Solutions Manual

CHAPTER 1

1.1 (a) 100 = ( )230 2

R; Eq (1.21)

or R = 529 �

(b) I = 230529

= 0.435 A

(c) W = 100 � 8 = 800 Wh

1.2 (i) (b) i = 12

v dt�

Element is inductance; L = 10 2

5�

= 4 H

(c) i = C dvdt

Element is Capacitance; C = 4 F(d) v = Ri

Element is resistance; R = 105

= 2�

(ii) (b) Peak energy = 12

Li2 = 12

� 4 � (5)2 = 50 J

Peak power = 10 � 5 = 50 W

� ����������������� �����

(c) Peak energy = 12

Cv2 = 12

�� 4 � (5)2 = 50 J

Peak power = 10 � 5 = 50 W(d) Energy storage is zero.

Peak power = 10 � 5 = 50 W(iii) 0 < t < 2s

v = 5t, i = 52

t

W1 = vi dt0

2

� = 252

2

0

2 ����� t dt = 33 1

3 J

2 < t < 6s

v = 10 – 104

(t – 2)

W2 = 50 1 14

2 1 14

20

� � � ���

��

��

��� ( ) ( )t t dt

t = 66 2

3 J

W = 3313

+ 66 23

= 100 J

1.3 (a) i = C dvdt

i = 100 � 10 – 6 � ddt

(200 2 sin 314t) = 8.88 cos 314t

(b) 90°

i

v p

314 t

i vleads by 90°

(c) p = vi (see figure)instantaneous power at double frequency (628 rad/s)

Average power = 0

p = 200 2 � 0.888 sin 314t cos 314t

= 100 2 � 0.888 sin 628t = 125.56 sin 628t W

1.4 (a) i = 1L

v dt�= 103

500 � 200 2 � sin 314t dt = – 1.8 cos 314t

i C

v+ –

i L

+ v –

��������������� �

+ + +

–

–+

–+

– –Vs = 4V V1 V244

4

22V Vs – 1

V V1 2–

21

(b) i lags v by 90°

90°

i

v

p

314 t

(c) p = vi = – 200 2 � 1.8 sin 314t cos 314t

= – 180 2 sin 628t WPower oscillates at double frequency 628 rad/s with zero average value

1.5 (a) v = 200 2 sin 314t V

i = 200 21000

sin 314t A

= 2 /5 sin 314t A

(b)

i

pv

314 t

pav

p = 200 2 � 2 /5 sin2 314t

= 40 (1 – cos 628t)Pav = 40 W; Frequency of oscillating component of power = 628 rad/s

1.6 p = i2 R = I02 R e–2R t/L

W = p dt0

�

� = I02 R e Rt L��

� 2

0

/ dt

= 12

LI02

1.7 (a)(b) At node 1

V V V V V V Vs1 1 1 2 1 2

4 2 4 2�

��

��

� = 0

or 32

V1 – 34

V2 = Vs

2(i)

i R

+ –v

i L

+v

–

� ����������������� �����

At node 2

V V V V V2 2 1 2 1

4 4 2�

��

� = 0

or – 34

V1 + V2 = 0 (ii)

(c) 6V1 – 3V2 = 2Vs = 8 (iii)– 3V1 + 4V2 = 0 (iv)

Solving Eqs. (iii) and (iv)

V1 = 3215

V, V2 = 1.6 V

1.8(a) I1 = 4 – 1 = 3 A

V1 = 1 � 3 = 3 V(b) KVL for mesh 1

3 + 1 � 1 + V2 = 0or V2 = 2V (voltage across R3)

I2 = 21

– 1 = 1 A

Voltage drop across R4 = 1 � 1 = 1 V(c) KVL for mesh 2

– 2 – 1 + Vs = 0or Vs = 3 V

1.9 (a)(b) At node 1

V VR

V VR

V VR

s1

1

1 2

2

1 3

4

��

��

� = 0

or 1 1

1 2R R�

���

���

V1 – 1

2R V2 – 1

4R V3 =

VR

s

1(i)

At node 2

– 1

2R V1 + 1 1 1

2 3 5R R R� �

���

���

V2 – 1

3R V3 = Is (ii)

At node 3

– 1

4R V1 – 1

3R V2 + 1 1 1

3 4 6R R R� �

���

���

V3 = 0 (iii)

Observe the symmetry in these equations

+

–

+

– –

+

–

–

–

+ –

+

+ +

VsIs

R5

R4

R1R6

( – )V V1 3

( – )V V1 2

( – )V V1 s

( – )V V2 3

312

V2 V3

+

–4A

( )R1 ( )R3

( )R4( )R2

1 11 2

1A

1 121

I1

I2

Vs

��������������� �

1.10

v6 W2F1His

iL+

–

iL = e–2t

v = 1 � ddt

e–2t = –2e–2t

iC = 2 dvdt

= 2 ddt

(–2e–2t ) = 8e–2t

iR = v2

= – e–2t

Henceis = iL + iC + iR = e–2t + 8e–2t – e–2t = 8e–2t

1.4

v 2 W 4 W1F

1H

is

i+

–

i = sin 2t

vH = 1 � ddt

sin 2t = 2 cos 2t

vR (4�) = 4 sin 2t� v = VH + vR = 2 cos 2t + 4 sin 2t

iR (2�) = v2

= cos 2t + 2 sin 2t

iC (IF) = 1 � dvdt

= – 4 sin 2t + 8 cos 2t

is = i + iR + iC = 9 cos 2t + 2 sin 2t

1.12

V1 V1 V06 W 8 W

4 W

Is

+ +

– –

14

4 168

14 1��

���V + 16 = V1 or V1 = 12 V

Is = 126

+ 214

12� ���

�� = 1 A

� ����������������� �����

1.13 Vab = 12V, Vac = 20 VVbc = Vba + Vac = – 12 + 20 = 8 V

Ibc = 88

= 1 A

Vab = 2 � 1 + V1 = 12 or V1 = 10 VVac = V2 – 1 � 4 = 20 or V2 = 24 V

1.14

V2 W 4 W

+

–

8A4A

I1 I2

V V2 4

� = 4 + 8 = 12 or V = 16 V

I1 = 162

= 8 A, I2 = 164

= 4 A

Power supplied by 4 A source = 16 � 4 = 64 WPower supplied by 8 A source = 16 � 8 = 128 W

1.15

10 W

+

–

i t( ) v t( )100 mF100 mF

5 mH

v = 10 2 sin 314t Vi (10�) = 1.414 sin 314t A

i (100 �F right) = 100 � 10–6 ddt

(10 2 sin 314t)

= 0.444 cos 314t Ai (through as 5 mH) = 1.414 sin 314t + 0.444 cos 314t

= 1.48 sin (314t + 17.4°) A

v (as 5 mH) = 5 � 10–3 ddt

[1.48 sin (314t + 17.4°)]

= 2.32 cos (314t + 17.4°) V

v (across 100 �F left) = 10 2 sin 314t + 2.32 cos (314t + 17.4°)= 13.43 sin 314t + 2.21 cos 314t

= 13.63 sin (314t + 9.3°) V

i (100 �F left) = 100 � 10–6 ddt

13.63 sin (314t + 93°)

��������������� �

= 0.428 cos (314 t + 9.3°) Ai(t) = 1.48 sin (314 t + 17.4°) + 0.428 cos (314 t + 9.3°)

= 1.59 sin (314 t + 32.8°) A

1.16

4 W 4 W

6 W2 W

Vs Is

+

–

A B

N

21

(a) Vs = 12 V, I = 0

VBN = 122 4�

� 4 = 8 V

VBN = VAN or 8 = 4 Is or Is = 2 A(b) Vs = 14 V, I = 1 A

KVL loop 1 – clock wise– 14 + 2 (1 + I1) + 4 I1 = 0

or I1 = 2A

VAN = VAB + VBN

or 2 � 4 = 1 � 6 + (1 + Is) � 4

or Is = – 12

A

1.17Voltage across 3 � resistance = 3 � 1 = 3 V

� V (2 �) = 3 V

I (2 �) = 32

A

I (4/9 �) = 1 + 32

= 52

A

V = 52

49

� + 3 = 4.1 V

CHAPTER 2

2.1 I

V2

+

–

–++

–

25 W

10 W10 V 50 W 20 W

2.5 W

V1

I2Req

3 W2 W

4/9 W

V+

–

1A

� ����������������� �����

1Req

= 110

150

120

� � or Req = 5.88 �

I = 1025 2 5 5 88� �. .

= 0.3 A

V1 = 0.3 � 2.5 = 0.75 VV2 = 0.3 � 5.88 = 1.764 V

I2 = 1 76450. = 0.035 A

2.2All resistances are 1 �

By series-parallel combinations

2 12 1

�

�= 2

3 + 1 = 5

3 �

53

1

53

1

�

�= 5

3 + 1 = 13

8 �

I1 = 13 138

= 8 A, I2 = 8 � 152

1� = 3 A

I3 = 3 � 13

= 1 A

V1 = 1 � 1 = 1 V2.3

Req = 3 73 7

6 46 4

�

��

�

� = 4.5 �

2.4(i) RAB

Converting star to delta

rab = 4 5 5 6 6 4

6� � � � �

= 746

|| 10

rbc = 744

|| 25, rca = 745

|| 5

By series-parallel combination

746

10

746

10

�

�= 5.522,

744

25

744

25

�

� =

1850174

,

745

5

745

5

�

� =

37099

I1 I2 I3

+

–13 V V1

+

–

7

6

3

4

25

105

56

4

A

C

N

B

��������������� �

1850174

37099

� = 14.37

RAB = 5 522 14 375 522 14 37. .. .

�

� = 3.99 4 �

(ii) RAN

Converting star with star point c to delta

ran = 5 6 6 25 25 5

25� � � � �

= 12.2 �

rbn = 3055

= 61 �, rcn = 3056

= 50.83 �

By series-parallel combination

12 2 412 2 4

.

.�

�= 3.01 �

10 50 8310 50 83

8 36

5 615 61

4 62

�

��

�

�

�

.

..

.

=

=

�

= 8.36 + 4.62 = 12.98

RAN = 3 01 12 983 01 12 98. .. .

�

� = 2.44 �

2.5

7A 12V 7A∫ 2A5 510 10 6

2 26

+

–

V1 V2

Node 1: – 7 + V1

5 +

V V1 2

2�

= 0

or 0.7 V1 – 0.5 V2 = 7 (i)

Node 2: – 2 + V V V V2 2 2 1

10 6 2� �

� = 0

or – 0.5 V1 + 0.767 V2 = 2 (ii)Solving

V1 = 22 2

1

. VV V= 22

, V2 = 171

1712

..

