Bakshi Basic Electrical

8/16/2019 Bakshi Basic Electrical

1/196

(i)

®

(One day Revision Book)

TECHNICALPUBLICATIONS

TM

An Up-Thrust for Knowledge

®

www.technicalpublications.org

[email protected]

As per Revised Syllabus of

Visvesvaraya Technological University

Basic ElectricalEngineering - at a Glance

Uday A. BakshiM.E. (Electrical)Formerly Lecturer in Department of Electronics Engg.

Vishwakarma Institute of Technology

Pune

Mrs. Varsha U. BakshiB.E. (Electronics)

Assistant Director,Noble Institute of Computer TrainingPune


2/196

®Module - 1

Chapter - 1 D.C. Circuits (1 - 1) to (1 - 22)

Chapter - 2 Electromagnetism (2 - 1) to (2 - 18)

Module - 2

Chapter - 3 D.C. Machines (3 - 1) to (3 - 28)

Chapter - 4 Measuring Instruments (4 - 1) to (4 - 8)

Module - 3

Chapter - 5 Single Phase A.C. Circuits (5 - 1) to (5 - 40)

Chapter - 6 Domestic Wiring (6 - 1) to (6 - 12)

Module - 4

Chapter - 7 Three Phase Circuits (7 - 1) to (7 - 24)

Chapter - 8 Synchronous Generators (Alternators) (8 - 1) to (8 - 12)

Module - 5

Chapter - 9 Transformers (9 - 1) to (9 - 16)

Chapter - 10 Three Phase Induction Motor (10 - 1) to (10 - 14)

(ii)

T able of Contentsable of Contents

T


3/196

Chapter at a Glance

1. Relation between Charge and Current

I = Q

t Amperes

Where I = Average current flowing while Q = Total charge transferred

t = Time required for transfer of charge.

2. Electric Potential and Potential Difference

Electrical Potential = Work done

ChargeWQ

3. Resistance

4.186 joules = 1 calorie and 1 joule = 0.24 calorie

So finally,R =

l a

Where l = Length in metres, a = Cross-sectional area in square metres

= Resistivity in ohms-metres, R = Resistance in ohms

4. Ohm's Law

Ohm's Law is, I = VR

amperes or V = IR volts or VI

= Constant = R ohms

5. Electrical Work

Electrical work W = V Q J But I = Q

t

W = V I t J Where t = Time in seconds

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM

1 D.C. Circuits

(1 - 1)


4/196

6. Electrical Energy

Electrical energy E = Power Time = V I t joules

1 Wh = 1 watt 1 hour = 1 watt 3600 sec = 3600 watt-sec i.e. J

and 1 kWh = 1000 Wh = 1 103 3600 J = 3.6 106 J

7. Current Division in Parallel Circuit of Resistors

I

2 =

R

R R

1

1 2

I

T

I

1 =

R

R R

2

1 2

I

T

Important Theory Questions and Answers

Explain the ideal and practical voltage source. VTU : Mar.-02, Marks 4

Ideal voltage source is defined as the energy source which gives constant voltage acrossits terminals irrespective of the current drawn through its terminals.

But practically, every voltage source has small internal resistance shown in series withvoltage source and is represented by R se as shown in the Fig. 1.3.

Basic Electrical Engineering 1 - 2 D.C. Circuits


V

IT

I1 I2

R1 R2+ –+ _

Fig. 1.1

+

–

Vs Vs

IL

VL

+

_

Load+

–

Vs

VL

IL0

V = VL s

(a) Symbol (b) Circuit (c) Characteristics

Fig. 1.2 Ideal voltage source


5/196

Because of the R se , voltage across terminals decreases with increase in the load current

and it is given by expression,

VL

= – (R se) IL + Vs = Vs – IL Rse

For ideal voltage source, Rse = 0 and for practical voltage source it is as small aspossible.

Explain the ideal and practical current source. VTU : Aug.-01, 03, Marks 5

Ideal current source is the source which gives constant current at its terminalsirrespective of the voltage appearing across its terminals.

But practically, every current source has high internal resistance, which is in parallelwith current source and it is represented by R

sh. This is shown in the Fig. 1.5.

Because of R sh

, current through its terminals decreases slightly with increase in voltage

at its terminals.

For ideal current source, R sh

= while for practical current source it is as high as

possible.



+

–

IL

VLRL

Rse

+

_

Internalresistance

Vs

Ideal

VL

IL0

Practical

When there is noload, = 0 and

= V

IL

sVL

(a) Circuit (b) Characteristics

Vs

Fig. 1.3 Practical voltage source

Is Is

IL

VL

+

_

Load Is

IL

VL0

I IL s=

(a) Symbol (b) Circuit (c) Characteristics

Fig. 1.4 Ideal current source


6/196

State Ohm's law and its limitations. VTU : Jan.-08, July-08, Marks 6

The Ohm's law gives relationship between the potential difference (V), the current (I)and the resistance (R) of a d.c. circuit.

It states that, the current flowing through the electric circuit is directly proportional tothe potential difference across the circuit and inversely proportional to the resistance of

the circuit, provided the temperature remains constant.

Mathematically, I V

R Where I is the current flowing in amperes, the V is the

voltage applied and R is the resistance of the conductor, as

shown in the Fig. 1.6.

The Ohm's law can be defined as, the ratio of potentialdifference (V) between any two points of a conductor to the

current (I) flowing between them is constant, provided that

the temperature of the conductor remains constant.

The limitations of the Ohm's law are,1) It is not applicable to the nonlinear devices such as diodes, zener diodes, voltage

regulators etc.

2) It does not hold good for non-metallic conductors such as silicon carbide. The law for

such conductors is given by,

V = k Im where k, m are constants.

State and explain Kirchhoff's laws. VTU : July-07, 11, Marks 6;Jan.-09, 11, 13, June-13, Marks 5

There are two Kirchhoff's laws.

1. Kirchhoff's Current Law (KCL)

The law can be stated as,

The total current flowing towards a junction point is equal to the total current

flowing away from that junction point.

Another way to state the law is,The algebraic sum of all the current meeting at a junction point is always zero .



I I I

I

I

I I

L sh s

sh

L

L s

+ =

Thus as

increases,

decreases.

<

Is Load

IL

VLRsh

+

_

nternalresistance

Ish Is

IL

VL0

deal

Practical

(a) Circuit (b) Characteristic

Fig. 1.5 Practical current source

R

V

+ –

I

Fig. 1.6 Ohm's law


7/196

The word algebraic means considering the signs of various currents.

I at junction point = 0

Sign convention : Currents flowing towards a junction point are assumed to be positive

while currents flowing away from a junction point assumed to be negative.

Consider a junction point in a complex network as shown in theFig. 1.7. The currents I

1 and I

2 are positive as entering the

junction while I3

and I4

are negative as leaving the junction.

Applying KCL, I at junction O = 0I I I I

1 2 3 4 = 0 i.e. I I

1 2 = I I

3 4

The law is very helpful in network simplification.

2. Kirchhoff's Voltage Law (KVL)

"In any network, the algebraic sum of the voltage drops across the circuit elements of any

closed path (or loop or mesh) is equal to the algebraic sum of the e.m.f. s in the path"

In other words, "the algebraic sum of all the branch voltages, around any closed path or closed loop is always zero."

Around a closed path V = 0

Important Solved Examples

Example 1.1 Find the equivalent resistance across the terminals PQ of the network shown in the

Fig. 1.8. VTU : June-83



I1

I3

I2

I4

O

Fig. 1.7 Junction point

100

100

100

100

50

50

50

Q

Fig. 1.8


8/196

Solution : Replacing the lowest parallel combination of 100 we get,

Example 1.2 Two voltmeters A and B, having resistances of 5.2 k and 15 k respectively are

connected in series across 240 V supply. What is the reading on each voltmeter ? VTU : June-81

Solution : The arrangement

is shown in the Fig. 1.9.

Req = R RA B

= 5.2 + 15 = 20.2 k



50

100

100

50

50

50

P

Q

100

100

100

= 50 100 100100+100

Parallel

50

50

P

Q

Series

Fig. 1.8 (a)

50

100

50

50

P

Q

100 100

50

P

Q

= 50

100 100100+100

Parallel

Series

Fig. 1.8 (b)

50

50

RPQ

P

Q

R = 50 + 50 =PQ 100

Series

Fig. 1.8 (c)

Voltmeter A

240 V

Voltmeter BI

5.2 k 15 k

R A

V A

RB

VB

240 V

I

Series

Fig. 1.9


9/196

I = VReq

= 240

20.2 10 3= 0.01188 A

According to Ohm's law, VA = I RA = 0.01188 5.2 103 = 61.7821 V

and VB

= I RB

= 0.01188 15 10 3 = 178.2179 V

Thus reading on voltmeter A is 61.7821 V and that on B is 178.2179 V.

