national diploma in electrical engineering technology

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UNESCO-NIGERIA TECHNICAL & VOCATIONAL EDUCATION REVITALISATION PROJECT-

PHASE II

YEAR I- SEMESTER II

THEORY

Version 1: December 2008

NATIONAL DIPLOMA IN

ELECTRICAL ENGINEERING TECHNOLOGY

ELECTRICAL SCIENCE II

COURSE CODE: EEC125

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TABLE OF CONTENTS

Department Electrical Engineering Technology

Subject Electrical Engineering Science II

Year 1

Semester 2

Course Code EEC 125

Credit Hours 2

Theoretical 1

Practical 2

CHAPTER 1 :

• To Study the Characteristics of an Inductor Week 1

CHAPTER 2 :

• Determination of Series and parallel Resistors

connection Using Kirchoff’s Law

Weeks 2

CHAPTER 3 :

• Determination of Series and parallel Resistors

connection Using Kirchoff’s Law

Weeks 3

CHAPTER 4 :

To demonstrate Faraday’s and Lenz’s Laws

Weeks 4

CHAPTER 5 :

• Demonstration of Self and Mutual Inductance Weeks 5

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CHAPTER 6 :

• Determination of inductance of a coil by Voltage

and Current Measurements

Weeks 6

CHAPTER 7 :

• Magnetic Effect of Current Carrying Conductor Weeks 7

CHAPTER 8 :

• Determination of Mutual Inductance and Polarity of Magnetically coupled coils.

Weeks 8

CHAPTER 9 :

• Determination of Mutual Inductance and Polarity

of Magnetically coupled coils

Weeks 9

CHAPTER 10 :

• The Effects of Saturation in a Magnetic Circuit Weeks 10

CHAPTER 11 :

• Determination of Hysteresis loop of an iron

sample

Weeks 11

CHAPTER 12 :

• To Study the Effect of LR and Undriven LRC Circuits

Weeks 12

CHAPTER 13 :

• To Study Free Oscillations of the RLC Circuit Weeks 13

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CHAPTER 14 :

• To study the charging and discharging of a capacitor.

Weeks 14

CHAPTER 15 :

• Demonstrate the application of EM in Transformer Weeks 15

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Week 1

On completion of this Topic the student should be able to:

• Define magnetic flux, magnetic flux density, magnetomotive force, magnetic field strength, reluctance, permeability of free space (magnetic constants), relative permeability.

• State the symbol, units and relationships of the terms

• State analogies between electrical and magnetic circuits.

• Draw the electrical equivalent of a magnetic circuit with or without air-gap.

• Solve simple magnetic circuit problems.

• Distinguish between soft and hard magnetic materials

• State examples of soft and hard magnetic materials.

• Explain the B - H Curves for soft and hard magnetic materials.

1.1 Basic definitions in magnetism and magnetic Circuits

1.2 Introduction For the benefit of readers who will need to study and understand well materials in this chapter and the subsequent two chapters, it is advisable to revise some of the basic facts that have been previously learnt in physics under magnetism. in this connection, we do know that simple experiments on magnets and magnetism have revealed the following facts:

(a) the magnetic effects of a magnet appear to emanate from its poles which,

in the case of a bar magnet, are located near each end. For this reason, iron filings cling mainly round the ends of a bar magnet as shown in Fig. 6.1 (a)

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(a) bar magnet with cluster of (b) bar magnet suspended by a thread iron filings

Fig. 1.1

(b) a bar magnet suspended freely by a thread in an horizontal place, always comes to rest with its axis in a north - south direction- The pole which points towards the north is called the north - seeking (N) pole; while the pole which points to the south is called south - seeking (S) pole, as shown in Fig. 1.1(b). (c) If a N pole of a magnet (say, bar magnet) is brought near to the N pole of another magnet repulsion of the magnets takes place. However, if the N pole of a magnet is brought near to the S pole of another magnet, attraction of the magnets takes place.

In other words, like poles (that is, two N poles or two S poles) repel, and unlike poles (that is, an N pole and S pole) attract.

(d) the influence of a magnet can pass both through space and through non-magnetic media. If a magnet, for example, is placed on one side of a sheet of paper, it can attract an iron pin or a steel pin positioned on the other side of the paper. This shows that the influence of the magnet passes through the paper. In conclusion, the region of space within which the influence of the magnet can be detected is called the magnetic field of the magnet.

1.3 Magnetic fields A region in space around a magnet where magnetic influence is felt is called a magnetic field. A compass needle is a device which can be used to find the direction of the magnetic field at different points. Conventionally, a magnetic field consists of lines of m force.

Studies of magnetic fields produced by magnets under various conditions reveal the following:

(a) lines of magnetic field never cross one another, (in other words, every line of magnetic field is a closed line). (b) the magnetic lines of force (which may be called magnetic flux) can continue through the body of the magnet which produces the field, and can consequently form a complete loop.

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(c) part of the magnetic field which happens to be outside the magnet normally traces its path from a north - to a south - pole.

Week 2

Fig- 2.2 shows different patterns of magnetic fields produced by some arrangements of magnets (bar magnets and a Horse - shoe magnet)

Fig. 2.2 Different magnetic fields At this juncture, it may interest readers to observe that in all the arrangements in Fig. 2.2 above the magnetic lines are oval in shape and come out of the N pole into the air and enter the S pole. Since the lines arc continuous, one

Should imagine them inside the magnet, passing from the south (S) pole to the north (N) pole through the magnetic material. 2.3 The definitions Magnetic flux,Φ (sounds phi-in Greek) is the number of lines of magnetic force which emanate from or entering into a magnet. It is therefore not surprising that a magnet with a strong magnetic field produces more magnetic lines of flux than does a magnet with a weak field. The SI unit of magnetic flux is the weber (symbol iVh), named after a German physicist, Weber (W).

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2.4 Magnetic flux density Magnetic flux density, B is a measure of the magnetic flux passing through a unit area in a plane at right angles to the flux. The SI unit of magnetic flux density is Tesla (T) or. Wb/m2, That is. Magnetic flux density, B = •'-.[^/m2] or[Tesla,T] (2.1)

where A is the area (in m2) through which flux <)> (Wb) passes- •i. Example 6.1 Calculate the value of the magnetic flux density when a flux of 50p,^& passes through an area of 2cm2.

2.5 Magnetomotive Force (m.m.f)

This is the force which drives magnetic flux through a magnetic circuit (i.e. the route or path which is followed by magnetic flux) and corresponds to electromotive force (e.m.f.) in an electric circuit. It is usually measured in ampere - turns. -—^>^

From our basic knowledge of magnetism in the study of physics, we do know that when a current flows through a wire conductor it causes a magnetic flux to be established around the wire. Besides, when the current through the conductor wire is increased, the magnetic field around it also increases.

Furthermore, if current. /. passes in a coil of N turns wound on a solenoid (or with an air core), it causes not only magnetic field (magnetic flux) to be established but it causes an increase in the magnetic flux.

For this reason, we can express mathematically that magnetomotive force (m.m.f) is given by

m.m.f.. F = !\. Ampere - turns or Ampere [A] \ (6.2)

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where /"is the current flowing in the conductors of the coil and iVis the number of turns on the coil. Here, we note that since the m.m.f. is equal to the product (current x number of turns) its unit is, therefore Ampere - turn (AT}. However, since the number of turns is merely a figure (which is a dimensionless quantity), then the m.m.f can just have the dimension of current (i.e. Ampere).

Example 2,2 Determine the m.m.f produced by a coil of 600 turns if a current of 5A flows in it.

Solution 7=5.4, N =600 turns.

From Eq. (6.2), m.m.f, F 5x600=3000/<

2.6 Magnetic Field Intensity The magnetic field intensity is the m.m.f. per unit length, / (in metre) of the magnetic circuit. Mathematically, it can be expressed as

IN F Magnetic field intensity H = — = —, Ampere/metre [A/m]. (6.3)

Other names by which magnetic field intensity are known are magnetic field strength or magnetising force.

Example 2.3 A coil of length 0 25m is wound with 1000 turns of wire and carries a current of 5A. Determine the magnetic field intensity at the centre of coil.