VV V=

; I12 = 22 2 171

2. .�

= 2.55 A

1050.8312.2 4

5

61

A

N

B

����������������� �����

2.6

4A5A

6A3 W

2 W 4 W21 3

W

W W

Node 1: – 4 + 2 (V1 – V2) = 0or 2V1 – 2V2 + 0V3 = 4 (i)Node 2: – 5 + 3 V2 + 2(V2 – V1) + 4 (V2 + V3) = 0or – 2 V1 + 9V2 – 4 V3 = 5 (ii)Node 3: – 6 + 4 (V3 – V2) = 0or 0V1 – 4 V2 + 4 V3 = 6 (iii)Generalized conclusions:(aii) Diagonal element = sum of all admittances

connected to node ii

(aij) Off-diagonal element = – (sum of all admittancesconnected between nodes i and j)

Solving

V1 = 16856

= 3V, V2 = 5656

= 1V, V3 = 14056

= 2.5 V

2.7

6 W6 W

8 W 4 W

16 V30 V

+

+–

–I1 I2

V+

–

(i) Mesh analysisMesh I: 14 I1 – 6 I2 = –14 (i)Mesh II: – 6 I1 + 16 I2 = 30 (ii)Solving

I1 = – 0.234 A, I2 = 1.787 AV = 6 � 1.787 = 10.72 V

(ii) Nodel analysisConverting practical voltage sources to current sources

Node 1: – 2 – 5 + 18

16

110

� ���

�� V1 = 0 or V1 = 17.87 V

V = 17.87 � 6

10 = 10.72 V

2A5A

1 10

8 6

���������������

2.8Writing mesh equations

260I1 – 10I2 – 150I3 = 0 (i)– 10I1 + 310I2 – 200I3 = 0 (ii)

– 150I1 – 200I2 + 360I3 = 0 (iii)Solving

I1 = 0.0529 A, I2 = 0.05834 A,I3 = 0.0878 AI = I1 – I2 = 0.0054 A

2.9 Converting voltage source to current source

Node 1: – 5 + V1

1 +

V1

2 – 5 = 0

Node 2: 5 + V2

2 +

V2

3 = 0

or V1 = 203

V, V2 = – 6V

i(1�) = 20 3 5

1/ �

= 53

A (away form node 1)

i(2� left) = 203 2�

= 103

A

i(5�) = 5 A

i(2� right) = �62

= – 3 A

i(3� right) = �63

= – 2 A

2.10

Node 1: –1 + V V V V1 2 1 5

2 5�

��

= 0

Node 2:V V V V2 1 2 2

2 25

4�

� ��

= 0

or 0.7 V1 – 0.5 V2 = 2 (i)– 0.5 V1 + 1.25 V2 = 1.25 (ii)

Soiving V1 = 5V, V2 = 3 V

(a) i12 (5�) = 15

(V1 – 5) = 15

(5 – 5) = 0 A

(b) Power output current source = 5 � 1 = 5 W

15010

100 100

200

12 V

+ –

I2I1

I3

I

1A 5 V2 W

2 W

5 W

4 W

+

–

5A

5A

1 W 2 W

5 W

2 W 3 W

21

� ����������������� �����

Current out of voltage source =5 3

45 5

5�

��

= 12

A

Power output voltage source = 12

� 5 = 2.5 W

2.11 Converting current sources to voltage sources

60 W

15 W

40 W20 W 100 W

30 V 60 V10 V

+ +

+– –

–I2I1

Writing mesh equations95I1 – 15I2 = 20 (i)

– 15I1 + 155I2 = – 50 (ii)Solving I1 = 0.162 A, I2 = – 0.307 A

I3 = I1 – I2 = 0.469 A0.5 A source

i(60�) = 0.5 – 0.162 = 0.338 Av(current source) = 0.338 � 60 = 20.28 Vp(current source) = 20.28 � 0.5 = 10.14 W

0.6 A sourcei (100 �) = 0.6 – 0.307 = 0.293 A

v(current source) = 0.293 � 100 = 29.3 Vp(current source) = 29.3 � 0.6 = 17.58 W

2.12

2A 4A10 W 10 W6 W

20 WV1

I1

V2

2I1

(i) Nodal method of analysis:Converting voltage sources to current sources

Node 1: – 2 + 2I1 + V1 16

110

120

���

�� � (V1 – V2) = 0

Node 2: – 4 + 110

V2 + 120

(V2 – V1) = 0

or 0.317 V1 – 0.05 V2 = 2 – 2I1, (i)

��������������� �

– 0.05 V1 + 0.15 V2 = 4 (ii)

But I1 = 120

(V1 – V2) (iii)

Substituting (iii) in (i)0.417 V1 – 0.15 V2 = 2 (iv)

Solving (ii) and (iv)V1 = 16.35 V, V2 = 32.12 V

I1 = 16 35 3212

20. .�

= �15 77

20.

(ii) Mesh method of analysis:V(10�) = 16.35 – 15.77

= 0.58 VWriting mesh equationMesh 1: 20I1 + 10 (I1 – I2) + 30I1 + 40 = 0Mesh 2: – 12 + 6I2 + 10 (I2 – I1) – 20I1 = 0or 60I1 – 10I2 = – 40 (i)

– 30I1 + 16I2 = 12 (ii)Solving

I1 = – 0.788 A, I2 = – 0.727 A12.13 V(10�) = (0.788 – 0.727) � 10

= 0.61 V

Node 1: – 5 + V1

101

15� (V1 – V2) + 4I1 = 0

Node 2: – 4I1 + 4 + V2

51

15� (V2 – V1) = 0

Rearranging0.167 V1 – 0.067 V2 = 5 – 4I1 (i)

– 0.067 V1 + 0.267 V2 = – 4 + 4I1 (ii)

But I1 = V2

5 = 0.2 V2 (iii)

Substituding Eq. (iii) is Eqs (i) and (ii)0.167 V1 + 0.733 V2 = 5 (iv)

– 0.067 V1 – 0.533 V2 = – 4 (v)Solving

V1 = – 6.675 V, V2 = 8.325

I1 = V2

5 = 0.2 � 8.325 = 1.665 A

5A 4A10 W 5 W

15 W

4I1

I1

1 2

10 W 10 W

6 W 20 W

12 V20I1 40 V

+

+

+–

–

–I2 I1

� ����������������� �����

12.14

Node 1: V V1

31

1 10

0 5500�

�� .

= 0

Node 2: 0.3 V1 + V V2

32

350 10 5 10��

� = 0

Rearranging3V1 = 1 or V1 = 1/3 V

300 V1 + V2 150

15

���

�� = 0

or 0.22 V2 = – 100 or V2 = – 454.5 V

(a) I1 = 1

3 1 103� � =

13

mA

I2 = �

�

454 5

5 103

. = – 90.9 mA

I2/I1 = – 272.7

(b) Power supplied by 0.5 V source = 0.5 � 13

= 16

mW = Pi

Power coursumed by 5 k� resistence = (90.9)2 � 5103

= 41.3 mW = P0

P0/Pi = 6 � 41.3 = 247.9

2.15

I1

I2h11

V1 h V12 2

h I21 1 1h22

– ––

+ + +RL1

V2

Mesh 1: I1 = V h V

h1 12 2

11

�(i)

Mesh 2: h21 I1 + V2 hRL

221

����

���

= 0 (ii)

Substituting I1 from Eq. (i) in Eq. (ii)

V2 hR

h hhL

2212 21

11

1� �

���

���

= – hh

21

11 V1 (iii)

I1

500 W

1 kW 5 kW50 kWV1

I2

––

+ +

0.5 V0.3V1

21

��������������� �

(ii) AV = VV

2

1 = –

hh h h h h RL

21

11 22 12 21 11� �

���

���/

(iv)

(i) I1 = 1 12 2 1

11

����

���

h V Vh( / )

V1

I2 = –VRL

2

AI = II2

1 = –

VV

Rh V V

hL

2

1 12 2 1

11

11

������ ��

�����

( / )(v)

Substituting for V2/V1 from Eq. (iv) in Eq. (v)

AI = h h

R h h h h h R h hL L

11 21

11 22 12 21 11 12 21[( / ) ]� � �

= h h R

h h h h h RL

L

21 11

11 22 12 21 112/

/� �(vi)

(iii) Zi = VI1

1 =

hh V V

11

12 2 11 � ( / )

or Zi = h h h h h h R

h h h h h RL

L

11 11 22 12 21 11

11 22 12 21 112( / )

/� �

� �(vii)

2.16(1) Open Is

I� = Vs

44 144 14

44 14

��

�

��

= Vs

32 A

(2) Shortcircuit Vs

I � = – Is � 44 12�

= –Is

4 A

� I = I � + I� = V Is s

32 4�

(a) I = 0; Vs = 16 V

0 = 1632 4

�Is or Is = 2 A

(b) I = 0; Is = 16 A

0 = Vs

32164

� or Vs = 128 V

I+

–

Vs Is4 W 4 W

4 W 10 W

� ����������������� �����

2.17(1) Short 40 –V (battery)

0 1 551

..�

= 0.098 + 0.6 = 0.698 �

0 698 1010 698.

.�

= 0.652 + 0.1 = 0.752 �

I� = 600 752

1010 698. .

� = 74.58 A

V � = 600 752

0 69810 698.

..

� � 10 = 52.06 V

(2) Short 60 –V battery

0 1 1010 1.

.�

= 0.099 + 0.6 = 0.699 �

0 699 55 699

..

�= 0.613 + 0.1 = 0.713 �

I� = –40

0 7135

5 699. .� = – 49.22 A

V� = 40

0 7135

5 69901

101. ...

� � � 10 = 4.87 V

By superposition theoremI = 74.58 – 49.22 = 25.36 A

V = 52.06 + 4.87 = 56.93 V2.18

(1) Open circuit BD, i.e. remove the galvanometer

I (ABC) = 1025

= 0.4 A

I(ADC) = 1028

= 0.357 A

VOC = VBD = VBA + VAD

VBA = – 0.4 � 10 = – 4 VVAD = 0.357 � 12 = 4.284 VVOC = 4.284 – 4 = 0.284 V

(2) Short-circuit battery

R 0 = 10 15

2512 16

28�

��

= 12.85 �

Thevenin equivalent

IO = 0 28412 85 5

.. �

= 0.0159 A

I

V

+ ++

– ––60 V 40 V

0.1 W 0.6 W 0.1 W

5 W10 W

I05 W

12.85 W

+

–0.284 V

+ –

G

B

A C

10 W 15 W

12 W

D

16 W

5 W

��������������� �

2.19

Node 1: 0.5 V2 – I + V V V1 1 2

8 4�

� = 0

Node 2: – 4 + V2 14

16 4

2 1���

�� �

�V V = 0

But V2 = 0Then, 0.25 V2 = I

0.667 V2 = 4 or V2 = 6 V� I = 0.25 � 6 = 1.5 A

2.20

Node 1: – 1 + V V V V1 1 1 2

4 12 30� �

� = 0

Node 2: – 2 + V V V2 2 1

12 30�

� = 0

or 0.367V1 – 0.033V2 = 1– 0.033V1 + 0.117V2 = 2

Solving V1 = 4.36 V, V2 = 18.26 VVOC = V1 – V2 = – 13.9 V

Open circuiting current source

4 1216�

= 3 + 12 = 15; 15 30

45�

= 10 �

Thevenin equivalentFor maximum power output

R = 10 �

Pmax (out) = 13 920

2.��

�� � 10 = 4.83 W

2.21

11 V 2 W

1 W

11/8 W

3 W

+

–

x

y

Converting star (1�, 3�, 2�) into delta

VOC = 11 � 11

11 1110

� = 10 V

4A 4 6

4

0.5V2 8 I

V2 V1

4 W 12 W 12 W30 W

R0a b

+

–13.9 V R

a

b

11/3

11/2

11/8

11

+

–11 V

x

y

1A 2A4 W 12 W 12 W

30 WVOC

V1

a b

V2

� ����������������� �����

short circuiting voltage source

R0 = 11 11

10

11 1110

�

� = 1�

2.22

I ¢1+

–6 V 6 3 2

11.5

3 36�

= 1.5, 1.5 + 1.5 = 3, 3 6

9�

= 2�

I �1 = 62

66 3

��

= 2 A

3 1 54 5� ..