Example 1.3 Find the equivalent resistance

between the two points A and B shown in the

Fig. 1.10.

Solution : Identify combinations of series

and parallel resistances.

The resistances 5 and 6 are in series, asgoing to carry same current.

So equivalent resistance is 5 + 6 = 11

While the resistances 3 , 4 , and 4 are inparallel, as voltage across them same but current divides.

Equivalent resistance is, 1

R =

13

14

14

= 1012

R = 12

10

= 1.2

Replacing these combinations redraw the figure as shown in the Fig. 1.10 (a).

Now again 1.2 and 2 are in series so equivalent resistance is 2 + 1.2 = 3.2 while11 and 7 are in parallel.

Using formula R R

R R1 2

1 2 equivalent resistance is

11 711 7

= 77

18 4 277. .

Replacing the respective combinations redraw the circuit as shown in the Fig. 1.10 (b).



B A

1

5

7

2

6

4

3

4

Fig. 1.10

B A

1

11

7

2 1.2

1

3.2

4.277

A B

ParallelSeries

Parallel

Fig. 1.10 (a) Fig. 1.10 (b)


10/196

Now 3.2 and 4.277 are in parallel.

Replacing them by 3 2 4 277

3 2 4 277

. .

. .

= 1.8304

RAB = 1+ 1.8304 = 2.8304

Example 1.4 Find the current in all the branches of the network shown in the Fig. 1.11.

VTU : Aug.-95

Solution : Let current through the branch AB be

I amperes.Hence applying KCL at various nodes, the

various branch currents can be obtained as

shown in the Fig. 1.11 (a).

Applying KVL to the loop ABCDEFA,

I 0.02 I 60 0.01 I 0.03 I 120 0.01 I 50 0.01 I 8 0 0.02 = 0

I [0.02 0.01 0.3 0.01 0.01 0.02] 0.6 1.2 0.5 1.6 = 0

0.1 I 3.9 = 0

I = 39 A



80 A

120 A

60 A

A

B

C

D

E

F

60 A

30 A

0. 0 2

0 .0 2

0 .0 1

0.01 0.01

0. 0 3

70 A

Fig. 1.11

80 A

120 A

60 A

AI

I

I –60I –50

I –80

I –120

B

C

D

E

F

60 A

30 A

70 A +

+

+

+

+

+

–

–

–

–

–

–

Fig. 1.11 (a)


11/196

Hence the various branch currents are,

Branch Current Direction

AB 39 A from A to B

BC – 21 A from C to B

CD 39 A from C to D

DE – 81 A from E to D

EF – 11 A from F to E

FA – 41 A from A to F

Example 1.5 Find the value of R and the current flowing through it in the network shown in

the Fig. 1.12, when the current in the branch OA is zero.

VTU : Oct.-85; Mar.-94

Solution : Step 1 : The circuit diagram is given.

Step 2 : Mark all the branch currents

Step 3 : Mark all the voltage polarities

The step 2 and 3 are combined and shown in the

Fig. 1.12 (a).

Step 4 : Apply KVL to various loops.

Loop AOCA, 1.5 I I I R 01 2 2 = 0 i.e. 1.5I I 1.5 R1 2 = 0 …(1)Loop AOBA, 0 I 1 4 I I2 1 2 = 0 i.e. 4 I 5 I1 2 = 0 …(2)Loop BOCB I 1 I R 1.5 I 102 2 1 = 0 i.e. 1.5I I 1 R1 2 = –10 …(3)

From equation (2), I1 = 5

4 I 2 = 1.25 I 2 …(4)



O

1.5

R

1 . 5

4

1

10 VB C

A

0 A

Fig. 1.12

O

1.5

R

1 . 5 4

1

10 VB C

A

0 A

2

2 2

1

1 2

1 2 1 2

1

Fig. 1.12 (a)


12/196

Substituting in equation (1) we get,

1.5 1.25 I I 1.5 R2 2 = 0 i.e. 1.875 I I 1.5 R2 2 = 0

1.875 I 2 = I 1.5 R2 i.e. 1.5 R = 1.875 R = 0.375

Substituting in equation (3) we get,

1.5 1.25 I I 1 0.3752 2 = –10 i.e. 3.25 I 2 = –10

I 2 = + 3.0769 A … Current through R

Example 1.6 Find the V CE and V AG for the circuit shown in Fig. 1.13.

Solution : Assume the two currents as shown in the Fig. 1.13 (a)

Applying KVL to the two loops,

6 I 5 I 9 I 201 1 1 = 0 and 8 I 5 I 7 I 402 2 2 = 0

I 1 = 1 A and I 2 = 2 Ai) Trace the path C-E, VCE = – 5 V = 5 V with C negativeii) Trace the path A-G,

VAG = 30 V with A positive



5

6

9

20 V

8

40 V5

7

10 V

A B E F

D C H G

+

–

+

–+ –

Fig. 1.13

5

6

9

20 V

8

40 V5

7

10 V

A B E F

D C H G

I1

I1

+

–

+

–

+ –

– +

+

–

I1 I2

– +

I2

I2

+ –

I3

+

–

+

–

Fig. 1.13 (a)

C EHB

– + + –

5

5 V

(5 )I1 (5 )I210 V 10 V

5

– +

Fig. 1.13 (b)


13/196

Example 1.7 Find the current in the branch A - B in the d.c. circuit shown in the Fig. 1.14,

using Kirchhoff's laws.

Solution : The various branch currents are shown in the Fig. 1.14 (a).

Applying KVL to loop ADBA

–I2 – (I2 – 5) + I1 = 0

I1 – 2I2 = –5 … (1)

Applying KVL to the loop ACBA,

– (16 – I1 – I2) – (12 – I1 – I2) + I1 = 0



1 1

1

1

1

A

B

4 A

7A

16 A

5 A

Fig. 1.14

1 1

1

1

1

4

7A

16 A

5 A

16 – –I I1 2I2

I2 –5

I2 –5

I1

I1

I

2 16 – –I I

1 2D

B

A

C

(16 – – – 4) = (12 – – )I I I I1 2 1 2

+

+

–

+

++

– –

– –

Applying KCL atvarious nodes

Fig. 1.14 (a)


14/196

– 16 + I1 + I2 – 12 + I1 + I2 + I1 = 0

3I1 + 2I2 = 28 … (2)

Add (1) and (2), 4I1 = 23

I1 = 5.75 A ... This is the current through branch AB.

Example 1.8 Find the value of 'R' so that 1 A would flow in it, for the network in the Fig. 1.15.

Solution : The various branch currents are

shown in the Fig. 1.15 (a).

Loop ABGH, – I2 – 12 – 10 I1 = 0

i.e. 10 I1 + I2 = – 12 … (1)

Loop BCEFGB,– 6 (I1 – I2 – 1) – 2 (I1 – I2) + 12 + I2 = 0

i.e. – 8 I1 + 9 I2 = – 18 … (2)

Solving, I1 = – 0.9183 A and I2 = – 2.8163 A

Current through 6 = I1 – I2 – 1 = 0.898 A

Drop across 6 = 6 Current through 6 = 6 0.898 = 5.388 V

Same is drop across R = R 1 = 5.388 as

R = 5.388 Example 1.9 For the circuit shown below, find the current

through each of the three resistors.

VTU : Dec.-12, Marks 10

Solution : Step 1 : Show the branch currents.

Step 2 : Apply KVL to the two loops.

–20I 20 – 100(I I )1 1 2 = 0



1

10

6

2

+

–

12 V

R 1A

Fig. 1.15

DC

E

I I1 2

–I I

1 2 –

I2

I1

I1

I1 2

– I

2

R

F

6 –1

1

1

+

– 12 V

G

B A

H

10

+ –

+ –

–

+

+

–

+

–

Fig. 1.15 (a)

100

20 V

60 V10

20

Fig. 1.16


15/196

i.e. 120I 100I1 2 = 20 … (1)

100(I I ) – 60 10I1 2 2 = 0

i.e. 100I 110I1 2 = 60 … (2)

Step 3 : Solving, I 1 = – 1.1875 A, I 2 = 1.625 A

Current through 10 = I 2 = 1.625 A

Current through 100 = I + I1 2 = 0.4375 A

Current through 20 = I 1 = – 1.1875 A i.e 1.1875 A

Example 1.10 Find the current in the 8 resistor in the following circuit using Kirchoff's laws.

VTU : May-13, Marks 8

Solution : Step 1 : Show the branch currents.

Step 2 : Apply KVL to loop I and II.