2.6 Permeability The permeability of a medium (such as air, magnetic materials and non-magnetic materials) is a measure of how easy it is to set up a magnetic field in

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that medium. At any point in a magnetic field, the ratio between the magnetic flux density B and the magnetic field strength H is called the permeability (\\) of the medium in

which the field exits.

we note that the above unit can also be deduced to mean Henry/metre, (i.e. H/m), since 1 ATfWb = \j Henry . The permeability of free space or a vacuum (with symbol, o )

has the value 4n x 10~7 ///m; that is, Ug =4n x 10~7 ///M. All non - magnetic

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Week 3

3.1 Symbols, units and analogy between magnetic and electric circuits The magnetic terms so far encountered are presented in Table 6.1 with their symbols, units and their electrical equivalents.

Table 3.1 Analogy between magnetic and electric circuits

S/N

Magnetic Circuit

Electric Circuit

Qty. Symbo Unit 1

Qty. Symbol Unit

1. 2. 3.

m.m.f. F Ampere-turn

flux 4> Weber, (Wb)

Reluctance S Amp./weber 1 / (A^b)

e.m.f. £ Volt(^) current / Amp.( ) resistance R Ohm

( Q {n}

4.

H.Ho A

Magneti- H Ampere-sing force turn/metre

Is ^J

Elec. field E VolV strength metre (F7m)

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5.

•

Flux density B Webers/sq. metre (1 tesia3S

1 Wb/m2 )

current f) .4/m2 density

6.

Permeability H webers/Amp .-turn-metre

Conduc. CT Siemens/ metre

RELATIONAL STATEMENTS /EQUATIONS

1.

Flux = m.m.f./reluctance

Current = e.m.f./resistance

2.

Once the magnetic flux is set up it requires no further supply of energy

Energy supply is required continuously to maintain the flow of current.

3.

Total m.m.f.= ΦS1+ Total e.m.f. = . IR.1 +IR2+IR3+...

3.4 Electrical Equivalent of Magnetic Circuits

3.4.1 Introduction In section 3.3 above, we discussed the analogy between magnetic and electric circuits. Now, we are in a better position to draw electrical equivalent diagram of any given magnetic circuit. In the succeeding sections we shall discuss the electrical equivalent diagram for the following magnetic circuits^ (i) series-connected magnetic circuits (ii) parallel-connected magnetic circuits (iii) composite magnetic circurt-

6.4.2 Series Magnetic Circuits Consider the magnetic circuit shown in Fig. 6.3(a) below.

Fig. 3.3(a) is a series-connected magnetic circuit because the same flux is established in all parts of the circuit, assuming there is no magnetic leakage. The circuit consists of an iron part of length ^ and air-gap of length /^. We note the iron part of length /[ = AB + BC + CD + DE + EF and the air-gap is

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made up of FA = l-i. A coil is uniformly wound on the iron part, and produces an m.m.f. ofF=NI. The equivalent electric circuit diagram is shown in Fig. 6.3(b). The procedure for calculating the equivalent reluctance of a series-connected magnetic circuit is the same as for calculating the equivalent resistance of a series-connected electric circuit, which is as follows: Equivalent reluctance, S, =Si +S,[A/Wb], (3.8)

where Si and S are the respective reluctances of the iron path and the air gap. Hence,

following the Kirchhoffs law for a series-connected magnetic circuit, we have;

F=F,+F,, (3.9) where, F == <|)S, == m.m.f required to produce flux <j> in the iron

path, F-i = ^2 == m-m-i required to produce flux in the air-gap. FJ- = ^S = m.m.f required to produce flux in the whole circuit.

i.e. FT = F + F == <t>5i + <fr5z

or, <t>5 =({>5i +

or, Sy. = S + S^.

This final result explains the proof of Eq.(6.8) above.

N.B. Example 3.5 provided ahead illustrates this concept.

The assumption that there is no flux leakage in the air-gap is not accurate. In practice there is some leakage flux, and the method of coping with this problem is discussed immediately below.

Magnetic Leakage and Fringing

Two of the most disturbing phenomena in magnetic circuits are magnetic leakage and fringing. Magnetic leakage has to do with magnetic flux which leak away from its normal magnetic path in a magnetic material. This is illustrated clearly in Fig. 6.4. Fringing concerns the phenomena whereby some of the flux produced by the wi\ fringes or bypasses the air-gap at the edges of the gap. This is also illustrated in Fig. 6.4. The ratio of the total flux to useful flux is given by the magnetic leakage coefficient, which is

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From Eq.(3.10) it is clear that leakage coefficient X , has a value greater than unity, since > <t>u. In pratice, the typical value of the leakage coefficient ranges between 1.05 to 1.4.

3.4 ParaIlel-connccted Magnetic Circuits Fig. 6.5 is a parallel magnetic circuit consisting of two parallel magnetic paths, acted upon by the same m.m-f. The parallel magnetic circuit can be dealt with in much the same way as a parallel electric circuit. The magnetic circuit in Fig. 6.5(a) can be represented by the equivalent electric circuit in Fig. 6.5(b).

Fig. 6.5 Parallel-connected Magnetic Circuit

Leakage coefficient, = total flux AT

useful flux 4>u

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3.6 Example of Soft and Hard Magnetic Materials 1. Materials with relative permeabilities very close to 1 in value are sometimes classified as non-magnetic. For this reason, both paramagnetic and diamagnetic materials can be classified as non-magnetic. Glass, air, aluminium, wood, and many other materials are non-magnetic. For practical purposes, they are neither attracted to nor repelled by a magnetic field.

2. Ferromagnetic materials are commonly used in the construction of motors, transformers, relays and in other devices where a strong magnetic flux is needed.

3.7 B - H Curves for soft and hard magnetic materials

6.8.1 Introduction If a graph of the magnetic flux density (B) is plotted against the magnetising field strength. (//) for a magnetic material, the resulting curve is known as the B - H curve. Fig. 6-9 shows a typical graph of the BH curve or magnetisation curve. Although for any magnetic material B = uH . however this will not lead us to a straight line graph since p is not a constant number. In practical terms, it is evident that p.r (relative permeability of. say, iron) is not constant. From the graph it can be observed that at the initial stage (between the origin 0 and the "Knee" of the curve), as the magnetic field strength (//) increases gradually the flux density (B) increases rapidly- The knee of the curve marks the onset of saturation (see position of the knee marked on the curve).

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Week 4

On completion of this chapter the student should be able to:

• Explain the magnetic effect of electric current. • Draw magnetic fields around straight conductors, adjacent par all el

conductors and solenoids. • Demonstrate by experiment the magnetic effect of a current carrying

conductor in a magnetic field.

• Explain the force on a current carrying conductor in a magnetic field.

• State the direction of the force • Derive the expression for the magnitude of the force (i.e. F = BIL.

Newtons}.

• Explain the concept of electromagnetic induction

• State Faraday's laws of electromagnetic induction.

• State Lenz 's laws of electromagnetic induction.

• Verify by experiments Faraday's law and Lenz's law.

• Derive the expressions for magnitude of e.m.f. induced in a conductor or a coil.

• State the applications of electromagnetic induction.

4.1 ELECTROMANETISM

4.1.1 Magnetic effect of electric current

A conductor carrying an electric current produces a magnetic field around

itself (conductor). This relationship between electricity and magnetism is known as electromagnetism. This phenomenon was discovered in 1820 by Oersted in Copenhagen, (now the capital city of Denmark). Oersted discovered that if a wire (conductor) carrying an electric current is placed above a magnetic compass needle (as shown in Fig. 4.1) and in line with the normal direction of the latter, the needle deflected is clockwise or anticlockwise, depending upon the direction of the current.

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4,2 Magnetic fields around straight conductors, adjacent parallel It is discovered that if we look along the conductor, and if the current is flowing away from the reader into the paper (as marked by the symbol ® ) inside the conductor (as shown in Fig. 4.2) the magnetic field has a clockwise direction and the lines of magnetic flux can be represented by concentric circles around the wire (as shown in Fig. 4.2)

If the current is reversed, the magnetic field will remain as concentric circles but in anticlockwise direction. In that case, the conductor with current flowing away from the paper towards the direction of the reader is represented usually by concentric circles with a dot in the centre. In other words, the direction of the magnetic field previously shown in Fig. 4.3 becomes anti-clockwise. Suppose we have a small rectangular cardboard pierced about its centre point, with a straight conductor carrying current passing through the centre point, If

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we sprinkle some iron filings fairly uniformly on the cardboard around the conductor, we see that the magnetic field pattern (formed by the iron filings) round the straight current -carrying conductor consists of concentric circles with the conductor as centre. We notice the concentric arrangement of the iron fillings tend to be most pronounced in the vicinity of the conductor and the intensity of the field decreases as the distance from the conductor increases.