= 1, 1 + 1 = 2

I �1 = – 4 � 24

34 5

�.

= –43

A

I = 2 –43

= 23

A

2.23

9A

9 V

9 V

45 V

3 W

3 W

5 W

5 W

1 W

1 W

+

+

–

–

–

+

x

x

y

y

VOC

VOC = 45 9

9�

� 3 = 12 V

R0 = 3 63 6

�

� = 2 �

11/10

11/3 11 R0

y

x

I ¢¢16 3 2

11.5

4A

3 W5 W

1 Wx

y

R0

��������������� �

CHAPTER 3

3.1 vc(0+) = vc(0–) = 12 � 12

8 = 8V

Using superposition

ic1(0+) = – 8

4 = – 2 A, Req = 4 �

ic2(0+) = 6

6 = 1 A

ic(0+) = – 2 + 1 = – 1 A

ic = C dvdt

or – 1 = 14

ddt

(vc(0+)) or

ddt

vc(0+) = – 4V/s

3.2

0.01 F

20 W50 W30 W

ic

20A

S t = 0

Natural responseAfter switch is closed, open circuit current source (20 A)

Req = 50 50

100�

= 25 �

� = Req C = 25 � 0.01 = 0.25 sicn = Ae– 4t

Forced response

icf = 0 (cap acts as open circuit)� ic = icn + icf = Ae–4t; t > 0 (i)At t = 0+

Cap acts as a short circuit

ic(0+) = 20 � 30

30 20� = 12 A

substituting in Eq. (i)12 = A

� ic = 12e–4t, t > 0

+ +

– –6 V 8 V

6 W

12 W

ic(0 )+

5030

20

20A

ic(0 )+

At = 0t +

50 W30 W

20 W

0.01 F

� ����������������� �����

3.3Natural response

� = 224

= 1

12 s

in = Ae–12t

Forced response

L acts as short circuit. Using superposition theroemif = [32/(16 + 8)] + [1 � 8/(16 + 8)] = 2 A

� � (t) = Ae–12t + 2; t > 0i (0+) = 0 0 = A + 2 or A = – 2

Hence i(t) = 2(1 – e–12t ); t > 03.4 Time Constant

Req = 12

32

� = 2�, � = Req C = 2 � 12

= 1s

Natural response

icn(t) = Ae– t

Forced responseC acts as open circuit.

vcf = 4 � 12

= 2 V

� vc(t) = Ae– t + 2; t > 0 ;vc(0

+) = 0 0 = A + 2 or A = – 2Hence Vc(t) = 2(1 – e– t) ; t > 0 or vc(t) = 2 (1 – e– t) u(t)To find vR(t):

vRn(t) = Be– t, vRf = 4 � 12

= 2V

� vR(t) = Be– t + 2 ; t > 0

vR(0+) = 4 � 1 32

1 1 32

|| ||��

�� ���

��

� �

���

= 32 V;

Cap acts as open circuit

Then B = – 12

Hence vR(t) = – ( 12 ) e– t + 2; t > 0

3.5

i(0+) = i(0–) = 1012

= 56

A

Time constant

Req = 6 + 6 126 12

�

� = 10 �, � = L/Req = 1

2 10� = 1

20 s

2 H

16 W 8 W

1 W 1/2 F

1 W 3/2 W

12 W 1/2 H

6 W 6 W

��������������� �

Natural responseix(t) = A e–20t

Forced response

if = 10

66 126 12

1218

��

�

� = 23

A

� i(t) = Ae–20t + 23

; t > 0

56

= A + 23

or A = 16

Hence

i(t) = 16

e–20t + 23

; t > 0

3.6

v1 = v + 12 1

ddt

v����

= v + 12

dvdt

I u(t) = 12

1dvdt

+ i

= 12

12 1

ddt

v dvdt

v���

��

��

�� �

or 14

12

2

2d vdt

dvdt

� + v = I u(t)

Source free equation is

14

12

2

2d vdt

dvdt

� + v = 0 ; Let = v = Aest

As

As A

4 22 � ��

��� est = 0

Ch. equation to

s s2

4 2� + 1 = 0

or s = (– 1 � j2)Resonant frequency

� = 1

��

= � �d2 2� = 2

or �0 = 5 rad/s

12 W

6 W 6 W

+

–10 V

if

1/2 F

1/2 H

1 W

V1 i

Iu t( )

+

–

v

�� ����������������� �����

3.7

Time constant � = L/R = 12

s

Natural response in(t) = A e–2t

Forced response if (t) = 12

A

Hence

i(t) = A e–2t + 12

; t > 0

0 = A + 12

or A = – 12

� i(t) = 12

(1 – e–2t); t > 0

3.8 iL(0+) = iL(0 –) = 10

26 612

612�

�� = 1 A

Natural response

Req = 2 6

8�

+ 6 = 152

�

� = L/R = 3 215�

= 25

s

iLn(t) = Aet� 5

2� �

Forced response

iL f = 5

26 612

612�

�� =

12

A

Hence

iL(t) = Aet� 5

2 + 12

; t > 0

1 = A + 12

or A = 12

� iL(t) = 12

152�����

�e

t; t > 0

vL(t) = 3 � didt

L = 32

52

52� �

�e

t

= – 154

52e

t�

6 W

2 W

6 W

3 H

2 W

2 W

1 H

��������������� ��

v(t) = – 154

52e

t� + 6 � 1

21

52�����

�e

t

= – 34

52e

t� + 3 ; t > 0

3.9 i (0+) = i(0–) = 22

= 1 A

After switch closure

� = L/R = 12

s

in(t) = Ae–2t

if = 21

= 2 A

� i(t) = Ae–2 t + 2; t > 01 = A + 2 or A = – 1

� i(t) = 2 – e–2t; t > 0

3.10 i(0+) = i(0–) = 248

= 3 A

With switch at “b the governing differential equation” is

8i + 1 didt

i dt� �10 05.

= 0 (i)

Differentiating once

d idt

didt

2

2 8� + 20i = 0 (ii)

Let i = Aest

Then A(s2 + 8s + 20)est = 0Ch. Eqn is

s2 + 8s + 2s = 0or s = – 4 � j2Hence

i(t) = e–4t (B1 cos 2t + B2 sin 2t) (iii)i (0 +) = 3 = B1

� i(t) = e–4t (3 cos 2t + B2 sin 2t) (iv)

vL(t ) = 1 � didt

or vL(0+) = di

dt( )0�

vc(0+) = 0

8i(0+) + vL(0+) + vc(0+) = 08 � 3 + vL(0+) + 0 = 0 or vL(0+) = – 24

�� ����������������� �����

�didt

(0+) = – 24 A/s

From Eq. (iv)

didt

= – 4e–4t (3 cos 2t + B2 sin 2t) + e–4t

(– 6 sin 2t + 2B2 cos 2t) At t = 0+

– 24 = – 4 � 3 + 2B2 or B2 = 6Hence,

i(t) = e–4t (3 cos 2t + 6 sin 2t); t > 0= 3.6e–4t sin (2t + 56.3°); t > 0

3.11

iss = 52 3�

= 1 A

vss = 5 � 35

= 3 V

3.12

(a) i = 14

dvdt

c + vc (i)

10 = 4i + 2 didt

+ vc (ii)

Substituting Eq. (i) in Eq (ii)

4 14

dvdt

vcc�

��

�� + 2 d

dtdvdt

vcc

14

���

�� + vc = 10

ord v

dtc

2

2 + 6 dvdt

c + 10 vc = 20 (iii)

(b) vc(0+) = vc (0–) = 0

From Eq. (iii)

vc(�) = 2010

= 2 V

i(0+) = i(0–) = 0� i(0+) = 0

ic(0+) = 1

4dvdt

c (0+) ordvdt

c (0+) = 4 � 0 = 0

(c) Ch. Eqn(s2 + 6s + 10) = 0

or s = (– 3 � j1)vcn(t) = e–3t (B1 cos t + B2 sin t) (iv)

++

––

5 V v

i

4 W 3 W

2 W

��������������� ��

vcf = 10 � 15

= 2 V (v)

� vc(t) = e–3 t (B1 cos t + B2 sin t) + 2; t > 0 (vi)At t = 0+

0 = B1 + 2 or B1 = – 2� vc(t) = e–3t (B2 sin t – 2 cos t) + 2; t > 0 (vii)

dvdt

c (t) = – 3e–3t (B2 sin t – 2 cos t) + e–3t (B2 cos t + 2 sin t)

0 = 6 + B2 or B2 = – 6� vc(t) = e–3t (2 cos t + 6 sin t) + 2 ; t > 0

= 2 – 6.32 e–3 t sin (t + 18.4°) ; t > 03.13

(a) iL = 1 � dvdt

+ 3v (i)

vs = 12

didt

L + v (ii)

or d vdt

2

2 + 3 dvdt

+ 2v = 2 (iii)

(b) Ch. eqn iss2 + 3s + 2 = 0

or s = – 2, – 1Natural frequencies are: – 2, – 1

(c) iL(0+) = iL(0–) = 0 (iv)v(0 +) = v(0–) = 0 (v)

ic(0+) = 0 = 1 � dvdt

(0+) ordvdt

(0+) = 0 (vi)

(d) vn = A1e–2 t + A2e– t (vii)

vf = 1 V� v = 1 + A1e–2 t + A2e– t; t > 0 (viii)

A1 + A2 = – 1

dvdt

= – 2A1e–2 t – A2e– t (ix)

0 = 2A1 + A2 (x)Solving Eqs. (ix) and (x)

A1 = 1 ; A2 = – 2Hence

v(t) = 1 + e–2t – 2e– t; t > 0 (xi)