–5I – 15 – 8I – 5I – 10I 151 2 1 1 = 0 i.e. 20I 8I1 2 = 0 … (1)

–10(I – I ) 25 – 10(I – I ) 8I 151 2 1 2 2 = 0 i.e. –20I 28I1 2 = – 40 … (2)

Step 3 : Solving equation (1) and (2),

I 1 = 0.444 A, I 2 = – 1.111 A

Current through 8 = – 1.111 A i.e. 1.111 A



10

100

20 V20 – +

–+

–+

– +

I1

I260 V

I1

I2

I

II

– +

I I1 2+

Fig. 1.16 (a)

5

5

10 8

15 V

10

10

15 V25 V

Fig. 1.17

I2

I2

+

–

–

+

+ –

+ –

+

–

10

15 V25 V

10 5

+ –

10 5

I1

I1

15 V

8

I1 (I1 – )I2

I1 (I1 – )I2

+ –

+ –

I III2

+ –

Fig. 1.17 (a)


16/196

Example 1.11 A 20 V battery with an internal resistance of 5 is connected to a resistance of

‘x' . If an additional 6 resistance is connected across the battery, find the value of ‘x' so

that the power supplied externally by the battery remains the same. VTU : Aug.-95

Solution : Case i] Consider the resistance x alone as

shown in the Fig. 1.18 (a) .

Now I = 20

5 x

P = I R2 = 205

2

x

x( ) …(1)

This is power supplied by battery, to x .Case ii] Now 6 resistance is connected in addition to x as shown in the Fig. 1.18 (b).

Combining x and 6 which are in parallel we get circuit as shown in the Fig. 1.18 (c).

Now I =

20

5 6 || x =

20

5 6x6 x

= 20 6 x

30 11x

Hence power supplied by battery is,

P =

20 6 x

30 11x 6 || x

2

as P = I R

2

P = 20 6 x

30 11 x

6 x

6 x

2

…(2)

Power supplied must remain same, so equating equations (1) and (2),

20

5 x x2

=

20 6 x

30 11 x

6 x

6 x

2

400 x

5 x 2=

400 6 x 6 x

30 11x 2

i.e. 30 11 x 2 = 6 x 6 x 5 x 2

900 660 x 121 x2 = 36 6 x 25 10 x x2 900 660 x 121 x2 = 900 360 x 36 x 150 x 60 x 6 x2 2 3



Parallel

5

x

6

20 V

5

(6 || x)

20 V

I

I

I

Fig. 1.18 (b) Fig. 1.18 (c)

5

x

20 V

I

I

I

Series

Fig. 1.18 (a)


17/196

6 x 25 x 150 x3 2 = 0 i.e. x 6 x 25 x 1502 = 0The value of x cannot be zero.

x =

25 25 4 6 150

2 6

2

x = 25 6512 = 7.5 ...Neglecting negative value

Example 1.12 A circuit consists of two parallel resistors having resistance of 20 and 30

respectively connected in series with 15 . If current through 15 resistor is 3 A, Find :

i) Current in 20 and 30 resistors ii) The voltage across the whole circuit iii) The total

power and power consumed in all resistances.

Solution : The arrangement is shown in the Fig. 1.19.

Total current I = 3 A

Req = 20 30|| + 15 = 20 3020 30

15

= 27

I = VReq

i.e. 3 = V27

V = 81 V ... Voltage across each circuit

I 1 = I 3020 30

3 3

5

= 1.8 A ... Current through 20

I 2 = I 2020 30 3 25 = 1.2 A ... Current through 30

P = V I 81 3 = 243 W ... Total power

P20 = I 20 1.8 2012 2 = 64.8 W

P30 = I 30 1.2 3022 2 = 43.2 W

P15 = I 15 3 152 2 = 135 W

Cross check is P = P P P20 30 15

Example 1.13 A particular battery when loaded by a resistance of 50 gives the terminal

voltage of 48.6 V. If the load resistance is increased to 100 , the terminal voltage is

observed to be 49.2 V.

Determine, i) E.M.F. of battery ii) Internal resistance of battery

Also calculate the load resistance required to be connected to get the terminal voltage of

(49.5V) VTU : Feb.-96



V

30

15

20

I2

I1

I = 3 A

Fig. 1.19


18/196


19/196

Important Multiple Choice Questions with Answers

Resistance

Q.1 The resistance is _________ proportional to length and _________ proportional to areaof cross-section.

a) directly, directly b) directly, inversely

c) inversely, directly d) none of these [Ans. : b]

Q.2 1 joule = _________ calories

a) 0.21 b) 0.24 c) 0.28 d) 0.22 [Ans. : b]

Q.3 A wire of resistance R is stretched to double its length. The new resistance of the wire is

_________. VTU : June-10

a) R

2

b) 2 R c) 4 R d) R

4 [Ans. : c]Q.4 The resistance of a conductor having length l , area of cross section a and resistivity is

given as _____ . VTU : June-13

a) R a

l

b) R

l

a

c) R la d) R

l

a

[Ans. : b]

Q.5 Resistance of a wire always increases if _____ . VTU : June-13

a) temperature is reduced b) temperature is increased

c) number of free electrons available become less

d) number of free electrons available become more [Ans. : b]

Energy Sources

Q.1 A circuit without any energy source is called _________ .

a) passive b) active c) linear D) distributed[Ans. : a]

Q.2 For ideal voltage source, internal resistance is _________ ohms.

a) infinite b) zero c) constant d) none of these[Ans. : b]

Q.3 For ideal current source, internal resistance is _________ ohms.a) infinite b) zero c) constant d) none of these

[Ans. : a]

Q.4 A practical voltage source is represented by _________ . VTU : Dec.-11

a) a resistance in parallel with an ideal voltage source

b) a resistance in series with an ideal current source

c) a resistance in series with an ideal voltage source

d) none of the above [Ans. : c]




20/196

Ohm's Law

Q.1 Validity of Ohm's law requires that the_____.

a) voltage should remain constant b) current should remain constant

c) resistance must remain constant d) power must remain constant

VTU : Feb.-09

[Ans. : b]

Q.2 For application of Ohm's law, _________ of circuit must remain constant.

a) voltage b) current c) resistance d) inductance[Ans. : c]

Q.3 The Ohm's law cannot be applied to _________.

a) resistance b) inductance c) capacitance d) diode [Ans. : d]

Q.4 The Ohm's law can not be applied to _____ . VTU : Jan.-13

a) resistance b) inductance c) capacitance d) diode [Ans. : d]

Q.5 The condition for the validity under Ohm's law is that the _____ . VTU : June-13

a) temperature should remain constant

b) current should be proportional to voltage

c) resistance must be wire wound type

d) all of the above [Ans. : a]

Q.6 A linear resistor is one which obeys _____ . VTU : June-13

a) Ampere's law b) Lenz's law c) Ohms law d) Kirchhoff's law

[Ans. : c]

Series Circuit

Q.1 A series circuit consists of 4.7 k 5.6 k 9 k , , and 10 k resistors. Which resistor hasthe most voltage across it ?

a) 4.7 k b) 5.6 k c) 9 k d) 10 k

VTU : Aug.-09

[Ans. : d]

Q.2 In a series circuit, _________ remains same.

a) current b) voltage c) resistance d) none of these

[Ans. : a]

Q.3 In a series circuit, the equivalent resistance is ___ of all the individual resistances.

a) smallest b) same as c) largest d) none of these

[Ans. : c]




21/196

Q.4 The voltage drop across 8 resistance is ___ .

a) 100 V b) 80 V c) 220 V d) 120 V [Ans. : d]

Q.5 The voltage across the short circuit is _________ .

a) infinite b) one c) zero d) none of these[Ans. : c]

Parallel Circuit

Q.1 The load increases means load resistance _________.

a) increases b) decreases c) remains constant d) none of these[Ans. : b]

Q.2 In a parallel circuit, _________ remains same.

a) current b) voltage c) resistance d) none of these

[Ans. : b]

Q.3 In a parallel circuit, the equivalent resistance is _________ of all the individual

resistances.

a) smallest b) same as c) largest d) none of these

[Ans. : a]Q.4 If 20 resistances, each of 1 are connected in parallel then the equivalent resistance is

_________ .

a) 0.01 b) 0.05 c) 20 d) 0.02 [Ans. : b]

Q.5 The total current drawn by the circuit shown from the supply is _________ .

a) 10 A b) 5 A c) 1 A d) 15 A [Ans. : d]

Q.6 If the 3 resistance is removed from the circuit shown in the Fig. 1.21 the current drawnby the circuit is _________ .

a) 14.285 A b) 9.185 A c) 2 A d) 1.185 A [Ans. : a]



1 8 4 2

+ –225 V

2 6 150 V

9

3

Fig. 1.21


22/196

Q.7 The voltage across the parallel circuit shown in the Fig. 1.22 is _______ .

a) 1.8 V b) 4.8 V c) 8.4 V d) 2.8 V [Ans. : b]

Q.8 Two resistors R1 and R2 give combined resistance of 4.5 when in series and 1 whenin parallel, the resistances are _________ . VTU : Dec.-11

a) 2 and 2.5 b) 1 and 3.5 c) 1.5 and 3 d) 4 and 0.5 .