4.2 Direction of the Magnetic Field due to an Electric Current in a straight conductor

Several rules are known for the determination of the direction of the magnetic field around a straight current-carrying conductor.

A good rule of representing the relationship is to grip the conductor with the right hand, with the thumb pointing in the direction of the current, the fingers then point in the direction of the magnetic field around the conductor. This rule may be referred to as the right-handgrip rule. This is illustrated in Fig. 4.4 below.

Fig. 4.4 Illustrating the right-hand grip rule

There is an alternative way of representing the relationship between the direction of current in a conductor and that of the magnetic field produced by it (the current) alongside the conductor carrying the current. Imagine that the screw is to travel inside the cork in the same direction as the current (i.e. towards the right as shown in Fig. 7.5), it must be turned clockwise when viewed from the left-hand side. By similar reasoning, the direction of the magnetic field is clockwise around the conductor, as shown by the curved arrow.

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(a) Screw moving in current direction

Fig. 4.5 Right-hand cork-screw rule

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Week 5

5.1 Magnetic Field of a solenoid For a start, let us define a solenoid as a number of rums of a wire wound in the same direction. All the turns in a solenoid are in the same direction so that they can assist one another in producing a magnetic field. We note that if a rod of iron is inserted inside the solenoid, only the magnetic field becomes intensified. When an electric current is passed through a solenoid or long cylindrical coil, the magnetic flux produced is very similar to that of a bar magnet. In this case, one end of the solenoid acts like a N pole and the other a S pole. Consequently, magnetic lines (shown in dotted lines in Fig. 7.6) move from the N pole to the S pole. The rule for the polarity of a solenoid carrying a current can be stated as follows: WJien viewing one end of the solenoid, it will be of N polarity if the current is flowing in an anticlockwise direction, and of S polarity if the current is/lowing in a clockwise direction.

Fig. 5.1 Magnetic effect of a solenoid

Fig 5.1 illustrates an easy method of remembering the rule. by indicating arrows on the letters N and S. At end A of the solenoid, the current direction is anticlockwise and so, that end serves as the N pole. At end B, the reverse is the case.

5.2 Magnetic Field around Two long Parallel Conductors In order to understand the resulting magnetic field around two long parallel conductors carrying current we need first to draw the magnetic fields produced by each conductor and then combining these fields.

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Fig. 5.1 shows two parallel conductors, A and B, each carrying current towards the paper- The magnetic field due to current in A .ilone is represented by the dotted circles in Fig. 7.7(a), and that due to B alone is represented by the fairly uniformly spaced curved dotted lines.

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Fig. 5.3 Magnetic fields due to two long parallel current – carrying conductors As a result of the influence of magnetic flux of A on B, and vice - versa we produce a resultant magnetic field configured as shown in 1-iy- 7.7(b). In this case, the two fluxes tend to neutralise each other in the space between conductors A and B, but they tend to assist each other in the space outside A and B. In addition, the resultant magnetic flux behaves like a stretched elastic material. This gives the impression that conductors A and B are attracted towards each other. In other words, there is a force of attraction between A and B. If the current in A is reversed, i.e. if the two conductors carry currents in opposite direction, as shown in Fig 7.7(c). The two fields assist each other in the space between the conductors, A and B, resulting in a lateral pressure between the lines of magnetic force. The circular line offerees are no longer symmetrical about each conductor, but appear displaced as shown. Consequently, there is a strengthened field between the conductors A and B, and a weakened field to the left of A and the right ofB. We notice that the lateral pressure referred to above is repulsive in nature, therefore there exist in the space between the two conductors lines offeree of repulsion.

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Week 6

6.1 Force on a Current Carrying Conductor in a Magnetic Field

Under this topic we shall consider the interaction between magnetic field due to the current in a conductor and the magnetic field in which the conductor is placed. For this purpose, we have the cross-section of conductor carrying current towards an observer together with its magnetic field shown in 1. A 1 so shown m Fig. 7.9(b) is another magnetic field considered to be uniform.

Fig. 6.1 If the conductor is placed in the field, it will be seen that the resultant magnetic flux has been distorted so much so that it partially surrounds the conductor (wire), as shown in Fig. 6.1(c). This distorted field acts like a stretched elastic string bent out of the straight (like a catapult) and the flux exerts a force F urging the conductor out of the way. In this case the conductor A will move from left to right, which is the direction from the strong part of the field (where the lines are very dense) to the weaker part. The brief explanation of the phenomenon is that if we compare the field in Fig. 6.1(a) with that on the left hand side of A in Fig. 6.1(a), we see that on the left side

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the two fields (i.e. the arrows) are in the same direction, whereas on the lower side they are in opposite directions. Consequently, the combined effect is to strengthen the magnetic field on the left side and weaken it on the right side, thus producing the distribution shown in Fig. 6.1(c). However, reversing the direction of the current reverses the direction of the resultant force, as shown in Fig. 6.1(d). 6.2 State the Direction of the Force on a Current-carrying

Conductor in a Magnetic Field. In the previous week a brief mention is made about the force F which acts on a current-carrying conductor in a magnetic field without much emphasis on how to determine the direction of the force, F.

In this section we shall discuss the direction of the force on a current-carrying conductor placed in a perpendicular magnetic field. Generally, the direction of the force can be determined using Fleming's left-hand rule. It states that: If the left hand is held with the thumb, first finger, second finger held mutually perpendicular to each other with the first finger in the direction of the field B and the second (middle) finger in the direction of the current /, then the motion or force on the conductor is in the direction of the thumb M.

Field (B)

(b) Fleming's left hand rule

Fig. 6.2 Fleming’s left hand rule to determine force direction on a conductor

The rule is applicable only if the magnetic field (B) and current (I) are perpendicular or inclined, to each other. We should observe that if the field and current are parallel to each other, no force acts on the conductor. As an exercise, we can use the above rule to verify the direction of the force acting on the current-carrying conductor in Fig- 7.10(b). Furthermore, we can also verify both by Fleming's left hand rule and by drawing magnetic flux patterns, the direction of the force on the current-carrying conductors shown in Fig. 6.1(c) & (d).

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7.6 Derive the Expression for the Magnitude of the Force in a Current-carrying Conductor in a Magnetic Field

Experiment has shown that the force F (Newton s)ac ting on a conductor carrying a current / (Amperes) at right angles to a magnetic field of flux density B (Tesla) is directly proportional to (a) the magnitude of current in the conductor (b) magnetic flux density and (c) length of conductor [metre]. Hence, F = 5/?[Newtons]

Furthermore, we recall (already treated in chapter 6) that if for a magnetic

field

having a cross-sectional area A [metre 2 ] and uniform flux density B (Tesia), then the total flux can be represented

Example 6.1 Draw the resultant magnetic flux structure and determine the direction of the force acting on a current-carrying conductor placed perpendicular to the magnetic field in the following diagrams shown in Fig. 7.11 (a) and (b) below.

Fig. 6.3 Solution (a) The direction of the current/and that of the magnetic flux are known and shown. Therefore using Fleming left hand rule the direction of the force F^ is perpendicularly upward as shown in Fig 6.3(a).

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Week 7

60 turns. It is situated in a radial magnetic field of 0.47'. Determine the force (in milli newtons) on the coil when the current is \OmA.

Hint: Using usual notations', F = Nx BIl, -where I = n x D.

ANSWERS 1. 75 iWb 2. 200 N/m 3. \AT A. 0.57- 5. 7.54mN

7.1 ELECTROMAGNETIC INDUCTION

7.2 Explain the Concept of Electromagnetic Induction Electromagnetic induction is a method of obtaining an electric current with the aid of magnetic flux. It is on record that Michael Faraday, one of the greatest British scientists made the discovery of electromagnetic induction in 1831. hi otherwords, he discovered that electric current could be produced without batteries but solely by using magnets or magnetic fields.

In explaining the concept of electromagnetic induction, we will find that when a permanent magnet NS is moved towards a coil C as shown in Fig. 7.13(a) galvanometer G deflects in one direction and when the magnet is moved away from the coil the galvanometer deflects in the opposite direction (sec Fig. 7.13(b)).