�� ����������������� �����

3.14iL(0+) = iL(0–) = 1 A, v(0+) = 0

(capacitance acts as short circuit when switch is closed)

12

dvdt

+ 32

v + v dt� = 0 (i)

or d vdt

2

2 + 3 dvdt

+ 2v = 0 (ii)

v(0+) = 0 (iii)

ic(0+) = –1 =

12

dvdt

(0+) ordvdt

(0+) = – 2 (iv)

Ch eqn iss2 + 3s + 2 = 0 or s = – 2, – 1

vn(t) = A1e–2 t + A2e

– t (v)vf = 0

Hencev (t) = A1e

–2 t + A2e– t; t > 0 (vi)

dvdt

(t) = – 2A1e–2t – A2e–t (vii)

Substituting initial conditions is Eqn (vi) and (vii)0 = A1 + A2 (viii)2 = 2A1 + A2 (ix)

SolvingA1 = 2, A2 = – 2

Hence,v(t) = 2e–2 t – 2e– t; t > 0 (x)

3.15i(0+) = 0

v2(0+) = 6v = 2 didt

(0+)

or didt

(0+) = 3 A/s

3.16 Open circuiting the current generator and short circuiting voltage generator.

1 H 1 H

8 W

8 W 4 W ∫ 2 W

+

–6 V 3 W

i (0 )+

vs (0 )+

��������������� ��

� = L/Req = 12

s

vn(t) = Ae–2 t (i)Forced responseInductance acts as a short circuit

vf = 0 (ii)� v(t) = Ae–2t (iii)Initial conditionAt t = 0+ inductance acts as open circuitAs per nodal eqn

v v v( ) ( ) ( )08

04

0 168

� � �

� ��

= 2

or v(0+) = 8V (iv)Substituting in Eq. (iii)

8 = A; v(t) = 8e–2 t; t > 0 (v)

CHAPTER 4

4.1 (a) (5)2 + (�L)2 = 24010

2��

��

or L = 74.7 m H(b) Iron loss = I2R = (10)2 � 5

= 500 W

(c) pf = cos tan–1 23 47

5.

= 0.208 lagging

4.2(a) XL = 314 � 0.1 = 31.4 �

(b) XC = 10314 160

6

� = 19.9 �

(c) Z = (8) ( . . )2 231 4 19 9� � = 14 �

I = 23014

= 16.43 A

pf = cos tan–1 31 4 19 9

8. .�

= 0.57 lagging

(d) Vcoil = (8) ( . )2 231 4� � 16.43 = 532.4 V

VC = XCI = 19.9 � 16.43 = 327 V

+

–16 V

+

–

2A 4 W8 W

8 W

v (0 )+

+

–16 V

+

–

vf2A 4 W8 W

8 W

5 W j Lw

+ –240 V, 50 Hz

10 A

8 W

+ –230 V, 50 Hz

I

0.1 H 160 Fm

�� ����������������� �����

4.36 W

+ –240 V

r xL

120 V 200 V

q fI = 5Ax

y240200

120

V(6�) = 20� � 6 = 120 V

cos � = ( ) ( ) ( )240 120 200

2 240 120

2 2 2� �

� �

or � = 56.3°x = 240 cos 56.3° – 120 = 13.2 Vy = 24 sin 56.3° = 199.7 V

(a) Z(coil) = 2005

= 40 �

5 � r (coil) = 13.2, r (coil) = 2.64 �

5 � � (coil) = 199.7, � (coil) = 39.94 �

(b) p (coil) = 52 � 2.64 = 66 W(c) pf (coil) = cos tan–1 (199.7/13.2) = 0.066 lagging

4.4

45� c

= 35 or �c = 1.286 = 1050

6

� �

or f = 101286 2 50

6

. � �� = 2.475 kHz

cos � = ( ) ( ) ( )25 50 40

2 25 50

2 2 2� �

� � ; � = 52.4°

x = 50 cos 52.4° – 25 = 5.5 Vy = 50 sin 52.4° = 39.61 V

35 � r = 5.5 or r = 0.157 �

35 � 24 � 2475 L = 39.61 or L = 0.073 mH

V(applies) = ( . ) ( . )25 55 5352 2� � = 30.97 � 31 V

R = 2536

= 0.714 �

4.5�L = + 250 tan cos–1 0.866 = 250 kVAR

Q (total) = + 250 tan cos–1 0.866 = 144.4 kVAR

qx

y5040

25

25 V 5.5 V

v

(45 – 39.61)= 5.35 V

��������������� ��

�c = 250 – 144.35 = 105.65 kVAR

( )2300 2

� c= 105.65 � 103 or �c = 50.07 � =

1314C

or C = 63.6 �F

4.6 Z = R + 10 210 2

�

�

jj

= R + 0.385 + j 1.923

(R + 0.385)2 + (1.923)2 = 10040

2��

��

or R2 + 0.77 R – 2.4 = 0or R = 1.21 �

Z = 1.21 + 0.385 + j 1.923 = 1.595 +j 1.923

pf = cos tan–1 19231 595..

= 0.638 lagging

4.7 � = 314 rad/s0.0255 H � 314 � 0.0255 = 8 �

318 �F � 10314 318

6

� = 10 �

0.038 H � 314 � 0.0382 = 12 �

57

8 – 10j

j 8j 12

V –+

10A

Z = ( ) (8 )5 8 10

13 2� �

�

j jj

+ (7 + j12) = 21.44 42.3° �

V = ZI = 21.44 � 10 = 214.4 Vpf = cos 42.3° = 0.74 lagging

4.8 (i) IL = 42 4 4( )� j

= 0.5 – 45° A

(ii) IC = 4 2

2

� j C = j 8

2 C A

For Is to be in phase with Vs

88

C = 0.354 or C = 0.0626 F

4

2 0° V

+

–

4 W

j 4 W

Is

Io

IL

12j C

�� ����������������� �����

4.9 (a) ZAB = 3 16 9

2

9 92

� �

�

j

j

�

�

= 2 3

2

12

�

�

j

j

�

�

V2 = V1 2 3

21

2

2 32

12

����� ���

��

� ����� ���

��

j j

j j j

� �

�� �

orV

V2

1

= 4 3

4 5 2

�

� �

j

j

�

� �

(b) � = 2

V

V2

1

= 4 6

4 10 4�

� �

jj

= 4 6

10� jj

= 0.6 – j0.4

V2 = 10° � (0.6 – j0.4) = 0.721 –33.7° V

v2(t) = 2 � 0.721 sin (2t – 33.7°)= sin (2t – 33.7°)

4.10

(a) I = 1 0

1

j = – j1 A

V (1�) = 1 � – j1 = – j1 V

V = (1 – j1) = 2 – 45° V

I (1F) = 1 1

12

� j

j

= 8 45° A

I 12

���

�� =

1 112

�

����

j = 8 – 45° A

Is = – j1 + (2 + j2) + (2 – j2) = 4.123 – 14° A

(b) is(t) = 2 � 4.123 cos (2t – 14°) = 5.83 cos (2t – 14°) A

(c) Ss = V Is* = 2 – 45° � 4.123 14°

= 5.83 – 31° = 5 – j3

+

–

–

+

jw

j

A

B

3

6

V1 = 1 V

92w

V2

12

12j

W WIs

I

V

VL

1 W

j1 W

–+

��������������� ��

Ps = 5 W, Qs = – 3 VARS (leading)(d) p(1�) = 12 � 1 = 1 W

p 12

���

�� = ( 8 )2 � 1

2 = 4 W

4.11

V = ( ) ( )2 2 22 2� = 10 V

� = tan–1 22 2

= 26.6°

� = 45° – 26° = 18.4°

V = 10 18.4°

v(t) = 20 cos (2t – 18.4°)

4.12

Z = ( ) ( )

( ) ( )2 2 2 1

2 2 2 1� �

� � �

j j jj j j

= � �

�

2 22 3

jj

V = Z I = � �

�

2 22 3

jj

�� 1 0°

= – � �

�

2 22 3

jj

= – 0.783 – 101.3°

= 0.783 78.7°

v(t) = 2 � 0.783 sin (2t + 78.7°) = 1.11 sin (2t + 78.7°)pf = cos � = cos 78.7° = 0.196 lagging

P(source) = 1 � 0.783 � 0.196 = 0.153 W4.13

Node 1: V

j

V V

j

1 1 223

43

��

� = 1 or j 3

2 V1 – j 3

4 V1 + j 3

4 V2 = 1

Node 2: V V

j

V V

j

2 2 2 1

1 12

43

��

��

= 0 or V2 + j2 V2 – j 34

V2 + j 34

V1 = 0

Rearranging

j 34

V1 + j 34

V2 = 1 (i)

+

–

j 2

j2 – 1j

1 0°A– V

2

V2

V1

V

1 0°A–

45°

45°

f

q

j22 2

2

2

+

–2 W

2 W

j2 W

– 2j W

V2

V1

V

1 0°A–

�� ����������������� �����

j 34

V1 + 154 2�����j V = 0 (ii)

Solving

V1 = 1.91 – 65.3° ; V2 = – 0.894 – 26.6°

4.14

(a) IL = 1 0° A

V1 = j2 V

I1 = j2 A

I3 = IL + I1 = 1 + j2

= 2.236 63.4° A

V2 – V1 = – j1 (1 + j2) = (2 – j1) V

or V2 = j2 + (2 – j1) = 2 + j1 = 2.236 26.0° V

I2 = 2(2 + j1) = 4.472 28.6° A

Is = I2 + I3 = (4 + j2) + (1 + j2) = 5 + j4 = 6.403 38.7° A

(b) Ps = Re [2.236 26.6° � 6.403 –38.7°] = 14 W

(c) P 12

���

�� = Re [2.236 26.6° � 4.472 – 26.6°] = 10 W

P (1�) = 2 � 2 = 4 W

cleck: P 12

���

�� + P (1�) = Ps

Phasor diagram

4.15 SLB = 12 + j12 = 16.97 45° kVA

ILB = 16 97 1000

220. �

= 77 A

B+

–

Vg

0.15 0.35 0.3 0.7

P jQG G+ P jQB B+P jQL L+

SLA

ILB

PB = 12 + (77)2 � 0.3/1000 = 13.78 kWQB = 12 + (77)2 � 0.7/1000 = 16.15 kVAR

SLA = 10 + j7.5, SB = 13.78 + j16.15 = 21.23 49.5° kVA

26.6°

63.4°

38.7°

1

2

2

4.472

6.403

2.23

6

IL

I2

V1

Is

I3I1

12 WIs

ILI1I2 I3

V2 V1

1 W j 2

– 1j

��������������� ��

SL = (10 + 13.78) + j (7.5 + 16.15)

= 23.78 + j 23.65 = 33.54 44.8° kVA

VLA = 21 23 1000

77. �

= 275.7 V

IL = 33 54 1000

275 7.