[Ans. : c]

Q.9 The total resistance of parallel circuit is ___________.

a) less than the smallest resistance b) more than the smallest resistance

c) more than the highest resistance d) none of these VTU : June-12

[Ans. : a]

Kirchhoff's Laws

Q.1 The algebraic sum of all the currents at a junction point is always zero is the statement of

_________ law.

a) KVL b) Lenz's c) Faraday's d) KCL [Ans. : d]

Q.2 The Fig. 1.23 shows a part of a closed electrical circuit. The potential drop between A

and B is _________ . VTU : June-10

a) 18 V b) – 18 V c) 4 V d) – 4 V [Ans. : a]

Q.3 Kirchhoff's voltage law applies to circuit with _________. VTU : Dec.-11

a) linear elements only b) non-linear elements only

c) linear, non-linear, active and passive elements

d) linear, non-linear, active, passive, time varying as well as time invariant elements.

[Ans. : a]



1

2

3

4

10 A

Fig. 1.22

4 A

1 A B

2

6 V

Fig. 1.23


23/196

Q.4 The polarity of voltage drop across a resistor is determined by _________.

VTU : Jan.-14

a) the value of resistor b) the value of current

c) direction of current in resistor d) the polarity of source [Ans. : c]

Electrical Power

Q.1 If 100 V is applied across a 200 V, 100 W bulb, the power consumed will be,_____ .

a) 100 W b) 50 W c) 25 W d) 12.5 W VTU : Feb.-09

[Ans. : c]

Q.2 The power dissipation in each of three parallel branches is 1 W. The total power

dissipation of the circuit is _________ . VTU : Aug.-09

a) 1 W b) 4 W c) 3 W d) 9 W [Ans. : c]

Q.3 The resistance of a 200 W, 250 V lamp is ________ . VTU : Feb.-10

a) 625 b) 1250 c) 3125. d) 3125. [Ans. : c]

Q.4 The voltage applied across an electric iron is halved. The power consumption of the iron

reduces to _________ . VTU : Feb.-10

a) one half b) three fourth c) one fourth d) 0.707 times. [Ans. : c]

Q.5 Which of the following statements is true both for a series and a parallel circuit ?

VTU : Feb.-10

a) Resistances are additive b) Powers are additive

c) Currents are additive d) Voltage drops are additive [Ans. : b]

Q.6 Refer to the Fig. 1.24. Which of the following statement is true ?

a) Lamp 1 will be less brighter than Lamp 2.

b) Lamp 1 will be more brighter than Lamp 2.

c) Both the lamp will glow with equal brightness.

d) None of the above. [Ans. : b]

Q.7 The practical unit of electrical energy is _____ .

VTU : Jan-13

a) kWh b) Wh c) Watt -second d) Joule second [Ans. : a]

Q.8 Three resistors of 4 , 6 and 9 are connected in parallel in a network. Maximumpower will be consumed by ________ . VTU : Jan.-14

a) 4 resistor b) 6 resistor c) 9 resistor d) all resistor [Ans. : a]

Electrical Energy

Q.1 The practical unit of electrical energy is _________ . VTU : Feb.-10

a) kW-hr b) watt-hr c) watt-second d) joule-second [Ans. : a]



Lamp 160 W, 240 V

Lamp 2100 W, 240 V

240 V

Fig. 1.24


24/196

Q.2 A 2 resistor is connected in series with parallel combination of 10 and 15 resistors.Then heat dissipated in kWsec for 1 hour in circuit, when current of 2 A flowing in 2 resistor is _________.

a) 115.2 b) 1.152 c) 11.52 d) 115200 [Ans. : a]

Q.3 Energy consumed by a heater of rating 1000 W by operating it for a period of

2 hours will be _________ . VTU : Dec.-11a) 1 kWh b) 2 kWh c) 2.5 kWh d) 4 kWh [Ans. : b]

Current Division in Parallel Circuit of Resistors

Q.1 In a circuit shown, the current through 5 resistance is _________ .

a) 15 A b) 20 A c) 25 A d) 4 A [Ans. : b]

Q.2 The current drawn by the resistance of 8 in the circuit shown is _________ .

a) 1.555 A b) 2.555 A c) 5.5555 A d) 4.5555 A [Ans. : c]

Q.3 The current in 5 ohm resistor is ___________. VTU : June - 12

a) 2 A b) 3 A c) 1 A d) 1.5 A [Ans. : a]

Concept of the Terminal Voltage of a Cell

Q.1 The practical voltage source has e.m.f. of E volts and internal resistance is r ohms. If itsupplies a load current of I amperes, the terminal voltage is _________.

a) E b) E

r

c) Ir d) E - Ir

[Ans. : d]

Q.2 A practical voltage source of 100 V is connected across 10 resistance and the terminalvoltage across 10 is found to be 80 V, then the internal resistance of the source is

_________ .

a) 2.5 b) 10 c) 5 d) 8 [Ans. : a]



+ _ 100 V 20 5

12 8

10

100 V

+

–

10

5

3 A


25/196

Chapter at a Glance

1. Faraday's Laws of Electromagnetic Induction

e = Nd

dt

volts

2. Magnetic Flux and Magnetic Flux Density

1 weber = 108

lines of force.

B =

aWb

m2or Tesla

3. Magnetic Field Strength

H = Ampere turns

Length =

NIAT/ m

l

4. Magnetomotive Force (M.M.F.)

m. m. f. = N I ampere turns

5. Reluctance and Permeance

S = l a

= l

0 r a A/Wb

Permeance = 1

Reluctance

6. Permeability

= B

H i.e. B = H


2 Electromagnetism

(2 - 1)


26/196

0 = B

H in vacuum = 4 10 7 H/m

= 0 r H / m

7. Relation between Flux, M.M.F. and Reluctance

Flux = M.M.F.

Reluctance =

F

S

8. Nature of the Induced E.M.F.

e = B l v sin volts

9. Force on a Current Carrying Conductor in a Magnetic Field

F = B I sinl newtons

10. Self Inductance

L = N

I

e = LdI

dt volts

L = N NII S

= NS

2 henries

L = N a2

l =

N a2 rl

henries … S = l a

11. Mutually Induced E.M.F.

M =

N

I2 2

1

henries and e 2 = Md I

d t1 volts

M = N

I2 2

1

M = N K

I2 1 1

1

Basic Electrical Engineering 2 - 2 Electromagnetism



27/196

M = K N N

S1 1 2

M = N N

S1 2 … For K1 = 1

M =

N N

a

1 2

l

= N N a1 2

l =

N N a1 2 0 r l … S =

l a

M =

K N N

S2 1 2

M = N N

S1 2 … For K2 = 1

12. Coefficient of Coupling or Magnetic Coupling Coefficient

K = M

L L1 2

13. Energy Stored in a Magnetic Field

E = 1

2

L I2 joules


State and explain the Faraday's laws of electromagnetic induction. July-03, 04, Marks 5; July-06, Marks 8; July-07, June-13, Marks 6

1. First Law : Whenever the number of magnetic lines of force (flux) linking with a

coil or circuit changes, an e.m.f. gets induced in that coil or circuit.

2. Second Law : The magnitude of the induced e.m.f. is directly proportional to the

rate of change of flux linkages (flux turns of coil).

Flux linkages = Flux Number of turns of coil

Consider a coil having N turns. The initial flux linking with a coil is 1.

Initial flux linkages = N1

In time interval dt, the flux linking with the coil changes from 1 to 2 .

Final flux linkages = N2




28/196

Rate of change of flux linkages = N N

dt

2 1

Now as per the first law, e.m.f. will get induced in the coil and as per second law themagnitude of e.m.f. is proportional to the rate of change of flux linkages.

e

N N

dt

2 1

i.e. e = K

N N

dt

2 1

e = Nd

dt

(d = 2 1 )

Thus such an induced e.m.f. is mathematically expressed alongwith its sign as,

e = Nd

dt

volts

State and explain Lenz's law. VTU : July-03, 04, July-06; July-07, Marks 4

Statement : The direction ofan induced e.m.f. produced by

the electromagnetic induction

is such that it sets up a

current which always opposes

the cause that is responsible

for inducing the e.m.f.

Consider a solenoid asshown in the Fig. 2.1.

According to Lenz's law, thedirection of current due to induced e.m.f. is so as to oppose the cause producing it.

The cause is motion of bar magnet towards the coil.

So e.m.f. will set up a current through coil in such a way that the end of solenoidfacing bar magnet will become N-pole.

Hence two like poles will face each other experiencing force of repulsion which isopposite to the motion of bar magnet as shown in the Fig. 2.1.

If the same bar magnet is moved away from the coil, then induced e.m.f. will set up a

current in the direction which will cause, the end of solenoid facing bar magnet to behave as S-pole.

Because of this, two unlike poles will face each other and there will be force of attraction which is opposite to the direction of movement of magnet.