Fig. 7.1 Electromagnetic Induction illustrated When the magnet is placed stationarily (at rest) near the coil the galvanometer remains stationarily at zero position. We should also note that when the magnet is moved nearer more magnetic lines link the coil.

As illustrated and proved by Faraday himself, an induced e.m.f always produces an induced current when there is a relative movement between the magnet and the coil and it is always obtained when a change occurs in the number of magnetic lines linking the coil. Besides, Faraday also showed that the magnitude of the induced e.m.f. is proportional to the rate at which the

Magnet moved towards the coil —>

. ^""^s

Magnet moved away from the coil <—— (-

~' ,/rww

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magnetic lines passing through the coil is varied. Alternatively, we can deduce that when a conductor cuts or is cut hy magnetic lines, an e.m.f. is generated in the conductor and the magnitude of the generated e.m.f. is proportional to the rate at which the conductor cuts or is cut by the magnetic flux.

7.3 Faraday's Laws of Electromagnetic Induction All the foregoing facts gathered from the explanation above sum up to what is known as Faraday's Laws of Electromagnetic induction. The laws can be stated as follow:

First law states that whenever there is a change m the magnetic lines linking a circuit. there will always be an induced e.m.f. in the circuit, or, whenever a conductor cuts a magnetic flux or is cut by a magnetic flux. an e.m.f. is induced in that conductor.

Second Law states that the magnitude of the induced e.m.f. in a circuit is directly proportational to the rate of change of the magnetic flux (or magnetic lines) linking the circuit.

7.9 Lenz’s Laws of Electromagnetic Induction Two methods are known for finding the direction of the induced or generated e.m.f. They are: (i) Fleming's Right Rule, and (ii) Lenz's Law.

Fleming's Right-hand Rule This rule states that if the first finger of the right hand is pointed in the direction of the magnetic flux, (as in Fig. 7.14). and if the thumb is pointed m the direction of motion of the conductor at right angle to the magnetic field, then the second finger, held at right angles to both the thumb and the first, will represent the direction of the induced e.m.f-

Fig. 7.2 Fleming's Righthand Rule Illustrated

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NB The right-hand rule is used/or induced current or e.m.f. but the left-hand rule.

7.92 Lenz's Law In 1835 Heinrich Lenz, a German physicist stated a rule, now known as Lenz's law, which can be used to determine the direction of induced e.m.f. Lenz 's law states that the direction of an induced e. m.f. is always such that it tends to set a current opposing the motion or the change of flux responsible for inducing that e.m.f.

At this juncture we may wish to note that Fleming's righthand rule is normally used where induced e.m.f. is due to magnetic flux-cutting, say a conductor (i.e. when there is a dynamically induced e.m.f.) and Lenz's law is used when there is a change by magnetic flux-linkages (i.e. statically induced e.m.f.) In other words, induced e.m.f. can be either dynamically induced or statically induced. If it is dynamically induced, usually the field is stationary and the conductors cut across it (the field). For example, this is akin to the structure of d.c. generator. However, if the e.mf. Is statically induced, usually the conductors or the coil remains stationary and flux linked with it is changed by simply increasing or decreasing the current producing the flux, for example as in transformers.

We know that the direction of the e.m.f. induced in a moving conductor depends on its direction of motion and also on the direction of the magnetic field lines. Applying the right-hand rule, the direction of e.m.f. is as shown in Fig. 7.18, having known the direction of flux of density B, Tesla and direction of motion of the conductor. Suppose is the length of the conductor, lying within the field and let it move a distance x in a time;, then area swept by the conductor within the flux is, (length x breadth) = l.x. Hence, flux cut, F=Blx (7.3) Furthermore, we know from Faraday's law that whenever a conductor passes through magnetic flux, an e.m.f. is induced in the conductor and its magnitude depends on how fast the conductor passes through the flux. In other words, the magnitude of e.m.f. induced in this way is given.

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Week 8

Symbolically, this becomes

where, V •= velocity of the conductor.

11 the conductor AA" moves at an angle with the direction of flux, then the induced c.m.f. becomes.

E = B IV sin 9, [volt] (7.6)

I

Example 7.4 One pole of an electric generator produces a magnetic flux of 0.2 Wb. An armature conductor passes through this flux at a uniform rate, taking 0.05 seconds to pass completely through the flux. Calculate the e.m.f. induced in the conductor.

Solution In this case, <(>=0.2Wb. t= 0.05s. Using Eq.(7.4), we obtain the e.m.f. induced

as E^-^ T 0.05 E=4V.

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Example 7.5 A conductor of length 0.5 metre moves in a uniform magnetic field of flux density 2

Wfr/ffi 2 at a uniform velocity of 40 metres/second. Calculate the induced e.m.f, under

the following conditions: (a) The conductor moves at right angle to the magnetic field,

(b) The conductor moves at an angle of 30° to the direction of the field, When the coil is positioned initially perpendicular to 5, the flux linkage = NAB

(as stated above). However, when the coil is turned through 60° , the flux density normal to the coil is now B cos60°. And so, flux change through the

coil == NAB -NAB cos60°

^^xlO^-UxlO-^xO^ =6xl0-3

. , , „ flux change 6 x 10"3 .'. average induced e.m.t. == ——————— = ————

time 0.2

= 30xl0-3^.

7.12 Solve Problems on Electromagnetism and Electro-magnetic Induction

^ Example 7.8 A current-carrying conductor is situated at right angles to a uniform magnetic field having a flux density of 0.57. Calculate the current in the conductor if the force per metre length of the

conductor is 2QN.

Solution Data given are: 5=0.57', F/\=20N, /=? Using Eq.(7.1), F= BH

r- i i i.e. Current required /= — • — = 20 • — = 40/1. N

1 5 0.5

JT Example 7.9 Calculate e.m.f. generated in the axle of a car traveling at 90 k/h, assuming the length of the axle to be 1.8/n and the vertical component of the earth's magnetic field to be 50i T

Solution Data provided are:

Exar

37

V==90km/h, l=1.8m. B^SOx^T Expressing V in m/s, we have

90xl000fm1 V^————J-J=25OT/s. 60x60[^]

Using the data given in Eq.(7.5), we obtain the required e.m.f. generated as E = B I V == 50 x 10~6 x 1.8 x 15V,

E =2250 x 10'6 = 2250 V.

Example 7.10 A current-carrying conductor of length 400/nm is moved at a uniform speed at right-angles to its length and to a uniform magnetic field having a density of 0.5 T. If the e.m.f. generated in the conductor is 3 V and the conductor forms part of a closed circuit having a resistance of 0.5H , calculate; (a) amount of current in the conductor; (b) the velocity of the conductor; (c) the force acting on the conductor; (d) the work done (in Joules) when the conductor has moved 500mm.

Solution Using the usual notations, the data given are:

/=400fl»»=0.4m; B=0.57; E=3^; R=0.5 ?; d = 500w/n = 0.5m.

E 3 (a) Current in the conductor, / = — = — = 6A.

R 05 (b) Velocity of the conductor, can be obtained from the expression, E =

BIV. where

i.e. V = -£- = ——3—— = 15 m/s B\ 0.5 x 0.4

(c) Force acting on the conductor,

F= 5/1=0.5x6x0.4 =1.2 N (d) Work done by the conductor,

W= Force x distance = F x d ^=1.2x 0.5 =6 Joules.

Example 7.11 A six-pole motor has a magnetic flux of 0.05Wh per pole and the armature is rotating at 600 rev/min. Calculate the average e-m-f. generated per armature conductor.

Solution Using the usual notations, the data given are:

38

B = 0.05 Wh per pole, v = speed of armature (conductor). It should be noted that each time the armature conductor passes under a pole it cuts a flux of 0.05 Wb. Hence, the total flux cut in one revolution is 0.05 x 6 = Q3Wb .

V = speed of the armature (conductor) = 600 rev./min.

600 , = —— rev.s.

60 ' =70 rev./s.

Using the expression, E = BIV we can obtain the average e-m.f- generated per (length of a conductor) in one revolution, i.e. E/\=BV

£/l=0.3xl0=3y (with/=! assumed).

Example 7.12

A coil of 500 turns is wound on an iron core and a certain current produces a flux of 4000 \^Wb- When the current is opened, the remaining flux in the iron is 2900 uW&-Tfthis reduction process of flux takes 0.2?, find the average value of the induced e.m.f.

