load at constant voltage
• Voltage at the toad end of the feeder
• For none of the above purpose
D.C. series generator is used

## 7. DC MOTOR

### Exercise 1

• Burning out the armature
• Damaging the driven equipment
• cause the air gap in the core
Which one is not the risk of the high starting current in the armature ?

### Exercise 2

• Dynamic breaking stops the motor slower than Plugging
• When a dc motor is stopped by dynamic braking, the speed decreases exponentially with time.
• Plugging is when the armature induce voltage it suddenly open the switch and connect to the external resistor for breaking.
• Dynamic breaking is more popular than plugging because it can break smoother than Plugging.
Which one is wrong?

### Exercise 3

• Series speed control
• Armature speed control
• Shunt speed control
• Field speed control
• Compound speed control
• Dynamic speed control
What are two types of speed control for dc motor?

### Exercise 4

• Reverse resistance pole
• Add more resistance
• Reverse armature connection
• increase the number of coil
• Reverse the rheostat connection
• Reverse the shunt and series field connection
State 2 ways to reverse the direction of rotation of compound motor.

### Exercise 5

• Adding series motor to the system
• Placing the set of commutating poles
• Creating new magnetic flux by adding permanent magnet
What is the way to reduce armature reaction effect?

### Exercise 6

• a. 230A b.250V,375V c. 2A, 3A
• a. 175A b.150V,175V c. 1.85A, 4.53A
• a. 150A b.100V,146V c. 50A, 4A
• a. 160A b.50V,73V c. 25A, 2A

The armature of a permanent-magnet dc generator has a resistance of 1 Ohms and generates a voltage of 50V when the speed is 500 r/min. If the armature is connected to a source of 150V, calculate the following:

• a. The starting current
• b. The counter-emf when the motor run at 1000 r/min.At 1460 r/min.
• c. The armature current at 1000 r/min. At 1460 r/min.

### Exercise 7

• a. 225r/min, 24.7kN.m b. 225 r/min, 35.84kN.m
• a. 228r/min, 31.8kN.m b. 228 r/min, 47.8kN.m
• a. 220r/min, 26.8kN.m b. 220 r/min, 54.3kN.m
• a. 178r/min, 23.65kN.m b. 178 r/min, 27.98kN.m

A 2000 kW, 500 V, variable-speed motor is driven by a 2500 kW generator,using a Ward-Leonard control system , the total resistance of the motor and generator armature circuit is 10 mOhms. The motor turns at a nominal speed of 300 r/min, when Eo is 500 V, calculate

• a. The motor torque and speed at Es=400V and Eo=380V.
• b. The motor torque and speed at Es=350V and Eo=380V.

### Exercise 8

• a. 50A b.120V c.5750W
• a. 40A b.76V c.4578W
• a. 60A b.135V c.6750W
• a. 30A b.140V c.3250W

A shunt motor rotating at 1500 r/min is fed by a 120V source. The line current is 51A and the shunt-field resistance is 120 Ohms. If the armature resistance is 0.1 Ohms, calculate the following:

• a. The current in the armature
• b. The counter-emf
• c. The mechanical power developed by the motor

### Exercise 9

• a. 5s b. 50s c. 20 min
• a. 10s b. 55s c. 27 min
• a. 5s b. 45s c. 26 min
• a. 10s b. 60s c. 28 min

A 225 kW, 250V, 1280 r/min dc motor has windage, friction,and iron losses of 8kW.It drives a large flywheel and the total moment of inertia of the flywheel and armature is 177 kg.m^2. The motor is connected to a 210 V dc source and its speed is 1280 r/min just before the armature is switched across a braking resistor of 0.2 Ohms, calculate

• a. The mechanical time constant of the braking system
• b. The time for the motor speed to drop to 20r/min
• c. The time for the speed to drop to 20r/min if the only braking force is that due to the windage friction ,and iron loss

### Exercise 10

• a. 1250 A, 150.5 kW b.50s
• a. 1150 A, 225.5 kW b.10s
• a. 1050 A, 220.5 kW b.20s
• a. 1350 A, 530.5 kW b.30s

A 225 kW, 250V, 1280 r/min dc motor has windage, friction,and iron losses of 8kW.It drives a large flywheel and the total moment of inertia of the flywheel and armature is 177 kg.m^2. The motor is connected to a 210 V dc source which it has the net voltage acting across the resistor 420V its speed is 1280 r/min just before the armature is switched across a braking resistor of 0.2 Ohms is plugged,and the braking resistor is increased to 0.4 Ohms, so that the initial braking current is the same as before.

• a. The initial braking current and braking power
• b. The stopping time