What is reluctance ? State its units. VTU : Dec.-03, 05, Marks 2

In an electric circuit, current flow is opposed by the resistance of the material, similarlythere is opposition by the material to the flow of flux which is called reluctance.



NS

G

Coil

Repulsive force due toinduced e.m.f. andcurrent in coil

Direction of motion

Bar magnet

NN SS

Fig. 2.1 Lenz's law


29/196

It is defined as the resistance offered by the material to the flow of magnetic fluxthrough it. It is denoted by 'S'.

It is measured in amperes per weber (A/Wb) or ampere-turns per weber (AT/Wb).

State Fleming's right hand rule. Mention its application.

VTU : Jan.-03, July-03,04; Feb.-05; July-04,11, Dec.-11, Marks 3

The direction of dynamically induced e.m.f. is given by Fleming's right hand rule.

If three fingers of a right hand, namely thumb, index finger and middle finger areoutstretched so that everyone of them is at right angles with the remaining two, and if

in this position index finger is made to point in the direction of lines of flux, thumb in

the direction of the relative motion of the conductor with respect to flux then the

outstretched middle finger gives the direction of the e.m.f. induced in the conductor.

The rule is commonly used in the d.c. generators.

State Fleming's left hand rule. Mention its application. VTU : Jan.-03, 11; July-03, 04,11; Feb.-05, Marks 3

The rule states that, ‘Outstretch the three fingers of the left hand namely the first finger,middle finger and thumb such that they are mutually perpendicular to each other. Now

point the first finger in the direction of magnetic field and the middle finger in the

direction of the current then the thumb gives the direction of the force experienced by

the conductor'.

The rule is commonly used in the d.c. motors.

Define mutual inductance and state its unit. Derive the expression for mutual inductance. VTU : Jan.-03; Feb.-05; July-04, Marks 8

Let N1 = Number of turns of coil A, N2 = Number of turns of coil B

I1 = Current flowing through coil A

1 = Flux produced due to current I1 2 = Flux linking with coil B

According to Faraday's law, the induced e.m.f. in coil B is,

e 2 = N d

dt22

Now 2 = 2

11I

I

Rate of change of 2 = 2

11I

Rate of change of current I

d

d t2 =

21

1

I

dI

dt




30/196

e 2 = N22

1

1

I

dI

dt =

N

I

dI

dt2 2

1

1

Here N

I2 2

1

is called coefficient of mutual inductance denoted by M.

Coefficient of mutual inductance is defined as the property by which e.m.f. getsinduced in the second coil because of change in current through first coil. It is also

defined as the flux linkages of the coil per ampere current in other coil.

Derive an expression for the energy stored in a magnetic field. Jan.-03, 04, 06,13, 14, Marks 6; July-04, 06, 09, 12, 13; Feb.-05, Marks 8

Let the induced e.m.f. in a coil be,

e = LdI

dt

This opposes a supply voltage. So supply voltage ‘V' supplies energy to overcome this,which ultimately gets stored in the magnetic field.

V = – e =

LdI

dt = L

dI

dt

Power supplied = V I = LdI

dt I

Energy supplied in time dt is,

E = Power Time = LdI

dt I dt

= L dI I joules . Integrating above, the total energy stored is,

E = L dI I

0

I

= L dI I0

I

= L I

2

2

0

I

= L

I

2 0

2

E = 1

2L I2 joules

Important Solved Examples

Example 2.1 A coil is wound uniformly with 300 turns over a steel of relative permiability 900,

having a mean circumference of 40 mm and corss-sectional area of 50 mm 2. If a current of 5

A is passed through the coil, find

i) m.m.f. ii) reluctance of the ring and iii) flux

VTU : Dec.-04

Solution : Given : N = 300, r = 900, l = 40 mm = 40 × 10– 3m,




31/196

a = 50 mm2 = 50 × 10– 6 m2, I = 5 A

i) m.m.f. = NI = 300 × 5 = 1500 AT

ii) S = l 0 r a

= 40 10

4 10 900 50 10

3

7 6

= 70.7355 103 AT/Wb

This is reluctance of the ring.

iii) S = m.m.f.

= m.m.f

S =

1500

707355 10 3. = 21.2057 mWb … Flux

Example 2.2 A ring shaped core is made up of two parts of same material. Part one is a

magnetic path of length 25 cm and with cross sectional area 4 2cm , whereas part two is of

length 10 cm and cross sectional area of 6 2cm . The flux density in part two is 1.5 Tesla. If

the current through the coil, wound over core, is 0.5 Amp., calculate the number of turns of

coil. Assume r is 1000 for material. VTU : Dec.-05

Solution : The arrangement is shown in the

Fig. 2.2.

B2 = a 2

= 1.5 6 10 4 = 9 10 Wb4

Key Point The flux is same through both the parts,

as series circuit.

S = S SI II = l l1

0

2

0 r 1 r 2a a

= 25 10

4 10 1000 4 10

10 10

4 10 1000 6 1

2

7 4

2

7

0 4

= 629988.3164 AT/Wb

= m.m.f S

NIS

9 10 4 = N 0.5629988.3164

N = 1133.979 1134 ... Number of turns



I

a = 4 cm

= 25 cm

1

1

2

l

B = 1.5 T2

a = 6 cm

= 10 cm

2

2

2

l

r

= 1000I

0.5 AN

Fig. 2.2


32/196

Example 2.3 A coil of 300 turns wound on a core of non magnetic material has an inductance of

10 mH. Calculate i) flux produced by a current of 5 A. ii) the average value of the emf

induced when a current of 5 A is reversed in 8 millisecond. VTU : Aug.-03, Marks 5

Solution : N = 300, L = 10 mH, I = 5 A.

i) L = N

I

i.e. =

LIN

= 10 10 5

300

3 = 166.667 Wb

ii) Current is reversed i.e. becomes –5 A in dt = 8 ms.

e = LdIdt

=

10 10

5 5

8 10

33

[ ]= 12.5 V

Key Point dI is change in current i.e. [final current - initial current] which is [–5 – 5] = –10 A.

Example 2.4 Two 200 turns, air cored solenoids, 25 cm long have a cross-sectional area of 3 cm 2

each. The mutual inductance between them is 0.5 H. Find the self inductance of the coils andthe coefficient of coupling. VTU : Jan.-90

Solution : l = 25 cm = 25 10 2 m, a = 3 cm2 = 3 10 4 m2

M = 0.5 H and N1 = N2 = 200The self inductance of both the coils will be same.

L1 = L2 = N

S

2

where S = l

0 a as r = 1 for air core

= 25 10

4 10 3 10

2

7 4

= 6 6314 10 8. AT/Wb

L = ( )

.

200

6 6314 108

2

= 6 031 10 5. H = 60.31 H

Now K = M

L L1 2=

M

L

2=

ML

= 0 5 10

60 31 10

6

6

.

.

= 0.00828

Example 2.5 Two identical 1000 turn coils X and Y lie in parallel planes such that 60 % of the

flux produced by one coil links with the other. A current of 5 A in X produces a flux of

5 10 6 Wb in itself. If the current in X changes from +6 A to –6 A in 0.01 sec, what will

be the magnitude of the e.m.f. induced in Y ? Calculate the self inductance of each coil.

VTU : June-86

Solution : N1 = N2 = 1000, I 1 = 5 A, 1 = 5 10 6 Wb




33/196

Now M = N

I2

1

2

But 2 = 0.6 1 as 60 % flux link with other

M = 0 6. N

I

2

1

= 0 6 1000 5 10

5

6. = 6 10 4 H

So induced e.m.f. in Y will be, e2 = M d I

dt1 =

6 10

6 6

0 014

. = 0.72 V.

As coils are identical, both will have same value of self inductance as,

L1 = L2 = N

I1

1

1 = 1000 5 10

5

6 = 0.001 H = 1 mH

Example 2.6 A magnetic core is in the form of a closed ring of mean length 20 cm and

cross-sectional area 1 cms 2 . Its relative permeability is 2400. A coil of 2000 turns is

uniformly wound around it. Find the flux density set up in the core if a current of 66 mA is

passed through the coil. Find the energy stored in the magnetic field set up.

Find the inductance of the coil, if an air gap of 1 mm is cut in the ring perpendicular to the

direction of the flux. VTU : May-05

Solution : Given l = 20 cm, a = 1 cm2 , r = 2400, N = 2000, I = 66 mA

Case 1 : S = l 0 r a

= 20 10

4 10 2400 1 10

2

7 4

= 663.1455 10 3 AT/Wb

m.m.f = NI = 2000 66 10 3 = 132 AT

= NI

S =

132

663.1455 10 3= 1 9905 10 4. Wb

B =

a =

1.9905 10

1 10

4

4

= 1.9905 Wb/m2 i.e. T … Flux density

L = N

S

2=

(2000)

663.1455 10

2

3= 6.03185 H or L =

NI

E = 12

LI 2 = 12

6.03185 (66 10 )3 2 = 13.1373 mJ … Energy stored

Case 2 : New air gap is cut of length lg = 1 mm in the ring.

li = Iron length = l l g = 20 10 1 102 3 = 0.199 m

S = S Si g = l li

0 r

g

0a a … r = 1 for air gap




34/196

= 1

0 ai

rg

ll

= 1

4 10 1 101 10

7 43

0.1992400

= 8.6175 10 6 AT/Wb …Total reluctance

L = NS

2= (2000)

8.6175 10

2

6= 0.4641 H …New inductance.