Solution m this case, using the usual notations the data given are:

7v==500, •,=4000xlO-A^yb, 4*2 = 2900 xW-^Wb

t=Q.2s. ,1 The average value of the induced e.m.f. can be obtained using the expression in

- A^, - •,) - 500(2900 - 4000) x 10" 6 Eq.(7.7),as £=

/ 02 -6

. 500x1100x10 0-2

7.13 State applications of electro magnet ism and electromagnetic induction For the purpose-of slating applications of electromagnetism and electromagnetic induction, the following examples are possible: (a) Electric Bell. • (b) Magnetic circuits of generators and motors. (c) Telephone Receiver. (d) Moving - iron ammeter and voltmeter (e) Moving - coil loudspeeker. (f) Ignition coil.

N.A The principle of operations of none of the above-Hsted items will be discussed because they are found in details in other courses such as

39

Electrical/Electronics Instrumentation. Electrical Measurements, and Electrical Machines.

40

Week 9

On completion of this should be able to: Define self inductance and mutual inductance state the symbols and units of the terms stated above

• State the expression for the equivalent inductance in inductances connected in series and in parallel.

• State the expression/or the induced voltage across an inductor.

• Slate the expression for inductance in inductive coupled coils connected in series aiding or opposing.

9.1 Self - inductance and Mutual inductance 9.2 introduction Inductance is that property of an electric circuit that opposes any changes (increase or decrease) in current flow (and hence flux change). We note that physical components known as inductors exhibit the property of inductance. Inductors exist in different sizes and shapes. Most practical inductors are made up of conductor wire formed into a coil of several or many turns. Any circuit in which a change of current is accompanied by a change of magnetic flux and which consequently produces an induced counter e.m.f. is said to possess self' - inductance. It is quantitatively measured in terms of coefficient of self induction L. In other words, whenever there is an increase of current (and hence flux) through an inductor (i.e. a coil), it is always opposed by the instantaneous production of counter e.m.f. of self - induction. Whenever we have an electric circuit magnetically coupled to another circuit (as evident in transformers), as the current changes (in a transformer) so does the flux. The changing flux induces a voltage into the other electric circuit. Circuits that are linked by magnetic flux exhibit mutual inductance. Inductance The effect of self - inductance in an electrical circuit is to produce e.m.fs which always oppose any changes in current value. In an R - L d.c. circuit such as shown in Fig. 8.1 (a) these effects are observable mainly when switching ON or OFF the current. The d.c. circuit contains a resistor of resistance R and an inductor (i.e. coil) of self - inductance L in series. In the circuit when the switch S is closed, the current rises from zero (exponentially) to the maximum value expected according to Ohm's law (i =

41

E/R). While the current is rising, the self- inductance of the coil causes an e.m.f to be induced which opposes the current build - up. '

Fig. 9.1 Illustrating Current Rise and Induced E.M.F fall in a RL d.c. Circuit, when S is closed.

Consequently, the current does not immediately achieve its maximum (final) value, but rises gradually as shown in Fig. 8.1 (a). Specifically, we note that at time t) =0 (say), the current / = 0 while the induced e.m.f. value e = E. As time progresses (i.e. t] > 0), i rises (exponentially) while e falls. Fig. 8.2 shows what happens when the switch S is suddenly opened at time /y.

In this case, the current falls gradually from its previous maximum value / (= E/R) to zero, and this fall of current causes a sudden high induced e.m.f. hi other words as the switch S is opened at time / = ^ , current begins to decrease (exponentially) from its maximum value (/) while the induced e.m.f. immediately attains its maximum value e == E and immediately later starts to fall until it reaches zero value. We note especially in Fig. 8.2(a) that when the current is decreasing, the induced e.m.f. tends to prevent the decrease of the current and its direction is therefore the same as that of the current.

44

Week 10

Method I (a) Self inductance in terms of flux ' linkages per ampere

Self inductance of a coil can be defined as the weber - turns per ampere in the coil. We note that weber - turns, Mt> = flux - linkage. Suppose we have a solenoid having N turns and carrying a current of / ampere, then it produces a flux of^ webers. Thus, its weber - turns are /V^i , and its weber -turns per ampere are N^/I.

By definition, L = ^M (8-1) Its unit of dimension is Henry [//] in commemoration of the famous American Physicist, Joseph Henry (1797 - 1878) who discovered electromagnetic induction independently about a year after Michael Faraday's discovery. From the above relation, in Eq. (8.1), if A^i = 1 Wh-tum,I= 1 ampere, then L =^1 henry (H). m words, this can be defined as follows:

A coil is said to have self- inductance (L) of one henry (H) if a current of I ampere flowing through it produces flux - linkage off Wb - turn in it.

Example 8.1 A coil of 250 turns produces a flux of 0.01 Wb when carrying current of 5A. For this current, calculate the inductance of the coil.

Solution •• • In this case. A'=250, =0.02»^», /^5A, Z,=TT • c /o ^ r M^ 250x0.01 Using Eq. (8.1), L == —- = ————— = Q5H .

Method II (b) Self- inductance (L) terms of average induced e.m.f. and rate of

change of current. Generally speaking, it can be stated that if a coil has an inductance L, henry's and if the current through it increases from;/ to iy amperes in / seconds, then average rate of change of current = ———, amperes/second h and, average induced e.m.f. e = -L x rate of change of current .'. average induced e.m.f, e = -L x '-^——/, volts (10.2)

45

From Eq. (8.3), we can state that if— = 1 ampere/sound dt

and e^ == 1 .volt, then L = \H

Hence, in words, we can state that if a coil has a self- inductance of one henry if e.m.f. of one volt is induced in it when current through it changes at the rate of one ampere/second.

Corollary If the LHS and RHS ofEq. (8.1) are cross - multiplied and the resulting expression is divided by (on both sides, we obtain Consequently, we can state that

average e.m.f. induced in a coil = - Ll/t or, average e.m.f. induced in a coil = - N^/t. If E q.(8-2) is expressed in differential calculus form, then we can obtain.

-L^-N (8.5) dt

Example 8.2 A coil of 500 turns is wound on a non - magnetic core and a current of 2.5A through the coil produces a magnetic flux of 150 Wb. Calculate (a) the inductance of the coil, and (b) the average value of the induced e.m.f. if the given current is reversed in 0.1.?.

Solution (a) The data given are: 7v=500, /=2.5A. ifr=l50x lO"6^.t^OAs. Hence,

from Eq.(8.1), we have

Method I (b) We note that the current changes from 2.5A to - 25A in O.ls,

.-. average rate of current change, —=-(2.5x2)/0.1 dt = -50A/S Hence, from Eqn (8.3).

.-. average e.m.f. induced in coil, e = -L— = -0.03 x (-50) dt CL

= 1-5^.

Method II (Alternatively, using expression in Eq. (8.5)) From the given data, it implies the flux changes from 150 u^Z> to - 150uW& in O.ls..'. average rate of change of flux, -•= -(150x 10~6 x 2)/0.1 dt

Hence, from the expression, e = -N—' dt • average induced e.m.f. in the coil, e^ = -500 x (-0.003)

=5V .

46

N.B. Thee comes with a positive sign because of the e.m.f acts in the same direction as the original current, initially Hying to prevent the current decreasing to zero value and then opposing its growth in the reverse direction.

Method III (c) Self inductance <U in terms of the dimensions of the solenoid that if / be the length of a magnetic circuit (i.e. coil) (in metres), and A its cross - sectional area (rn ), then for a coil ofN turns with a current / amperes:

H = IN//{Amperes/metre] \ and, total flux, 4> = BA = pp// . (Remember, u == B/H and u = U (for non-magnetic core, u^ = 1).

\.e. 4=47i x 1Q-7 x (IN/l)A Wb (where, Ho =4rt x 10-7) Substituting for ()) in Eq. (8.1), we have inductance, L = (4n x 10~7 x /4A^2//, henrys (8.6) However, if the coil is wound on a magnetic core then 4 ^ 1 - In that case,

Inductance, L = (4n x 10~7 x u,. x AN2/I), henrys. (8.7)

inductance is proportional to the square of the number of turns and the

cross-sectional area, and is inversely proportional to the length of the magnetic material.