.�

= 121.7 A

PG = 23.78 + (121.7)2 � 0.15/1000 = 26 kW�G = 23.65 + (121.7)2 � 0.35/1000 = 28.83 kVAR

SG = 26 + j28.83 = 38.82 48° kVAR

VG = 38 82 1000

121 7.

.�

= 320 V

pfG = cos 48° = 0.67 lagging

4.16

Load++

––G1 G2V1 V2

0.3 0.7 0.2 0.4

P QG G1 1+

P QG G2 2+

P jQ¢ ¢L L+

P jQ¢¢ ¢¢L L+

P jQL L+IG1

ILB

IL

SL = 10 + j 7.5 = 12.5 36.9°

IL = 12 5 1000

230. �

= 54.34 A, IL = 54.34 –36.9°

P �L = 10 + (54.34)2 � 0.2 �/1000 = 10.59 kWQ �L = 7.5 + (54.34)2 � 0.4/1000 = 8.68 kVARS �L = (10.59 + j8.68) = 13.69 39.3° kVA

S �L = V IL2*

V2 = 13 69 39 3 1000

54 34 36 9. .

. . �

= 251.9 2.4° V

SG2 = PG2 + j�G2 = 5 + j5 = 7.072 45° kVA

P L = 10.59 – 5 = 5.29 kW� L = 8.68 – 5 = 3.68 kVAR

��SL = 5.59 + j3.68 = 6.69 33.4° kVAR

IG1 = 6 69 1000

2519.

.�

–33.4° + 2.1° = 26.56 –31.3° A

PG1 = 5.59 + (26.56)2 � 0.3/1000 = 5.80 kWQG1 = 3.68 + (26.56)2 � 0.7/1000 = 4.17 kVAR

� ����������������� �����

SG1 = 5.80 + j4.17 = 7.14 35.7° kVA

pfG1 = 0.812 lagging

V1 = 714 35 7 1000

26 56 3130. .

. . �

= 268.8 4.4°

4.17 Mesh Equations(200 + j240)I1 – (124 + j240) I2 = 16 (i)– (124 + j240) I1 + (245 + j180) I2 = 0 (ii)

or Z1I1 – Z12I2 = 16–Z12I1 + Z2I2 = 0

where Z1 = 312.41 50.2°Z12 = 270.14 62.7°Z2 = 304.01 36.3°� = Z1Z2 – Z12

2

= 59648.6 36.3°Solving I1 = 0.0815 0°

= 0.0815 + j0I2 = 0.0724 26.4°

= 0.0649 + j0.0322To determine VAB consider the mesh – source, 76 �, 120 � – A, B, 120 �, source.

76 I1 + 120 (I1 – I2) + VAB + 120 I2 = 1676 � 0.0815 + 120 (0.0815 – 0.0649 – j0.0322) + VAB

+ 120 (0.0649 + j0.0322) = 16Solving

VAB = 0.018 0° V

4.18

10 0° V–+

–

2 W 4 W

5 W+

––

+

+

–

vc2vc

– 2j

– 1j

V1

V2

V Vj

V Vj

1 1 1 2102 4 2 1�

��

��

� = 0 (i)

V Vj

V Vc2 1 2

12

5�

��

� = 0 (ii)

Vc = V

j1

4 2� � – j2 = (0.2 – j0.4)V1 (iii)

76 W

1 W

4 W

120 W

120 W

– 60j W

j 240 W

A

B

+

–

16 0° V–I1

I2

��������������� �

Rearranging Eq. (i)

0 5 14 2

1 1 2. ��

����

���

�j

j V jV = 5

or 1.304 57.5° V1 + 1 – 90° V2 = 5 (iv)

Substituting Eq. (iii) in Eq. (ii)

(0.08 + j0.84)V1 + (0.2 + j1) V2 = 0

– 0.844 84.6° V1 + 1.02 78.7° V2 = 0 (v)

Solving Eqs. (iv) and (v)

V2 = 0.201 13.5°

� v2(t) = 0.201 2 (cos 2t + 13.5°)4.19 Writing mesh equations

(5 + j10) I1 – j10 I2 = 15 (i)

– j10 I1 + (2 + j8) I2 = 4 60° (ii)

15 0° V–

– 2 2 30°= –4 60° Vj ¥ –-

–+

–

2 W

5 Wj10 W

– 2j

+

–

V1

I1 I2

V2

Solving

I1 = 2.089 29.3° A

V1 = 15 0° – 5 � 2.089 29.3°

= 7.8 –40.9° V

I2 = 2.816 34.1° A

V2 = –j2 � I2 – 4 60°

= 8.21 – 81.9° V4.20 Combining the parallel circuit part

Z = 12 10 8

22 8( )�

�

jj

= 6.22 – j2.11

B 80 30° A–-120 0° V– Vi

V1P

I1 I2

1 W

6.22 W

0.8 W j 1 W j 0.8 W

–j 2.11 W

�� ����������������� �����

Writing nodal equations

Vj

Vj

1 1

6 22 2 10120

0 8 1. . .��

�

�= 80 –30°

Vj j1

16 22 2 10

10 8 1. . .�

��

�

��

= 80 – 30° + 120 00 8 01

�. .j

Solving

V1 = 202 0° V

I1 = 120 0

0 8 11 �

�

Vj.

= 120 0 202 0

0 8 1 �

�. j = – 64.06 –51.3°

P = Re (V I1 1* )

= Re (202 0° � – 64.06 – 51.3°) = – 8 kW

P(V) = Re (120 0° I1* )

= Re [120 � – 64.06 51.8°] = – 4.754 kW

Vi = V1 + (1 + j0.8) � 80 –30°

= 202 + 1.28 38.7° � 80 – 30°= 298.3 2.8° V

P(I) = Re {298.3 2.8° � 80 – 30°} = 21.225 kW

4.21 Z0 = 0.644 – j0.786

RL = 0.644 �, XL = 0.786 �

Maximum power transfer = 10 12 0 644

2..�

���

���

� 0.644

= 39.6 WRL = 0.8 WXL = – XTH = – (– 0.786) = 0.786 �

Maximum power transfer = 10 10 8 0 644

2.

. .�

���

���

� 0.8

= 39.14 W

4.22 V0 = 1290° � 4 6

6 12 4 6�

� � �

jj j( ) ( )

= 7.42 2.7° V

Z0 = ( ) ( )

( )6 12 4 6

10 6� � �

�

j jj

= 7.6 – j3.36

I = 7 42 2 7

116 2 64. .. .

� j = 0.624 – 10.1° A

4 W

j 6 W

A

B

+

–

7.42 2.7°–

2.6 – 3.36j

��������������� ��

I 2R = (0.624)2 � 4 = 1.56 WI 2X = (0.624)2 � 6 = 2.34 VARS

Hence

SL = (1.56 + j2.34) = 2.81 56.3° VA

CHAPTER 5

5.1

(a) H ( j�) = V j

V j2

1

( )

( )

�

� =

Rj C

R Rj C

2

1 2

1

1

�

� �

�

�( )

= ( )( )

jj

2 110 1

�

�

�

�

Corner frequencies are:

12

= 0.5 rad/s, 110

= 0.1 rad/s

10.50.1

0

–45°

–90°

1020

40db

f

log w

(b) H ( j�) = V j

V j2

1

( )

( )

�

� =

R

R

Rj C

2

2

1

11

�� �

= R

R Rj CRj CR

2

1 2

1 11�

���

���

�

�

( )( )

�

�

where R = R R

R R1 2

1 2�

�� ����������������� �����

Substituting values

H ( j�) = 0.5 � ( )

( . )j

j�

�

�

�

10 5 1

Corner frequencies: � = 1, � = 220 log 0.5 = – 6db

2 10 log w

45°

20

1

0

40

60

db

f

5.2

0.22 0.5 1 0.28 10

40

20

0.1

45°

–45°

db

flog w

H ( j�) = V j

V j2

1

( )

( )

�

� =

2 2

2 2 114 2

����

���

����

���

�

���

��

j

jj

�

� �

= 12

� ( ) ( )

( ) ( )

j j

j j

� �

� �

� �

� �

1 2 152

12

2

��������������� ��

= 12

� ( ) ( )

. .

j j

j j

� �

� �

� �

���

�� ���

��

1 2 1

0 221

2 281

Corner frequencies are:

Upper � = 1, � = 12

= 0.5

Lower � = 0.22, � = 2.285.3

Y ( j�) = 11r j LC

� ���

���

�

At C = 400 pF, current is max.

� �L = 1

�C

or 2� � 106 L = 10

2 10 400

12

6� � �

or L = 0.0633 mH

At C = 450 pF, current is reduced to 12

of max.

�1

2 10 0 0633 10 102 10 450

6 312

6r j� � � � �

� �

�

���

����

�.

= 12 r

or1

44r j�= 1

2 r

or r2 + (44)2 = 2r2

or r = 44 �

� = �Lr

= 2 10 0 0633 10

44

6 3� � � � �.

= 9.04

5.4 r = 2300 35.

= 657 �

VC = I

C� = 300 or C = 0 35

314 300.�

= 3.72 �F

r j Lw

1j CwY j( )w

� ����������������� �����

�L = 1

�C

or L = 10

3 72 314

6

2. ( )� = 2.73 H

5.5 �0L = 1

0� c

or C = 1

02� L

= 12 100 7 52( ) .� � �

= 0.338 �FTo fend � at I = 0.848 A

140

� 1

40 7 5 100 338

6

� �j j..

��

= 1 20 848

..

= 2

or 1 0 18750 074 106

� ��

j j..

��

= 2

or 1 + 0 18750 074 106 2

..