Example 2.7 Two coils A and B, have self inductances of 120 H and 300 H respectively. A

current of 1 A through coil 'A' produces flux linkage of 100 Wb turns in coil 'B'. Calculate

i) mutual inductance between the coil.

ii) average e.m.f. induced in coil 'B' if current of 1 A in coil 'A' is reversed at a uniform rate

in 0.1 sec. Also find coefficient of coupling. VTU : Dec.-04

Solution : LA = 120 H, LB = 300 H

IA = 1 A produces NB B = 100 Wb

i) M = N

IB B

A

=

100 101

6 = 100 H …Mutual inductance

ii) eB = M dI

dtA

The current in coil A is reversed i.e. it is –1 A in 0.1 sec.

I = (New value – Original value) = (–1 –1) = –2 A

and t = 0.1 sec

dI

dtA =

It

= 20.1

= – 20 A/sec

eB = 100 10 206 ( ) = 2 mV … Induced e.m.f. in B

K = M

L LA B=

100 10

120 10 300 10

6

6 6

= 0.527 … Coefficient of coupling

Example 2.8 Two identical coils P and Q, each with 1500 turns, are placed in parallel planes

near to each other, so that 70% of the flux produced by current in coil P links with coil Q. If

a current of 4 A is passed through any one coil, it produces a flux of 0.04 mWb linking with

itself. Find the self inductances of the two coils, the mutual inductance and coefficient of

coupling between them. VTU : Dec.-03

Solution : N NP Q = 1500, Q P0.7 ... 70 % linking

Let I P = 4 A and P = 0.04 mWb




35/196

LP = N

I1500 0.04 10

4P P

P

3

= 15 mH

Let I Q = 4 A then Q = 0.04 mWb

LQ =N

I1500 0.04 10

4Q Q

Q

3

= 15 mH

M =N

I

N 0.7

IQ Q

P

Q P

P

=

1500 0.7 0.04 104

3 = 10.5 mH

And K =

M

L L

10.5 10

15 10P Q

3

3 2

= 0.7

Example 2.9 If a current of 5 A flowing in coil with 1000 turns wound on a ring of

ferromagnetic material produces a flux of 0.5 mWb in the ring. Calculate i) self inductance of

coil ii) e.m.f. induced in the coil when current is switched off and reaches zero value in2 millisec. iii) mutual inductance between the coils, if a second coil with 750 turns is wound

uniformly over the first one. VTU : May-03

Solution : = 0.5 mWb, N = 1000, I = 5 A

i) L = N

I

= 1000 05 10

5

3 .= 0.1 H

ii) e =

L

dIdt

0.1 0 5

2 10 3 250 V

iii) Let N2 = 750 of other coil

As other coil is wound on first, all the flux produced by coil 1 links with the second coil.

2 = K1 1 1 as K1 = 1

M = N

I2 2

1

=

N KI

2 1 1

1

=

750 0 5 105

3

.0.075 H

Example 2.10 Two windings connected in series are wound on a ferromagnetic ring having

cross-sectional area of 750 mm2 and a mean diameter of 175 mm. The two windings have

250 and 750 turns, while the relative permeability of material is 1500. Assuming no leakage of

flux, calculate the self inductances of each winding and the mutual inductance as well.

Calculate e.m.f. induced in coil 2 if current is coil 1 in increased uniformly from zero to 5 A

in 0.01 sec. VTU : Dec.-01

Solution : l = Length of magnetic circuit = dmean

l = 175 10 3 = 0.5497 m

a = 750 mm 750 10 m 7.5 10 m2 6 2 4 2




36/196

N1 = 250, N2 = 750, r = 1500

Self inductance,L = N

I

but =

N IS

L = N NI

I SN

S

2

We have, S = l a

S = l

r 0 a =

0 5497

4 10 1500 7 5 107 4

.

.

= 388833.2 AT/Wb

L1 = N

S

250

388833.212 2

= 0.1607 H

L2 = N

S

750

388833.222 2

= 1.4466 H

The mutual inductance between the two windings is given by,

M = N N

S1 2 =

250 750388833 20.

= 0.4822 H

M = 0.4822 H

E.M.F. induced in coil 2 is,

e2 = MdI

dt1 =

0 4822

5 0

0 01.

. = –241.1 V

Example 2.11 The winding of an electromagnet is wound with 96 turns and has a resistance of

50 ohms. The exciting voltage is 250 V and the flux linking the coil is 5 mWb. Find the

energy stored in the magnetic field. Then if the current is reversed in 0.1 sec, what emf is

induced in the coil ? VTU : Aug.-05, Marks 6

Solution : N = 96, R = 50 , V = 250 V, = 5 mWb.

I = V

R =

25050

= 5 A

L = N

I

= 96 5 10

5

3 = 96 mH

W = 1

2 LI 2 =

12

96 10 53 2 ( ) = 1.2 J

The current is reversed i.e. Ifinal = – 5 A, dt = 0.1 s




37/196

e = LdIdt

=

96 10 3

I I

dtfinal initial

= 96 10 [ 5 5]

0.1

3= 9.6 V

Example 2.12 An iron ring of 10 cm in diameter and 8 cm2

in cross-section is wound with300 turns of wire. For a flux density of 1.2 Wb/m

2and relative permeability of 500, find the

exciting current, the inductance and the energy stored. VTU : May-07

Solution : d = 10 cm, a = 8 cm2, N = 300, B = 1.2 Wb/m2, r = 500

l = d = 10 cm = 0.3141 m

S = l 0 r a

= 0 3141

4 10 500 8 107 4.

– – = 624 882 10 3. AT/Wb

= B a = 1 2 8 10 4. – = 9 6 10 4. – Wb

= NI

S

9 6 10 4. – = 300 I

624.882 10 3

I = 2 A

L = N

S

2=

( )

.

300

624 882 10

2

3= 0.14402 H

E = 12

LI2 = 12

014402 2 2 . ( ) = 0.288 J

Example 2.13 An air cored solenoid 1 m in length and 10 cm in diameter has 5,000 turns.

Calculate : i) the self inductance and ii) the energy stored in the magnetic field when current

of 2 A flows in solenoid. VTU : Dec.-06

Solution : l = 1 m, d = 10 cm, N = 5000, = 0 as air cored

a =

42d =

100

42 cm = 7 854 10 3 2. – m

S = l 0 a = 1

4 10 7 854 107 3 – –.= 101 3209 106. AT Wb

i) L = N

S

2=

(5000)

101.3209 10

2

6= 0.2467 H

ii) I = 2 A

E = 1

22LI =

12

0 2467 2 2 . = 0.4934 J




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Important Multiple Choice Questions with Answers

Magnetic Field and Magnetic Lines of Force

Q.1 The region around the magnet in which magnetic influence can be experienced is called _______ .

a) flux b) line of force c) strength d) magnetic field

[Ans. : d]

Q.2 The direction of flux internal to the magnet is from_____.

a) N-pole to S-pole b) S-pole to N-pole

c) circular d) none of the above [Ans. : d]

Magnetic Flux and Magnetic Flux Density

Q.1 1 weber = ____ lines of force.

a) 108

b) 106

c) 103

d)108

[Ans. : a]

Q.2 The unit of flux density is _____ .

a) weber b) AT c) tesla d) none of these

[Ans. : c]

Magnetic Field Strength

Q.1 The magnetic field strength H is given by_____ .

a) N

I

lb)

I

N

lc)

NI

l d) N I l

[Ans. : c]

Q.2 The unit of magnetic field strength H is_____ .

a) weber b) AT c) tesla d) AT/m [Ans. : d]

Reluctance and Permeance

Q.1 The S.I. unit of reluctance is________ .

a) AT/Wb b) AT c) AT/m d) Wb/AT [Ans. : a]

Q.2 The reluctance is _____ to the relative permeability of the magnetic circuit.

a) inversely proportional b) directly proportional

c) not dependent d) none of the above [Ans. : a]

Q.3 The reluctance in a magnetic circuit is analogous to ____ in an electric circuit.

a) voltage b) current c) resistance d) e.m.f. [Ans. : c]

Q.4 The _______ is reciprocal of the reluctance.

a) permeability b) permeance c) susceptance d) resistance[Ans. : b]




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Permeability

Q.1 The unit of permeability is _______ .