Example 8.3 A coil of 500 turns is wound uniformly on an iron ring which has a mean diameter of 10cm, cross - sectional area of 5cm2 and a relative permeability of 350, calculate the inductance of the coil

Solution The data given are as follows:

^o=4TtxlO'\ ^,.=350. D= 10cm = O.lm, A^xlO'^m 2 . Using Eq. (8.7), the value of the inductance of the coil is

given as, HpU.^v2 _47ixl0-7 x350x5xl0"4 x 5002

- 47ixl0-7 x350x5xl0-4x5002 „,,,., LK= U.I where nD=l

10.2 Determination of Mutual Inductance (M) If two coils X and Y are placed relatively close to each other as shown in Fig. 8.3 below then we can notice that when switch S is closed, current flows in X and produces a flux which becomes linked with Y and the e.m.f.

47

induced in Y causes a momentary current to flow through galvanometer G. (for the fig see week 11)

48

Week 11

Fig. 11 Illustrating mutual inductance We may also notice that when S is opened, the collapse of the flux induces an e.m.f. in the reverse direction in Y. In conclusion, a change of current in coil X is accompanied by a change of flux linked with coil Y and therefore by an e.m.f. induced in Y. Consequently the two coils are said to have mutual inductance.

Mutual inductance can also be determined in three ways as given for self-inductance. The three ways are: (a) Mutual inductance (M) in terms of average induced e.m.f. and rate of

change of current. (b) Mutual inductance in terms of flux - linkages per ampere. (c) Mutual inductance (M) in terms of the dimensions of the two coils.

Next, we shall discuss each of these three methods in turns.

Method I Mutual inductance in terms of average induced e. m.f. and rate of change of current If, say, two coils, X and Y, have a mutual inductance ofM henrys and if the current in coil X increases fi-onu'/ to i^ amperes in t seconds:

From Eq. (8.8) we can deduce the following: • the expression is similar to that of Eq. (8.2). • the minus sign shows that the e.m.f. induced in coil Y tends to make

current flow in such a direction as to oppose the increase of flux due to the growth of current in coil X.

• two coils have a mutual inductance of 1 Henry if an e.m.f. of 1 Volt is

.'. average e.m.f. induced in coil Y =

~M(i,-i,} volts (8.8)

49

induced in one coil when the current in the other coil varies uniformly at the rate of 1 ampere per second.

Example 11 Two coils A and B have a mutual inductance of 0.2H. If the current in coil A is varied from 4 to 2A in 0-2s, calculate the average e.m.f. induced in coil B,

Solution The data given are: M=0.2H, t/=4/t, i^=2A, t=0.2s. Using Eq. (8.8) we can obtain the average e.m.f. induced as

(2-4) average e.m.f. induced in coil B = -0.2 -0.2 =2V.

Method II Mutual inductance in terms of flux - linkages per ampere Suppose, and ((ie represent the flux in webers linked with coil Y due to currents and amperes respectively in coil X (sometimes called the primary) and if N^ represents the number of turns on coil Y (called the secondary)

average e.m.f. induced in r=(T2 '1) 2 , volts (8.9) t

Equating expression in Eq. (8,8) and Eq(8.9) we have: change of flux - linkages with secondary

change of current in primary

N..B. By comparison, the expression in Eq. (8.10) is similar inform to that obtained in Eq. (8.5) for self inductance except that the change influx takes place in the secondary coil Y and the change in current takes place in the primary coil X. Furthermore, if we know only flux - linkages with secondary and current in primary (but not change in flux - linkages and change in primary) as expressed in Eq. (8.10), then the mutual inductance

Example 8.5 Refer to the question in Example 8.4 and calculate the change of flux linked with coil B, assuming that coil B has a winding of 250 turns.

Solution In addition to the data provided in Example 8.4 we have N = 250. Using Eq.(8.9) we get the required change of flat (<(^ -((^)=({i , having known that

the average induced e.m.f. =1V. (see solution to Example 8.4)

2=i^0(^,) Q2 ~ '

^=OA-=0.00\6Wb 250

Method m Mutual inductance in terms of the dimensions of the two coils

50

For illustration let us refer to Fig.8.3 and assume that coil X and coil Y have windings TV") and N turns respectively.

Without bothering ourselves for proofs, we can state that mutual inductance in respect of the two coils can be given as,

M= NIN2 = Jv1^2 , W (8.12) //UoU^ reluctance L J

(N.B. From our studies of magnetic circuits in chapter 6 we know that reluctance =

//HoHrA).

Example 8.6 Two coils X and Y having 40 and 400 turns respectively are wound side - by – side on an iron core of cross - sectional area of 100c/n2 and mean length 160cin. Calculate the mutual inductance between the coils if the relative permeability of the iron is 1600.

•Solution The data given are; A =40 =400 A = 100 x 10~4m2 ^O'W. / = 1.6/n .

Using the expression of Eq. (8.10), mutual inductance can be obtained as JViJv; 40x400

1.6/4n x 10'7 x 1600 x 10.'. M= 0-20

11.1 Coefficient of Coupling or Coupling Coefficient If two coils, A and B, are magnetically coupled and each has self - inductances Li and L,2 respectively and if the mutual inductance between the coils is M, then we note that if the coils are placed close together, almost all the flux produced by current in one coil passes through the other coil and ihc coils are said to the tightly coupled. In that case K re 1. However, if the coils arc well spaced apart, only a small fraction of the flux in the primary is linked with the secondary, then the coils are said to be loosely coupled. Consequently, either K « 1 or K » 0.

Example 8.7 Two coils, X and Y, have self inductances ofl50u// and 300u// respectively. A current of 2A through coil X produces flux linkages of 80^ Wb - turns in coil Y. Calculate (i) the mutual inductance between the coils (ii) coefficient of coupling of the coils.

51

Solution The data given aie:7L, =150uff. N3 = 300»Aff . i, = 2A A^<t>,

^SOxlO'6^. (assuming that coils X and Y have windings/V, and N

respectively, and current;/

leads to the production of flux (^i initially in coil Y).

flux - linkage of coil Y yv,A,

(i) Using Eq. (8.10), M=-—————, „ ——^ current in

coil X t,

.. 80xl0-6 M = 40u//, 2M 40xl0-6

(ii) Coefficient of coupling, K=0.189.

11.2 Symbols of Inductors and units of Inductance

11.2.1 Symbols The common circuit symbols for inductors are shown in Fig. 8.4. The names given to the symbols are known as air - core lype, iron - core type and variable iron - core type as illustrated in Fig. 8.4.

(a) air - core (b) iron - core (c) variable iron - core Fig.

11.3 Circuit Symbols for Inductors

11.3.1 Units of Inductance We recall that there are two types of inductance: namely, self inductance and mutual inductance. Whether we refer to either self inductance or mutual inductance, the unit of measurement is the Henry [H]. However, we should note that sub - units of Henry are possible: such as microhenry) and milli henry {mH).

11.4 Expression for equivalent Inductance of inductances connected in Series and in Parallel

11.4.1 Equivalent Inductance of Series - connected Inductors Consider a circuit consisting of N inductors, connected in series, as shown in Fig. 8.5(a), with the equivalent circuit shown in Fig. 8.5(b). Like in the case of resistors connected in series, the inductors have the same current through them.

52

Fig. 8.5 Electric Circuit diagram of Inductors connected in series and their equivalent Circuit

FUNDAMENTALS OF AC. THEORY Week 12

50

Week 12

12.1 Describe the production of an alternating e.m.f. Alternating e.m.f. may be produced by rotating a coil in a magnetic field or by rotating a magnetic field within a stationary coil. A typical circuit arrangement whereby an alternating e.m.f- can be produced is shown in Fig. 9.1. Fig. 9.1 shows a loop DABC rotated at a constant speed in a clockwise direction in a uniform magnetic field due to poles NS. The ends of the loop are connected to two slip rings C| and C^. Bearing on these rings are carbon brushes £, and E- which are connected to either an electric lamp L or an external resistor R. In the present position of the rotating coil (loop) of Fig. 9.1, me plane of the coil is parallel to the field. In this case, sides DA and CB are moving at right angles to the field. According to Fleming's right hand rule, an e.m.f. is produced along DA and CB in the direction shown on the loop. In this case when the coil is horizontal, e.m.f. attains its maximum value. However, no induced e.m.f. is produced along the sides AB or DC because these sides do not "cut" the field lines as they rotate.