��

���

�����

= 2

� = 630.666, 625.334 rad/sor f = 100.37 Hz, 99.552 Hz

5.6 �0 = 1LC

= 10

140

250

3

�

= 400 rad/s

Hence excitation frequency is the resonant frequency

� IR = I = 10°; iR = cos 400 t

V = 500 � 10° = 500 0° V

IL = 500 0

400 140

�j = 50 – 90°

iL = 50 sin 400 t

Ic = 500 0

10 400 2506

�( / )j = 50 90°

ic = – 50 sin 400 t

I IR = = 1 0° A–

IL = 50 –90° A–

IC = 50 90° A–

V = 500 0°–

��������������� �

5.7 Circulating current = V�0C

= V C

LC

= V C L/

At resonant frequency

I = VL

RC

= V RCL

��

�� = V

1 10 10

10 10

4 12

6

� �

�

���

���

�

� = 0.001 V

�0 = 1LC

= 1

10 10 10 106 4 12� � �� � = 3.16 � 106 rad/s

� = 1.9 �0 = 2.844 � 106 rad/s

Z ( j�) = 11RC

Lj C

L� ���

���

�

; Eq. (ii) of Ex. 5.2

RCL

= 1 10 10

10 10

4 12

6

� �

�

�

� = 1 � 10–3 = 0.001 V

�C = 2.844 � 106 � 104 � 10–12 = 0.02844

1� L

= 12 844 10 10 106 6. � � � �

= 1.03516

�C – 1� L

= – 6.72 � 10–3

I = V | 1 – j6.72 | � 10–3

= 0.0068 V5.8

Find the series equivalent of the parallel circuit

Zs = 1

110

100 10512

0� � �j � = 10

1 10

5

50( )� �j �

= 101 10

5

1002�

�

���

��� � – j

�

�

010

021 10�

�

���

���

At resonant frequency

400 � 10–6 �0 = �

�

010

021 10� �

�0 = 5 � 106 rad/s or 0.796 MHz

I

C

R

L

+

–V–0°

� ����������������� �����

Req = 10

1 10 2500 10

5

10 10� � �� = 40 �

I = 10° m A

Zs = 40 – j 5 10

1 10 2500 10

6

10 10

�

� � ��

= 40 – j2000 = 2000 – 88.9°V0 = 1 � 10–3 � 2000 = 2V

� 0 = � 0 LReq

= 5 10 400 10

40

6 6� � � �

= 50

5.9 f0 = 1

2� LC = 1

2 5 10 50 103 13� � � �� � = 0.318 MHz

Q = �0 RC = 2� � 0.318 � 106 � 60 � 63 � 50 � 10–12

= 37.7

�b = � 0

Q =

2 0 318 106

6� � �. = 333 kHz

5.10 At resonant condition

314 = 1

0 75. � Cor C = 13.52 �F

Let the capacitance connected in parallel to the series RLC circuit be C1. Given � = 628rad/sec

15 + j628 � 0.75 – j 1013 52 628

6

. �

= 353 87.6°

Z ( j�) = 1628 2 83 10 87 61

3j C � � � �. .

= 1628 0119 10 2 83 101

3 3j C j� � � �� �. .

For the imaginary part to be zero628C1 = 2.83 � 10–3

or C1 = 4.51 �F

R = 10

0 119

3

. = 8404 �

I = 2008404

= 0.0238 A

C1

15 W

0.75 H

13.52 Fm

Z jw( )

��������������� �

5.11 Using the period from – � to �

v(t) =

0

0

2

2 2

2

� � � �

� � �

� �

�

��

��

� �

� �

� �

�

� �

�

t

V t t

tm cos

Because of even symmetry sine terms would be absent.

am = 12

2

�� � �

�

�

V t m t d tm cos cos ( )��

m = 1

a1 = 1 2

2

2

�� �

�

�

V t d tm cos ( )�� =

Vm

2

m > 1

am = 12

2

�� � �

�

�

V t m t d tm cos cos ( )��

= V

m t m t d tm

�� � �

�

� 12

1 12

2 [cos ( ) cos ( ) ] ( )� � ���

= V m t

mm tm

m

211

11

2

2

�

� �

�

�

sin ( ) sin ( )�

��

�

�

�

�� �

= 2

1 2

V

mm

� ( )� cos m�

2 ; m > 1

Hence

v(t) = V Vm m

��

2 cos cot +

23Vm

� cos 3 � t

– 215

Vm

� cos 4 �t +

235Vm

� cos 6 �t

5.12 From Eq. (5.33)

a0 = 12 0 3

22

2�� �

�

�

�

Vd t Vd t( ) ( )� ���

��

= V2

Because of even symmetry sine terms will be absent. From Eq. (5.34)

am = 10 3

22

2�� � � �

�

�

�V m t d t V m t d tcos ( ) cos ( )� ��

�

��

= � 2Vm�

; m = add; = 0, m = even

����������������� �����

Thus

v(t) = V V2

2�

� cos � t –

23V�

cos 3 � t + 25V�

cos 5 � t.

5.13

V0 ( j�) = V j

j Lj C

j CR

R

( )�

��

���

��1

1

1

1

= V ( j�) � 1

1 2( )� ���LC j LR

Substituting values

V0 ( j�) = V ( j�) � 11 10 106 2 3( )� �� �� �j

1 – DC component

V0 (dc) = V2

2 – Fundamental component� = 314

V0 (1) = 2 0

21

1 10 314 0 3146 2V

j

�� � ��� ( ( ) ) .

= 0.472 V –19.2°3 – 3rd harmonic

� = 3 � 314 = 942

V0 (3) = 23

10 113 0 942

Vj�

��. .

= 0.158 V –83.2°4 – 5th harmonic

� = 5 � 314 = 1570

V0 (5) = 25

11 455 1 57

Vj�

�� �. .

= 0.042 V –133°

Hence

v0(t) = V2

+ 0.472 2 V cos (� t – 19.2°) – 0.158 2 V

cos (3�t – 83.2°) + 0.042 2 V cos (5�t – 133°)

= V2

+ 0.667 V cos (� t – 19.2°) – 0.233 V cos (3wt – 83.2°)

+ 0.059 V cos (5 cot – 133°)

���������������

5.14

Fundamental frequency is � = 2�

T rad/s

The wave over one period is expressed as

i(t) = Im

�����

� ; 0 < � < �

= Im 21

������

� ; � < � < 2�

a0 = 12

22

� �� �

�� �

�

�

�Id I I dm

d m���� � ���

��

�

�� � � =

12

Im

Because of even symmetry sine terms would be absent

am = 1 2 10

2

� �� � �

�� � �

�

�

�Im d I m dm

mcos cos� �� �����

�

��

= 2

2 2

I

mm

� (cos m� – 1)

= 0 ; m even

= –4

2 2

I

mm

� ; m odd

5.15

� = 21�

� 103 = 2000� rad/s

V

I0 = 1

12000

2000 0 2 10 6j m( ) .� � � � ; m = harmonic number

Im = 10 mA or 0.01 A

1 – V0(dc) = 12

� 0.01 � 2000 = 10 V

2 – Fundamental

V0 (1) = – 0 012

4 20001 2 512

..

� ��� j

= – 2.123 – 68.3° V3 – 3rd harmonic

V0 (3) = 0 012

49

20001 2 51 32

..

� �� �� j

= – 0.083 – 82.4° V

� ����������������� �����

4 – 5th harmonic

V0 (5) = – 0 012

425

20001 2 51 52

..

� �� �� j

= – 0.018 – 85.4° VHence

vc(t) = 10 3 2000 68 3 0117 6000 82 4

0 025 10000 854� � � �

� �

cos ( . ) . cos ( . ). cos ( . )� �

�

t

5.16Fourier series are

v(t) = 1�

Vm + 23�

Vm cos �t – 215�

Vm cos 4� t + .......

Substituting value(Vm = 200 V, � = 314 rad/s)v(t) = 63.7 + 42.4 cos (2 � 314) t – 8.5 cos (4 � 314) t

V

V0 = 100

100 314� j n = 1

1 314� j n.

V0(dc) = 63.7 Vn = 2

V

V0 = 1

1 6 28� j . = 0.157 – 81°

v0(2) = 42.4 � 0.157 cos (628 t – 81°)= 6.66 cos (628 t – 81°)

n = 4

V

V0 = 1

1 12 56� j . = 0.074 – 85.4°

v0(4) = 8.5 � 0.079 cos (1256 t – 85.4°)= 0.672 cos (1256 t – 85.4°)

Hencev0(t) = 63.7 + 6.66 cos (628 t – 81°) – 0.672 cos (1256 t – 85.4°)

5.17(a) Part 2 open circuited, port 1 excited:

z11 = V

II

1

12 = 0

= 152 45

10 03

.

� = (1075 + j1075) �

z21 = V

II

2

12 = 0

= 2 29 28

10 03

. �

� = (2022 – j1075) �

��������������� �

Port 1 open circuited, port 2 shorted:

z22 = V

II

2

2 1 = 0

= 131 55

10 03

. �

� = (751 – j1073) �

z12 = V

II

1

2 1 = 0

= 1 075 90

10 03

. �

� = – j1075 �

(c) With reference to Fig. 5.16

z11 – z12 = (1075 + j1075) + j1075

= 1075 + j2150 = 1075 + j106 (2150 � 10–3)

z12 = – j1075 = – j 1

10 930 106 12� � �( )

z22 – z12 = 751 – j1073 – (– j1075) = 751

z21 – z12 = (2022 – j1075) – (– j1075) = 2022

The equivalent circuit is drawn below:

1075 W 751 W2.175 mH

930 pF

+ +

V1 V2

I2I1

– –

– +

2022 I1

5.18 (a) As the network is resistive all voltages and currents would be in phase and therefore wecan work in terms of magnitudes only.

V1 = 200 I1 – 800 I2 (i)V2 = 1200 I1 + 16 � 103 I2 (ii)

Given: RL = 8 � 103 �, I2 = –V2

38 10�(iii)

I1 = 1 – V1/800 (iv)Substituting for I1 and I2 in Eqs (i) and (ii)

V1 = 200 1800

1���

��

V + 800 �

V238 10�

V2 = 1200 (1 – V1/800) – 16 � 103 � V2/ 8 � 103

or 1.25 V1 = 0.1 V2 = 2001.5 V1 + 3 V2 = 1200

Solving V1 = 184.6 V; V2 = 307.7 V(b) To find Thevenin impedance, inject unit current at port 2 and open circuit current source

V1 = – 800 I1 or I1 = – 1

800 V1 (v)

1 0° A–

I1 I2

RL V2V1––

++8 W

� ����������������� �����

I2 = 1Substituting in Eqs (i) and (ii)

V1 = – 8001 25.

= – 640 V

V2 = –1200800

� – 640 + 16 � 103 = 16.96 � 103 V

or RTH = 16.96 � 103 �

� RL (for max power output) = 16.96 k�

(c) Solving Eqs (i) and (ii) for I1 and I2

I1 = 3.85 � 10–3 V1 + 0.192 � 10–3 V2

I2 = – 0.288 � 10–3 V1 + 0.048 � 10–3 V2

Hence

y11 = 3.85 m� y12 = 0.192 m�

y21 = – 0.288 m� y22 = 0.048 m�

5.19Port 2 short circuitedWriting nodal equation

V V V V V�� � �1 1

100 200 500 40 = 0

or V = 0.3 V1

I1 = V V1

100�

= V V1 10 3

100� .