a) H/m b) Hm c) weber d) tesla [Ans. : a]

Q.2 Higher value of

r means _______ .

a) difficult to pass the flux, through the material

b) easy to pass the flux, through the material c) none of these [Ans. : b]

Q.3 For a magnetic material, the value of r is _______ .

a) low b) zero c) high d) one [Ans. : c]

Q.4 For free space, the value of r is_______ .

a) low b) zero c) high d) one [Ans. : d]

Magnetic Field due to Circular Conductor i.e. Solenoid

Q.1 The direction of magnetic field due to straight current carrying conductor is given by___ .

a) Fleming's left hand rule b) Right hand thumb rule

c) Fleming's right hand rule d) None of the above [Ans. : b]

Q.2 According to right hand thumb rule applied to solenoid, the thumb points in the direction

of _____ .

a) S pole b) N pole c) positive of battery d) none of these[Ans. : b]

Q.3 A coil wound around a core to produce a magnet is called ______ .

a) pole b) transformer c) solenoid d) none of these[Ans. : c]

Q.4 An electromagnet with a circular core is called ______ .

a) compensating winding b) transformer c) solenoid d) toroid [Ans. : d]

Magnetic Circuit

Q.1 The reluctance in a magnetic circuit is analogous to ______ in an electric circuit.

a) voltage b) current c) resistance d) e.m.f. [Ans. : c]

Q.2 The flux is analogous to _______ of the electric circuit.

a) voltage b) current c) resistance d) e.m.f. [Ans. : b]

Q.3 The m.m.f. is analogous to ______ of the electric circuit.

a) voltage b) current c) resistance d) e.m.f. [Ans. : d]

Faraday's Laws of Electromagnetic Induction

Q.1 The e.m.f. is obtained from the magnetic flux by________ .

a) Ampere b) Faraday c) Oersted d) Ohms [Ans. : b]




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Q.2 A coil of 2000 turns, produces a flux of 1 mWb. The flux is reversed in 0.1 sec then

e.m.f. induced is ________ V.

a) 40 b) 20 c) 40 d) 60 [Ans. : c]

Q.3 The e.m.f. induced in a coil of N turns is _______ .

a) N

d

dt

b) N

d

di

c) – N

d

dt

d) L

d

di

[Ans. : c]

Q.4 The law that finds application in electrolysis _____ . VTU : June-13

a) Faraday's law b) Coulomb's law c) Ohm's law d) Lenz's law[Ans. : a]

Q.5 According to Faraday's law of electromagnetic induction an emf is induced in a conductor

whenever it _____ . VTU : June-13

a) lies in a magnetic field b) lies perpendicular to the magnetic field

c) cuts the magnetic flux d) moves parallel to the direction of magnetic field.[Ans. : c]

Lenz's law

Q.1 According to ________ the induced e.m.f. opposes the cause producing it.

a) Ohm's law b) Faraday's law c) Kirchhoff's law d) Lenz's law[Ans. : d]

Q.2 "In all cases of electromagnetic induction, an induced voltage will cause a current to flow

in a closed circuit in such a direction that the magnetic field which is caused by that

current will oppose the change that produces the current" is the original statement of

_____ . VTU : June-13

a) Lenz's law b) Faraday's law of magnetic induction

c) Fleming's law of induction d) Ampere's law [Ans. : a]

Q.3 Which law is synonymous to the occurrence of diamagnetism ? VTU : June-13

a) Ampere's law b) Maxwell's law c) Coulomb's law d) Lenz's law [Ans. : d]

Nature of the Induced E.M.F.

Q.1 When e.m.f. is induced due to physical movement of the coil then it is called ______ .

a) statically induced e.m.f. b) forcefully induced e.m.f.

c) dynamically induced e.m.f. d) magnetically induced e.m.f. [Ans. : c]

Q.2 The dynamically induced e.m.f. can be found in _______ .

a) generator b) transformer c) bulb d) none of these[Ans. : a]

Q.3 The magnitude of statically induced e.m.f. depends on _____. VTU : Jan.-14

a) the coil resistance b) the flux magnitude

c) the rate of change of flux d) all of these [Ans. : c]




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Self Inductance

Q.1 An e.m.f. of 7.2 volts is induced in a coil of 6 mH. Then the rate of change of current is :

_______ . VTU : Jan.-09

[Ans. : c]

a) 12 A/s b) 120 A/s c) 1200 A/s d) 12000 A/s.

Q.2 The self inductance L is given by _______ . VTU : Jan.-13

a) N I b) NI

c)

N

I

d)

I

N [Ans. : c]

Q.3 The self inductance L is _______ number of turns.

a) directly proportional to square of b) inversely proportional to square of

c) directly proportional to d) none of the above [Ans. : a]

Q.4 A current of 20 A is reversed in 0.1 sec through an inductor of 1 H then e.m.f. induced is

_____ volts. VTU : Jan.-13

a) 200 b) 200 c) 400 d) 400 [Ans. : d]

Q.5 Inductance opposes _______ in current in a circuit. VTU : June-12

a) only increase b) only decrease c) change d) none of these [Ans. : c]

Coefficient of Coupling or Magnetic

Coupling Coefficient

Q.1 The maximum value of coefficient of coupling is ________ . VTU : July-09, Jan.-11

a) 100 % b) more than 100 % c) 90 % d) none of these[Ans. : a]

Q.2 If the entire flux produced by one coil links with the other then its coefficient of coupling

is ________ .

a) zero b) unity c) very high d) 0.5 [Ans. : b]

Q.3 The unit of coefficient of coupling is ______ .

a) amperes b) AT/Wb c) unitless d) H/m [Ans. : c]

Q.4 The flux linkage between the coils is maximum when m = ? VTU : July-11

a) 1 L L1 2 b) L L1 2 c) L L1 2 d) L L1 2. [Ans. : b]

Q.5 If coefficient of coupling between two coils is increased, mutual inductance between the

coils ___________. VTU : June-12

a) is increased b) is decreased

c) remains unchanged d) none of these [Ans. : a]




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Chapter at a Glance

1. E.M.F. Equation of D.C. Generator

E = P Z

60 A

Ne.m.f. equation with A = P for Lap and A = 2 for Wave

2. Types of D.C. Generators

3. Shunt Generator

Ia = IL + Ish


3 D.C. Machines

Separatelyexcited

Self excited

D.C. generators

Shunt Series Compound

Cumulative Differential

Fig. 3.1 Types of d.c. generators

(3 - 1)

G Vt LoadRshR

a

Fig. 3.2

G E

LOAD Vt

+

–

IL

IaIsh

A1F1

A2F2

IL

Fig. 3.3 D.C. shunt generator


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Ish = Vt

shR

E = Vt + IaRa + V brush

4. Series Generator

Ia = Ise= IL where Ise = Current through series field winding.

E = Vt + IaRa + IaR se + V brush

5. Compound Generator

a) Long Shunt :

Ia = Ish + IL

Ish = V

R t

sh

where

Rsh = Resistance of shunt field winding

E = Vt + IaRa + Ia Rse + V brush where Rse = Resistance of series field winding

b) Short Shunt :

Ia = IL + Ish

Ish = E I R

R a a

sh

Basic Electrical Engineering 3 - 2 D.C. Machines


G E

LOAD Vt

+

–

IL

Ise

Ia

Ish

A1

S2

S1F1

A2

F2

IL

Fig. 3.5 Long shunt compound generator

GE

LOAD Vt

+

–

ILI

se

Ia

Ish

A1

S2

S1

F1

A2

F2

Ia

Fig. 3.6 Short shunt compound generator

G E

LOAD Vt

+

–

IL

Ia

A1

S1 S2

A2

Fig. 3.4 Series generator


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Power developed (E b0 Ia0) = Friction, windage and, iron losses.

where E b0 = Back e.m.f. on no load.

and Ia0 = Armature current drawn on no load.

10. Types of D.C. Motors

11. Torque and Speed Equations

T Ia

T Ia …For shunt motors

T Ia I a2 …For series motors

E b N i.e. N E b


Draw the neat sketch representing the cut section of a d.c. machine. Explain the important features of different parts involved there on.

VTU : Jan.-03, July-04, 06, 08, 11 Marks 5; June-10, Marks 8

Fig. 3.8 shows a cross section of typical d.c. machine (See Fig. 3.8 on next page)

It consists of the following parts :

Yoke

a) Functions :

1. It serves the purpose of outermost cover of the d.c. machine.

2. It provides mechanical support to the poles.



D.C. shunt motor D.C. series motor

D.C. motor

D.C. compound motor

Long shunt Short shunt

Fig. 3.7 Types of d.c. motors


47/196


48/196

As it requires a definite shape and size, laminated construction is used. The laminationsof required size and shape are stamped together to get a pole which is then bolted to

the yoke.