Fig. 12.1 Product of an alternating e.m.f.

When the coil has turned through 90°, that is: when the coil is in vertical position, e.m.f. produced is zero. At this instant, the sides DA and CB are both moving parallel to the field. Consequently no induced e.m.f. is obtained. When the coil has turned another 90° and its plane is now horizontal or parallel to the field but in an opposite direction to the first horizontal position we started with. So the e.m.f. along DA and CB are in opposite directions, according to

A coil of loop DABC C1C2 - two slip rings C1 and C2. E1]E2 - carbon brushes


51

Fleming's right angle rule. In summary, when the coil is turned 180°, the e.m.f. is reversed. 12,2 Equation of the Alternating Voltages

12.2 Stating relationship between instantaneous and peak values of a sinusoidal wave

in Fig. 12.1 let us imagine a rectangular coil (loop) of length / metres of each of the parallel sides DB and CB, having N turns and rotating in a magnetic field of flux density B W/m1. Suppose the peripheral speed of each side of the loop to be V metres per second. This velocity can be resolved into two components, perpendicular and parallel respectively to the direction of the magnetic flux, as shown in Fig. 9.2 below. When the coil has turned through angle 0, its velocity V can be resolved into two mutually perpendicular components, viz (i) VCosO component - which is parallel to the direction of the magnetic flux and (ii) VsinO component which is perpendicular to the direction of the magnetic flux

Fig. 9.2

e.m.f. generated in one side of the loop which contains N turns, e = Nv-BlV sin 6.And so, total e.m.f. generated in, both sides of the coil is

(9.1)

we note that when 0 = 90°, e has maximum value of

En, (say) =1BNIV (Volts) (9.2)

Equation (9.1) can be rewritten as e = E sin 6 .

If b = AB = width of the coil (in metres), and,

e -= 2BNI

V

sin 9


52

f~= frequency of rotation of coil (in Hz), then, V=nbf.

If A = / x b = area of loop in square metres, then

Em=2BNl x nbf=l7iBANf (volts) (9.3)

However, the instantaneous value of e.m.f. (generally at any value of 9) generated in the coil can be expressed as

(9.4)

The e.m.f. can be represented by a sine wave as in Fig. 9.3, where £„ represents the maximum value of the e.m.f. and e is the value after the loop has rotated through an angle 6 from the position of zero e.m.f. at =0°.

50=1x n .'. Speed = 50 revolution/second or. Speed = 50 x 60 rev./min == 3000 rev./min.

e=E,,, sin 8 = InBANf sin Q volts)

Fundamentals of A.C. Theory Week 13

53

Week 13

13.1 Another form of e.m.f. equation Apart from the form the e.m.f. equation was expressed in Eq. (9.4), yet another form could be as

e=En, srnQ^E sincotsE sin2nft=En, sin—t, (9.6)

where, 6= cot, o)=2Jcf and period, T==l/f.

From Eq. (9.6), we can deduce the following: (i) the peak value or amplitude of an alternating e-m.f. is given by the coefficient of die

sine of the time angle. (ii) the frequency/is given by the coefficient of time divided by Ix.

For example, if the equation of an alternating e.m.f. is given by e= 10 ^n314?.then

314 its peak value (i.e. maximum value) E,n == 10F, its frequency / = —— = 50Hz . and e is

In the instantaneous value, (e is the e.m.f. value at every time instant.)

13.2 Defining instantaneous, average, r.m.s, form factor and peak value alternating current or voltage

13.2.1 Instantaneous value of an alternating voltage or cuiirent. As explained earlier on, the instantaneous value is the value of the e.m.f. or current at any time instant.

For an alternating e.m.f., its instantaneous value e = Em sin wt, whereas for an alternating current its instantaneous value / = A,, sin wt. Example 9.1 A square coil of side 10cm, having 200 turns, is rotated at 1200 rev/min about an axis through the centre and parallel with two sides in a uniform magnetic field of density 0.57'. Calculate: (a) The frequency (b) The induced e.m.f.

Solution 60 sec sec (ii) Using Eq (9.3), the instantaneous e.m.f can be obtained as,

e - 27rBANn. In our case, B = 0.5T, A = 0.1 x Q.lm2 = 0.01m-'


54

N= 200, n = 20 (obtained form (i) above) Consequently, the r.m.s. value of the e.m.f. can be obtained as

Crms = 0-707e = 0.707 x 2w x 0.5 x 0.01 x 200 x20 e = 88.84

Example 9.2 Determine the number of poles of an alternator driven at a speed of 375 /mm an<* itg®"61^1^ an e.m.f having a frequency of50//z.

Solution

Using Eqn. (9.5). /^=——- =8pairs ["^J

P=8x2 = 16poles

Example 9.3 Ife= 125 sin 2nft is the instantaneous value of an alternating e.m.f. with periodic time 0.01s. (a) what will be its value 0.002s after passing through zero? (b) if the voltage is applied across a 20 - ohm resistor, what is its instantaneous current value at; = 0.002s

Solution

(a) Frequency, /=-=——=io0 c/s or Hz

.'. instantaneous value, e=125 sin27txl00xt i.e. e=125 sin2007Tt •At t= 0.002s.

e=125 sin 20071 x 0.002=125 sin0.4 bearing in mind that, 2n radians = 360° (i.e. K rad = 180°), . (. . 180 e = 125 x sin 0.47C x =125 sin 72° =125x0.951^119V (b) when the instantaneous voltage is applied across a 20 - ohm resistor, its instantaneous current,

. e 125 sin 2007rt

After t = 0.002s and using the same notation as before,

Example 9.4 (a) Find the amplitude, phase, period, and frequency of the sinusoidal waveform e=10cos(50t+20°) (b) An e.m.f. waveform is given by V = 200sin628t. How long does it take this waveform to complete one - half cycle?

Solution


55

(a) Amplitude, emf = 10V; phase, (& == 20°; angular frequency, ro = W^y Period,

22 f=l =7.95g Hz 50 T 0.1257

(b) Period, T = 10ms, which is the time for 1 cycle. w = 28

Consequently, the time for one-half cycle = .5 x 10= 5ms

13.3 Average value of an alternating voltage or current Suppose we consider the case of current which is a non - sinusoidal waveform as shown in Fig. 9.4.

Fig. 9.4 Illustrating how to find the average value

if we consider generally n equally - spaced mid - ordinates /,, i^ ..., in taken over

either the positive or the negative half- cycle then, average value of current over half a cycle This approach is known as the mid - ordinate method. This method may be used both for sinusoidal and non ~ sinusoidal waveforms. If we have a sinusoidal waveform i = !„, sin 9, it can be proved that (for half of a cycle):

average value of current, ;,„ = 0.637 x maximum value

As an illustration, we consider an alternating current i = /„ sin 6, shown in Fig. 9.5 and show how to determine its average current value.

(9.8)

We note that average value over a complete cycle for a sinusoidal waveform is zero. There is another approach known as the analytical method by which the average

value can be determined. This will be discussed immediately below. With the aid of integral calculus, the average value can be determined by the expression


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Fig. 9.5 Full - wave rectified sine wave

The average current value of the full - wave rectified sine wave can be determined as follows (bearing in mind that the period T =- n):

iav-0.6371,

In other words, the average current, i^ = 0.637 x maximum current value. If we consider voltage quantity we shall state that the average voltage V= 0.637 x maximum voltage value.

N.B. ffwe consider a whole cycle of a sinusoidal signal i = 1 sin cot d (o)t) (such as in Fig. 9.3} and the period T = In, we shall be able to prove according to Eq. 9.9(b) that its average current i^ = 0.

13.4 Root - Mean - Square (R.M.S) Value The r.m.s value of an alternating current is the d-c. current value flowing through a given circuit for a given time produces the same heat quantity (heating effect) as produced by the alternating current when flowing the same circuit for the same time duration.

For the purpose of computing the r.m.s value either the mid - ordinate method or analytical method may be used.