= 0.007 V1

or y11 = IV

1

1 = 0.007 �

Now

I2 = – V40

= – 0 3

401. V

or y21 = IV

2

1 = – 0.0075 �

Port 1 short circuited

I2 = V2

40 100 200� || = 0.009375 V2

or y22 = IV

2

2 = 0.009375 �

Now I1 = – I2 � 200300

= – 0.009375 � 200300

V2

= –0.00625 V2

200 W

100 W 40 WV

V1500

+V1

–

I1 I2

200 W

100 W 40 W

V1500

+V2

–

I1 I2

= 0V1 = 0

��������������� �

or y12 = IV

1

2 = – 0.00625 �

Nowy11 + y12 = 0.007 – 0.00625

= 0.75 m�

y22 + y12 = 0.009375 – 0.00625

= 3.125 m�

y21 – y12 = – 0.0075 + 0.00625

= – 1.25 m�

5.20 V1 = h11I1 + h12V2 (i)I2 = h21I1 + h22V2 (ii)

Also V2 = – RL I2 (iii)

Solving Eqs (i) and (ii)

V2 = h I h Vh h h h

11 2 21 1

11 22 12 21

�

�

Substituting for I2 from Eq. (iii)

V2(h11h22 – h12h21) = –hRL

11 V2 – h21V1

or GV = VV

2

1 = –

hh h h h h RL

21

11 22 12 21 11( ) /� �

5.21 V1 = z11I1 + z12 I2 (i)V2 = z21 I1 + z22 I2 (ii)

Also V2 = – RLI2 (iii)Solving Eqs (i) and (ii)

I2 = z V z V

z11 2 21 1�

det(iv)

det z = z z

z z11 12

21 22

V2 = –R

zL

det (z11v2 – z21V1)

orVV

2

1=

z Rz z R

L

L

21

11det �

5.22 The frequency domain form of the circuit isdrawn here.

0.75

mW

3.12

5 m

W

0.00625 mW

W W

W

1.25,mA

+ +

V1

I1 I2

V2

– –

+ ++

–– –

Vs 2V11 W

4 W2jw

2jw

V1 V0

� ����������������� �����

Converting voltage source to current source

+ ++

–– –

Vs 2V11 W

4 W

2jw2

jw V1V V0w2

j

Writing nodal equations

– j �

2 Vs + j �

2 V + (V – V1) j �

2 + 0.25 (V – V0 ) = 0 (i)

V2 = 2V1 (ii)

V1 = 1

1 2�

j�

V = j

j�

�2 �

���

���

V

or V = V1 2 ����

���

jj

�

�(iii)

Substituting Eq (iii) in Eq (i)

– j �2

Vs + j Vj

jV j V�

�

�

���

���

�� �

14

22

121 1 1

( ) = 0

j�

2 Vs =

( ) ( )( )

1 4 24

12

1� �

� ��

��

j jj

j V� �

��

Solving yields

V

Vs

0 = 2

12

1

2

2

( )

( ) ( )

j

j j

�

� �� �

= 2

13186

10 314

2( )

. .

j

j j

�

� ����

�� ���

��

0.314 1 103.186 log w

20

0.1

40

db

20 log 2 (0.1)= 0.06

¥ 2

db

�������������� ��

CHAPTER 6

6.1 Vp = 2303

= 132.8 V

Zy = 8 + j6 = 10 –38.9° W

IL = 132 8

10.

= 13.28 A

pf = cos 36.9° = 0.8 lag

P = 3 ¥ 230 ¥ 13.28 ¥ 0.8 = 4.232 kW

Q = 4.232 tan cos–1 0.8 = 3.174 kVAR

S = ( . ) ( . )4 232 31742 2+ = 5.29 kVA

6.2 Ip = 1003

= 57.74 A

ZD = 110057 74.

= 19.05 W

160 ¥ 1000 = 3 ¥ 1100 ¥ 100 cos for cos f = 0.84 lead, f = 32.9°

ZD = 19.05 –– 32.9° W= 16 – j10.34 W

R = 16 W, C = 10314 10 34

6

¥ . = 308 mF

6.3

ZD = 16 + j12 = 20 –36.9° W

Ip = 23020

= 11.5 A

IL = 11.5 3 = 19.9 Apf = cos 36.9° = 0.8 log

P = 3 ¥ 230 ¥ 19.9 ¥ 0.8 = 6.34 kW

Q = 6.34 tan 36.9° = 4.47 kVAR

S = ( . ) ( . )6 34 4 762 2+ = 7.93 kVA

6.4 pf = cos 37° = 0.8 lagging

IL = 30 3 = 52 A

P = 3 ¥ 400 ¥ 52 ¥ 0.8 = 28.82 kW

�� ���� ��� ��� �������������

37°Vab = 400 V

Iab = 30 A

Ia = 52 A

Vca

Vbc

Ibc

Ica

Ib

Ic

6.5 Z1(Y ) = 16 + j20 = 25.6 –51.3° W

Z2 (Y) = 27 18

3- j

= 9 – j6 = 10.8 ––33.7° W

Z (eq)/phase Y = (16 + j20) || (9 – j6)= 9.69 ––11.7° W

IL = 4003 9 69¥ .

= 23.83 A

pf = cos 11.7 = 0.979 leading

Q = 3 ¥ 400 ¥ 23.83 ¥ sin 11.7° = 3.15 kVAR

6.6 (a) IL = 10 3 = 17.32 A

f = 37°

(b) P = 3 ¥ 220 ¥ 17.32 ¥ cos 37° = 5.27 kW

(c) I 2pR = 5270

3 or R = 17.6 W

CHAPTER 7

7.1 (a) £u(t) – u(t – t) = 1s

e– st 1s

= 1 - -e

s

st

(b) cosh bt = e ebt bt+ -

2 ; t ≥ 0

£ cosh bt = 12

1 1s b s b-

+-

LNM

OQP =

ss b2 2-

(c) sin w0t = e ej t j tw w0 0

2- -

; t ≥ 0

�������������� ��

£ sin w0t = 12

1 1

0 0j s j s j--

+LNM

OQPw w =

ww0

202s +

(d) cos w0t = e ej t j tw w0 0

2+ -

; t ≥ 0

£ cos w0t = 12

1 1

0 0s j s j-+

+LNM

OQPw w = s

s202+ w

7.2

(a) 82s s( )+

= 82

1 8 12

0 2s s s ss s++

+-= =

= 4 42s s

-+

Taking inverse laplace transformf (t) = (4 – 4e–2t ) u(t)

(b)8 1

2( )( )s

s s++

= 8 1

21 8 1 1

20 2

( ) ( )ss s

ss s

s s

++

++

+-= =

= 4 42s s

++

or f (t) = (4 + 4e–2t ) 4(t)

(c)8 1

2 2

( )

( )

s

s s

++

= 8 1

21 8 1 1

220 2

2( )

( )

( )

( )

s

s sss ss s

++

+ ++-= =

+ dds

s

s ss

8 1 122

2

( )+LNM

OQP +-=

= 2 42

222s s s

++

-+( ) ( )

or f (t) = (2 + 4t e–2t – 2e–2t) u(t)

(d) F(s) = 12

4 82s

s s+ + = 12

2 2 2 2s

s j s j( ) ( )+ + + -

= As j

As j( ) ( )

*

+ ++

+ -2 2 2 2

A = 122 2 2 2

ss j s j+ - - -=

= 6(1 – j1)

�� ���� ��� ��� �������������

f (t) = 2Re [6(1 – j1) e– (2 + j2) t]

= 12e–2 t (cos 2t – sin 2t) = 12 2 e–2t cos (2t + 45°) u(t)

7.3 £ f dt

( )t t-•zLNM =

f d

sF s

s

( )( )

t t-•z +

0

f (t) = te–2t u(t) ´ 12 2( )s +

te u t dtt--•z 20

( ) = 0

Hence

£ te u t d ttt -

- •z 2 ( ) ( ) = 12 2s s( )+

7.4 v(t) = 10 u(t) – 20 u(t – 1) + 10 u(t – 2)

V(s) = 10 20 10 2

s se

ses s- +- -

= 10 1 2 2( )- +- -e e

s

s s

7.5 (i) es

s-

+ 1´ e–(t – 1) u(t – 1)

(ii) e

s

s-

+

2

22( )´ (t – 2) e–2 (t – 2) u(t – 2)

7.6 Ri + 1C

i dtt

-•z = u(t)

RI (s) + 1Cs

I(s) + u

sc ( )0

= 1s

or I (s) = 1 0

1-

+FH IK

v

s RCs

c ( )

i (0+) = lim( )

s

csv

s RCs

Æ •¥

-

+FH IK

1 01

= 1 0- v

Rc ( )

i(•) = lim( )

s

csv

s Rs

Æ¥

-

+FH IK0

1 01

= 0

t s( )

21

10

–10

v

�������������� ��

7.7

z(s) = 10 10 10

20

10 10 1020

36

36

¥ ¥

¥ +

s

s

= 1020 100

6

s + =

50 105

3¥+s

I (s) = £ cos 10t u(t) = ss2 100+

Vc(s) = z(s) I(s) = 50 10

5 100

3

2

¥+ +

s

s s( ) ( )

vc(t) = [– 2 ¥ 103 e–5 t + 4.46 ¥ 103 cos (10t – 63.4°)] u(t)

7.8

I(s) = V s

R sLsc

( )

+ +FH IK1

= 1012sR s Lc

+ +FH IK = 5

2 42( )s s+ +

i (t) = 5

3 e– t sin ( 3 t) u(t)

7.9

20050 106

+ ¥FHG

IKJs

I(s) – 9V1(s) = V(s) (i)

I(s) = V s V s( ) ( )- 1

200(ii)

Substituting Eq. (ii) in Eq. (i)

20050 10

200200

50 10200

6 61+ ¥F

HGIKJ - + ¥F

HGIKJs

V ss

V s( ) ( ) – 9

V1(s) = V(s)

or V1(s) = 0 25 10

10 0 25 10

6

6

. ( )

( . )

¥+ ¥

V s

s(iii)

V(s) = 1s

\ V1(s) = 250 10

250 10

3

3

¥+ ¥s s( )

(iv)+

+

–

++

––

–V s0 ( )

9 ( )V s1

I s( )

V s1 ( )V s( )

200 W 50 10 /¥ 6 s

�� ���� ��� ��� �������������

Solvingv1(t) = u(t) – e–25 ¥ 103t u(t)v0(t) = 9v1(t) = 9 (1 – e–250 ¥ 103t) u(t) (v)

7.10

Z(s) = ( )2 2

2 2

+

+ +

ss

ss

= 2 2

2 22

( )s

s s

++ +

V(s) = Z(s) ¥ 5 = 10 2

2 22

( )s

s s

++ +

= 10 2

1 1 1 1( )

( ) ( )s

s j s j+

+ + + -

A = 10 2

1 11 1

( )( )

ss j

s j

++ - - -=

= 5(1 + j1)

v(t) = 2 Re [5(1 + j1) e– t e–jt] u(t)

= 10 2 e– t cos (t – 45°) u(t)7.11

Z(s) =

1 1

1 1

ss

ss

( )+

+ +

= s

s s

++ +

1

12

Vc(s) = Z s

Z s s( )

( ) +¥

11

= s

s s s

++ +

1

2 22( )

= s

s s sA

s jA

s js

++ +

++ +

++ -

LNM

OQP

12 2

11 1 1 12

0=

*

A = s

s s js j

++ - - -

11 1

1 1( )

=

= - +1 1

4j

vc(t) = 12

u(t) + 2 Re - +F

HIK

LNM

OQP

- -1 14

je et jt u(t)

= 12

u(t) – 12

e– t cos (t + 45°) u(t)

s

V sC ( )Z s( )

+

–

1s

1s

1

+

–

2 W

s

s V s( )

+

–

2s

�������������� ��

7.12

Z(s) = 14

131

131+ +ss

= 134 132

ss s+ +

V(s) = 134 13

4 12s

s s s+ +¥ +FH IK =

13 4

4 132

( )s

s s

++ +

= As j

As j( ) ( )

*

+ ++

+ -2 3 2 3

A = 13 4

2 32 3

( )( )

ss j

s j

++ - - -=

= 13 12

13

+FH IKj

v(t) = 2 Re 13 12

13

2 3+FH IKLNM

OQP

- -j e et j t u(t)

= 15.6 e–2 t Re (e– j (3t – 33.7°) ) u(t)= 15.6 e–2t cos (3t – 33.7°) u(t)

7.13

Z(s) = 112

12 4

+ +s

s

= 42 22s

s s+ +

V(s) = 42 2 0 042 2s

s ss

s( ) ( . )+ +¥

+

= A

s jA

s jA

s jA

s j1 1 2 2

0 02 0 02 1 1 1 1++

-LNM

OQP

++ +

++ -

LNM

OQP. .