Field Winding (F1 - F2)

The field winding is wound on the pole core with a definite direction.a) Functions : To carry current due to which pole core, on which the field winding is

placed behaves as an electromagnet, producing necessary flux.

As it helps in producing the magnetic field i.e. exciting the pole as an electromagnet it is

called field winding or exciting winding.

b) Choice of material : It is made up of aluminium or copper.

Armature

The armature is further divided into two parts namely,

I) Armature core II) Armature winding

I) Armature core : Armature core is cylindrical in shape mounted on the shaft.

a) Functions :

1. Armature core provides house for armature winding i.e. armature conductors.

2. To provide a path of low reluctance to the magnetic flux produced by the field

winding.

b) Choice of material : It is made up of magnetic material like cast iron or cast steel.

II) Armature winding : Armature winding is nothing but the interconnection of the

armature conductors, placed in the slots provided on the armature core periphery.

a) Functions :

1. Generation of e.m.f. takes place in the armature winding in case of generators.

2. To carry the current supplied in case of d.c. motors.

3. To do the useful work in the external circuit.

b) Choice of material : It has to be made up of conducting material, which is copper.

Commutator

The basic nature of e.m.f. induced in the armature conductors is alternating. This needsrectification in case of d.c. generator, which is possible by a device called commutator.

a) Functions :

1. To facilitate the collection of current from the armature conductors.

2. To convert internally developed alternating e.m.f. to unidirectional ( d.c.) e.m.f.

b) Choice of material : It is also made up of copper segments.




49/196

Brushes and Brush Gear

Brushes are stationary and resting on the surface of the commutator.

a) Function : To collect current from commutator and make it available to the stationary

external circuit.

b) Choice of material : Brushes are normally made up of soft material like carbon.Bearings

Ball-bearings are usually used as they are more reliable. For heavy duty machines, roller bearings are preferred.

What is back e.m.f. ? State its significance. VTU : Feb.-05; Jan.-11; July-07, 08, 11, Marks 5

In a d.c. motor, electrical input i.e. the supply voltage is the cause for the armaturecurrent and the motoring action and hence this induced e.m.f. opposes the supply

voltage. This e.m.f. tries to set up a current through the armature which is in theopposite direction to that, which supply voltage is forcing through the conductor.

As this e.m.f. always opposes the supply voltage, it is called back e.m.f.

Significance of Back E.M.F.

Due to the presence of back e.m.f. the d.c. motor becomes a regulating machine.

Back e.m.f. is proportional to speed, E b N.

Back e.m.f. regulates the flow of armature current and it automatically alters thearmature current to meet the load requirement.

At start the speed N of the motor is zero hence the back e.m.f. is also zero.

Derive the expression of armature torque develped in a d.c. motor. VTU : Jan.-03, 05, 09, 13, 14; July-03, 07, Dec.-11, Marks 6

Consider a wheel of radius R meters acted upon by acircumferential force F newtons as shown in the Fig. 3.9.

Angular speed is,

= 2

60

Nrad/sec

Workdone in one revolution is,W = F Distance travelled in one revolution = F 2 R Joules

P = Power developed = Workdone

Time =

F 2 R

Time for 1 rev

=

F R

N

2

60

= ( )F R

N

2

60

P = T Watts

Let Ta be the gross torque developed by the armature of the motor.



F

RotationR

Fig. 3.9


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The gross mechanical power developed in the armature is E b Ia.

Power in armature = Armature torque i.e. E b Ia = Ta 2

60

N

But E b in a motor is given by, E b = P N Z

60 A

P N Z60 A Ia = Ta 2 60 N

Ta = 12

Ia P Z

A = 0.159 Ia .

P Z

A Nm

Explain the various characteristics of d.c. shunt motor. VTU : Jan.-04, 06, 07, 08, 10; Feb.-05; July-05, Marks 5

i) Torque - Armature current characteristics :

For a constant values of Rsh and supplyvoltage V, Ish is also constant and henceflux is also constant.

Ta Ia

As load increases, armature currentincreases, increasing the torque developed

linearly.

On no load Tsh = 0. The current required is

Ia0 on no load to produce Ta0 and henceTsh graph has an intercept of Ia0 on the

current axis.

ii) Speed - Armature current characteristics :

From the speed equation we get,

N V – Ia Ra as is constant.

So as load increases, the armature

current increases and hence drop IaRaalso increases.

Hence for constant supply voltage,V – Ia Ra decreases and hence speed

reduces.

But as Ra is very small, for change in Iafrom no load to full load, drop IaRa is

very small and hence drop in speed is



TTa Tsh

Ta0

0Ia0

Tf Losstorque

a

Fig. 3.10 T Vs Ia for shunt motor

Constant speed line

Ia

N

N0

Fig. 3.11 N Vs Ia for shunt motor


51/196

also not significant from no load to full

load.

iii) Speed - Torque characteristics :

This curve shows that the speed almostremains constant though torque

changes from no load to full load

conditions. This is shown in the

Fig. 3.12.

Explain the various characteristics of d.c. series motor. VTU : Jan.-04, 07, 08, 10; Feb.-05; July-05, Marks 5

i) Torque - Armature current characteristics :

For the series motor the series field winding is carrying the entire armature currenthence,

Ta Ia I a2

Thus torque in case of series motor isproportional to the square of the

armature current. This relation isparabolic in nature as shown in the

Fig. 3.13.

As load increases, armature currentincreases and torque produced increases

proportional to the square of the

armature current upto a certain limit.

Saturation means though the current through the winding increases, the flux producedremains constant. Hence after saturation the characteristics take the shape of straight

line as flux becomes constant, as shown.

ii) Speed - Armature current characteristics

From the speed equation we get,

N E b

V I R a a I RI

a se

aas Ia in case of series motor

The values of Ra and Rse are so small that the effect of change in Ia on speed overridesthe effect of change in V – Ia Ra – Ia Rse on the speed.

Hence in the speed equation, E b V and can be assumed constant.



TTa

2TIa

Ta Tsh

Tf

Point of

saturation

aIa00

Fig. 3.13 T Vs Ia for series motor

Constant speed line

T

N

Fig. 3.12 N Vs T for shunt motor


52/196

So speed equation reduces to,

N 1I a

So speed - armature current characteristicsis rectangular hyperbola type as shown

in the Fig. 3.14.

iii) Speed - Torque characteristics

In case of series motors,

T I a2 and N

1I a

Hence we can write,

N 1

T

Thus as torque increases when loadincreases, the speed decreases.

On no load, torque is very less andhence speed increases to dangerously

high value.

Thus the nature of the speed - torque

characteristics is similar to the nature of the speed - armature current

characteristics.

The speed - torque characteristics of aseries motor is shown in the Fig. 3.15.

State the applications of d.c. shunt and series motors. VTU : Jan.-04, 06, 07, 08, 10; Feb.-05; July-05, Marks 2

Types of motor Characteristics Applications

Shunt Speed is fairly constant andmedium starting torque.

1) Blowers and fans2) Centrifugal and reciprocating pumps3) Lathe machines4) Machine tools5) Milling machines6) Drilling machines



T

N

Fig. 3.15 N Vs T for series motor

Ia

N

0

Fig. 3.14 N Vs Ia for series motor


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Series High starting torque. Noload condition is dangerous.Variable speed.

1) Cranes2) Hoists, Elevators3) Trolleys4) Conveyors5) Electric locomotives

Explain the necessity of starter for a d.c. motor. With a neat sketch explain the working of three point starter for a d.c. motor.

VTU : Jan.-03, 04, 07; July-04, 05, 06, 09; Feb.-05, Marks 8

At the starting instant the speed of the motor is zero, (N = 0). As speed is zero, therecannot be any back e.m.f.

E b at start = 0

The voltage equation of a d.c. motor is, V = E b + Ia Raat start, V = IaRa as E b = 0

Ia = VR a… At start

As armature resistance is very small, the armature current at start is very high.

So at start, motor is showing a tendency to draw an armature current which is 15 to 20times more than the full load current.

Such high current drawn by the armature at start.1. It may affect the performance of the other equipments connected to the same line.

2. Such excessively high armature current, blows out the fuses.3. A large armature current flowing for a longer time may burn the insulation of the

armature winding.

To restrict this high starting armature current, a variable resistance is connected inseries with the armature at start. This resistance is called starter or a starting resistance.

So starter is basically a current limiting device.

The Fig. 3.16 shows three point starter. (See Fig. 3.16 on next page)

The starter is basically a variable resistance, divided into number of sections. Thecontact points of these sections are called studs and brought out separately shown as

OFF, 1, 2, … upto RUN.

There are three main points of this starter :1. ‘L’ Line terminal to be connected to positive of supply.

2. ‘A’ To be connected to the armature winding.

3. ‘F’ To be connected to the field winding.

The OLR and NVC are the two protecting devices of the starter.

Initially the handle is in the OFF position. The d.c. supply to the motor is switched on.



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