Let us consider a current having the waveform shown in Fig. 9.6 (a). If this current flows through a circuit having resistance R ohms, the heating effect of i, is i^R, That of is ij R, etc. as shown in Fig. 9.6(b). we observe that the heating effect is positive during both the positive and negative half cycles. In general, if there are n equally - spaced mid - ordinates in half a cycle, then:

average heating effect during half a cycle

Suppose we have a direct current of I amperes flowing through the same resistance R it will produce a d.c. heating effect equal to the average heating


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effect of the alternating current, and thus to produce the same quantity of heat in half a cycle:

Then,

In other words, it can be stated that the root- mean-square ( or r.m.s.) value of a sinusoidal current is measured in terms of the direct current that produces the same heating effect in the same resistance. Hence, if Im be the maximum or peak value of the sinusoidal current, the average heating effect over a cycle (or half a cycle ) is half the maximum heating effect,

Thus far, we have been discussing the mid - ordinate method. The next method is the analytical method of calculating the r.m.s. value. Analytical method involves of the application of the expression,

Y = r.m.s of current or voltage

y(t} = general (current or voltage) function

\y(t)\ = square of the (current or voltage) function

r= period of a cycle

•(a) A given current waveform

or,

where

(Current)


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(b) a resulting (heating effect) wavef Fig. 9.6 Illustrating the r.m.s value


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Week14

Example 9.5 (a) Determine the peak voltage value of a sinusoidal alternating voltage ofr.m.s

4.5V. (b) Determine the r.m.s value of a rectangular current wave with an amplitude of

8.M. (c) Determine the average value of a sinusoidal alternating current of ISA

maximum value.

Solution (a) From Eq.(9.10), ^=0.707

V_ = —1— x V. = -45- V = 6.36V m 0.070 "" 0.707

(b) From Eq.(9.11). /^ =0.707 /,

=0.707 x 8.8/4 =6.22A

(c) From Eq.(9.IO), /„== 0.637

=0.637x25/4=15.9.4

Example 9.6 A triangular current waveform has the following values over one - half cycle Current (A) 0 2 4 68 10 86 42 0 Time (ms) 0 10 20 30 40 50 60 70 80 90 100 Determine (a) the average current value, and (b) the r.m.s current value.

Solution (a) By applying Eq. (9.7), the average current value.

where n = number of time intervals (b) By applying Eq. (9.11), the r.m.s current value


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Example 9.7 Find the average and effective (rms) values of the rectangular voltage wave shown in

Fig.9.7.

Fig. 9.7 Diagram for Example 9.7

Solution

Method I

fT Looking at Eq. 9.9(a) this method is based on the fact that eff)dt stands for

J t)

the area under the graph from limits 0 to T. area under the graph from limits 0 to T

Time period (T)

This means that rinding (calculating) the average value of the waveform implies calculating the net area of the waveform over one period. T= 0.03s. If A == area of section abed and A^= area of section defg, then the net

area = A {-A,).

(a)


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Bearing in mind that Fig. 9.7 enables us to determine we need to draw the

graph of V^(t)to find mmf- Since both negative and positive voltages become positive when squared, therefore the negative portion in Fig. 9.7 becomes positive when (- 2V} is squared i.e. W1. Consequently the graph of V2(t) lS as shown in Fig. 9.8 below.

Method II With reference to Fig. 9.7, we have the following: (a) For the time interval a-d, Vs1 \QV

For the time interval d~g, V=~2V The period T = 7^ + 7^ = 0.01 + 0.02 = 0.03s.[1 f r 0.

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(b) In order to determine the mis value, we start by finding V2^) for each lime interval (a - d) and (d - I For the a - d interval. For the d - g interval, Next, we apply Eq. (9.13) to get the mis value of voltage,

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Week 15

N.B.

as result a is the same as obtained by method I.

15.1 Form factor, Kf Form factor

is the ratio of mis value to average value. Generally it can be expressed as,

15.2 Crest or Peak or Amplitude Factor, K, It can be defined as the ratio of the maximum value to the r.m.s. value.

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15.3 Explain phase lag or phase lead as applied to a.c. circuits 15.3.1 Phase difference Let us consider a general expression for the sinusoidal waveform,

e(t)=E,sin(cot+(()). where (cot + <t>) is the argument and (() is the phase angle. Both argument and phase can

be measured in radians or degrees. Next let us imagine we have two sinusoidal waveforms expressions as

C|(t)= E^sincdt and e^t) = E sin(cot +<^> ) respectively and shown in Fig. 9.9

(t)=e,sin(cot + 9) Fig. 9.9 Two sinusoidal waveforms with the same amplitude but different phases.

The starting point A, of e^t) occurs first in time. Consequently, we say that €3(1) leads e|(t)orthat e,(t) lags 63(1) by <|). Ifij)=0, then ei (t) and e;(t) are said to be in phase; in other words, they reach their zero point, minima and maxima at exactly the same time. Generally speaking, when comparing two sinusoidal waveforms the leading sinusoidalwaveform is one which reaches its maximum (or zero) value earlier than the other one. Similarly, the lagging sinusoidal waveform is one which reaches its maximum (or zero) value later than the other waveform. There are some general points to note when comparing two or more waveforms. They include the following: (a) a sinusoidal waveform can be expressed in either sine or cosine form. However, when comparing two sinusoidal waveforms, it is advantageous to express both as either sine or cosine with positive amplitudes. To achieve this, the following trigonometric identities are useful;

66

We can transform a sinusoidal waveform from sine form to cosine form or vice versa using these relationships stated above. (b) The phase difference between any pair of waveforms can be determined either from their respective instantaneous equations or from the drawing of their respective waveforms, (see example 9.8 as an illustration).

15.4 Differentiate between series and parallel resonance An ac circuit is said to be in resonance when the applied voltage V (with constant magnitude, but of varying frequency) and the resulting current / are in phase.

Series Resonance Consider we are given an RLC series circuit of Fig. 9.29

Fig. 9.29 An RLC Series circuit

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The RLC series circuit has a complex impedance, Z=R+j(X,-Xc)=R+J(coL-——}.

For resonance to occur, (X,-X,-)=Q This implies that Z = R (from Eq. (9.21) i.e. inductive reactance, X= capacitive reactance, X• Consequently, when X\ = X^ This also implies that or. Resonance frequency. /, =(9.22(a)) , (Hz)

2;r WLC In regard to series resonance, there are few things of interest to note. These

may include the following: (i) Graphical representation of impedance (Z) X^, X and R as functions of

frequency, co, is as shown in Fig. 9.30-

Fie. 9.30

From Fig. 9-30 we see that when series resonance occurs at w=cDr, the impedance

curve (Z) reaches its minimum value at X, and Z = R. In other words, the current; in

V y the circuit of Fig. 9.29 /=—=—, becomes maximum. z R

(ii) The phase angle, (0) relationship with frequency (o) is as shown in Fig. 9.31. It can be observed that before resonance occurs when, X > X\. and the phase angle (6) varies from above 0° to 90°. Conversely, when X > X the phase angle 0 varies from below 0° and -90°- However, at 9=0°, resonance occurs at &»,.. Phase angle.

* f^>\

N.B. The graphs of XL , Xc and R are shown in broken lines

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Fig. 9.31 Phase angle curve as a function ofo

(iii)The phasor diagram showing the current and all the voltages in a series resonant circuit is as shown in Fig. 9.32.

Fig. 9.32 Phaser diagram of current and the voltages in an RLC series circuit. At resonance, (see also Fig. 9.29) ^ + V = 0 and V = V = Rl. (iv)The graphical relation in a series RLC circuit between current I and frequency co of the applied voltage is shown in Fig. 9.33. The graph is known as the resonance curve.

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At resonance, (X - X ) = 0 and when this happens (9.22) R

Before we reach resonance frequency, w the impedance Z is greater than A, •therefore the current / flowing in the circuit is less than /^.

Example 9.25 A 120 - V ac source supplies a series circuit resistance and inductance of \0 and 25mff respectively. The generator frequency is the resonance frequency of the circuit. Determine (a) the resonance frequency and (b) the current.

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Solution (a) From Eq.(9.21), the resonance frequency,

/,=————=——, ' =225/fe. WLC 2 25 x 10-3 x 20 x 10-6

(b) At resonance, Z = R. Therefore the required current,

Example 9.26 An RLC series circuit consists of a coil having resistance of 10/3 and inductance of Q.5H and a variable capacitance. The circuit is connected across a 150^, 50C/S supply. Calculate the capacitance at which the circuit resonates and the voltage across the capacitor at this resonant frequency.

Solution N.B. A coil is an element consisting of a resistor, R and an inductor, each of a fixed resistance value and a fixed inductance value respectively.

Fig. 9.34 is the required circuit diagram

Fig. 9.34

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