* *

A1 = 0.04 – j0.196, A2 = – 2e– j88.9°

u(t) = 0 4 0 2 78 3 4 88 9. cos ( . . ) ( ) cos ( . ) ( )t u t e t u tt- ∞ - - ∞-

Forced response Natural response� ����� ����� � ����� �����

7.14V s V s

s( ) ( )

200 4 10 3+¥ - = I(s) + 101 ¥

V ss

( )4 10 3¥ -

or V(s) = 200

5 106

s I s

s

( )

( )- ¥

134

13s

4s W 1 s V s( )

+

–

4s2s2 V s( )

+

–

s

s2 + 0.04

�� ���� ��� ��� �������������

I(s) = £ t e–0.2t u(t) = 10 2 2( . )s +

\ V(s) = 200

0 2 5 102 6

s

s s( . ) ( )+ - ¥

= A

s

As

A

s12

211 3

60 2 0 2 5 10( . ) ( . ) ( )++

++

- ¥

A12 = 2005 106

0 2

s

s s- ¥ -= .

= 8 ¥ 10–6

A11 = dds

s

ss

200

5 1060 2

- ¥FHG

IKJ

-= .

= – 40 ¥ 10– 6

A3 = 200

0 2 2

s

s s( . )+ ¥= 5 106

= 40 ¥ 10–6

Hencev(t) = 8 ¥ 10–6t e–0.2t u(t) – 40 ¥ 10–6 e–0.2t u(t)

+ 40 ¥ 10–6 e5 ¥ 106 u(t)

7.15 With switched closed (for long time) the capacitor gets fully charged and ic = 0; thedependent current source is open. Hence

vc(0) = 20 3020 30

¥+

¥ 2 = 24 V

s-domain circuit after t = 0 is drawn.By KCL

V s sV s( ) ( )12 4

+ – 6 – 4 sV s( )

46-F

HIK = 0

or V(s) = 2419

s -

\ v(t) = 24et1

9 u(t)7.16 The transformed circuit is shown below:

Using nodal analysis at nodes i and 0.

V s V s V s V s

s

i s i( ) ( ) ( ) ( )-+

-2 2

0

++

V s

s

i ( )

4 2= 0 (i)

V0(s) = 4 V1(s) (ii)

4s

+

–

12 W6 V

sc(

)

4 ( )I sc

I s tc( )

+ ++

–– –

–

+

2s

23V ss( )

2 4

Vs

1()

V s0( )V si ( ) 4 ( )V s1

i 0

�������������� �

From Eqs (i) and (ii)Vi(s) = (2s + 1) V1(s) (iii)

Substituting for Vi(s) from Eq. (iii) are V0(s) from Eq. (ii) in Eq. (i)

V1(s) = 0 5

0 52.

( . )s + Vs(s) (iv)

(a) V0(s) = 4 V1(s) = 20 52( . )s +

Vs(s)

orV sV ss

0 ( )( )

= H(s) = 2

0 52( . )s +

(b) Vs(s) = 1s

\ V0(s) = 20 52( . )s +

= 212

12

s s j s j+FHG

IKJ -FHG

IKJ

v0(t) = 4 2 20 512e et j t- -

+ -FHGIKJ

FHG

IKJ

. Re u(t)

= 4et- 2 u(t) – 4 cos t

2 u(t)

(c) Vs(s) = ss2 1+

FHG

IKJ

V0(s) = 2

1 12

2 2

s

s s( )+ +FH IKSolving

v0(t) = 2 2 2 212Re ( ) Re- + F

HGIKJ

FHG

IKJ

- -e ejt j t

u(t)

= - +FHG

IKJ4 4 1

2cos cost u(t)

7.17 Transformed circuit is shown below.

21

521 V s( )

+

–

10

2s 4+–

V s sV s V ss

( ) ( ) ( )1 2

52

102 4

+ - + -+

= 0

� ���� ��� ��� �������������

or V(s) = 5 4

4 52

( )

( )

s

s s

++ +

= As j

As j+ +

++ -2 1 2 1

*

A = 5 4

2 1 2 1

( )ss j s j

++ - - -=

= 52

+ j5

Hence

u(t) = 2 Re 52

5 2+FH IKLNM

OQP

- -j e et jt u(t)

= 5e–2t (cos t + 2 sin t) u(t)7.18 Steady state with switch at ‘a’

I = 512

= 10 A

At t = 0 switch is thrown to ‘b’

iL(0) = 10 A Li (0+) = 15

¥ 10 = 2

Vi(0) = 10 ¥ 12

= 5 V Cvc(0+) = 1 ¥ 5 = 5

s-domain circuit is drawn below:Writing nodal equations

2V(s) + s V(s) + V(s) – 2) 5s

= 5

or V(s) = 5 2

2 52

( )s

s s

++ +

= 5 2

1 2 1 2( )

( ) ( )s

s j s j+

+ + + -

A = 5 12

14

+FH IKj

u(t) = 5e– t Re 1 12

2 2+FH IK -LNM

OQPj t j t(cos sin u(t)

= 5e–t cos sin212

2t t+FH IK u(t)

7.19 Switch s open; steady state current = 104

= 2.5 A

iL(0+) = 25; Li (0+) = 0.2 ¥ 2.5 = 0.5

1s

12

V s( )

+

–

+

–2

s/5

5

�������������� ��

s-domain circuit 5 drawn in Fig below:

I(s) =

10 0 5

2 0 2s

s

+

+

.

. =

2 5 5010

.( )

ss s

++

i(t) = 5u(t) – 2.5 e–10t u(t)= (5 – 2.5 e–10t) u(t)

7.20(a) y(t) = (1 – 3e– t + 4e–2t) u(t)

y(s) = 1 31

32s s s

-+

++

= ( )

( ) ( )s

s s s

2 21 2

++ +

x(s) = £ u(t) = 1s

\ H(s) = Y sX s

( )( )

= ( )

( ) ( )s

s s s

2 21 2

++ +

(b) For forced response to be zeros2 + 2 = 0

or s = ± j 2 or w = 2 rad/s7.21

(a) y(t) = (2t + 1) e–2t u(t)

Y(s) = 22

122( ) ( )s s+

++

= ( )

( )

s

s

++

4

2 2

x(t) = e–3t u(t)

x(s) = 13s +

H(s) = Y sX s

( )( )

= ( )

( )

s

s

++

4

2 2 ¥ (s + 3)

Poles: s = – 2, – 2Zeros: s = – 3, – 4

(b) x (t) = e–3(t – 1) u (t – 1)y(t) = (2(t – 1) + 1) e–2(t – 1) u(t – 1)

7.22

H(s) = V sV s

2

1

( )( )

= 2

2 112

0 5+

+ . s

= ss

++

12

v1(t) = 10[u(t) – u(t – 2)]

0.2s+–

0.52

+

–

10s

I s( )

�� ���� ��� ��� �������������

Response to 10u(t) ´ 10s

v2(t) = £–1 10 1

2( )

( )s

s s+

+ = 10 ¥

ss s

++

12

= 0

u(t) + 10 ¥ s

s s

+-

1

2= e–2t u(t)

= 5u(t) + 5e–2t u(t)Response to 10 u(t – 2)

v2(t) = 5u(t – 2) + 5e2(t – 2) u(t – 2)Hence total response is

v2(t) = 5[u(t) – u(t – 2)] + 5[e–2t u(t) – e–2(t – 2) u(t – 2)]= 5[u(t) – u(t – 2)] + 5e–2t [u(t) – e4 u(t – 2)]

7.23

V s( )10

+ s V(s) + V s V s( ) ( )- 1

200 = 0 (i)

sV s V s V s1 1

100 200( ) ( ) ( )

+-

= 0.5 (ii)

Solving

V(s) = 0 250 605 0 052

.( . . )s s+ +

= 0 250 506 0 1

.( . ) ( . )s s+ +

u(t) = 0 250 506 0 1

.( . ) .s s+ -=

e–0.1t u(t) + 0 250 1 0 506

.( . ) .s s+ -=

e–0.506t u(t)

= (0.616 e–0.1t – 0.5e–0.506t) u(t)

CHAPTER 8

8.1

BC = 3 10

25 0 9 10

3

4

¥¥ ¥

-

-. = 1.3 T

From Fig. 8.10HC = 350 AT/mAT = 350 ¥ 50 ¥ 10–2 = 175

8.2 HC = 350 AT/mATC = 350 (50 – 0.2) ¥ 10–2 = 174.3

ATair-gap = lg

m Ag

FHG

IKJ f =

2 10

4 10 25 10

3

7 4

¥¥ ¥ ¥

-

- -p ¥ 3 ¥ 10–3

= 1910ATtotal = 174.3 + 1910 = 2084

0.5100

s1s10

200V s1 ( ) V s( )

�������������� ��

I = 20841000

= 2.084 A

8.3 AT = BC

m0 lg + HClC

1000 ¥ 1.5 = BC

4 10 7p ¥ - ¥ 2 ¥ 10–3 + HC (50 – 0.2) ¥ 10–2

or 1500 = BC

4 10 4p ¥ - + 0.498 HC (i)

HC = 0, BC = 0.942BC = 0, HC = 3012

Eq. (i) and the BH curve are both plotted in Fig. Q 8.3, from whichBC = 0.89

f = 25 ¥ 2.9 ¥ 10–4 ¥ 0.89 ¥ 10–3 = 2 mWb

5000

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1000 1500 2000 2500 3000

H = AT/m

BT(

)

8.4 Ag = AC

cos 45∞ =

25 0 90 707

¥ ..

= 31.8 cm2

ATairgap = 2 10

4 10 318 10

3

7 4

¥¥ ¥ ¥

-

- -p . = 1501

45°

25 cm2

Ag

2 mm

Fig. Q.8.3

�� ���� ��� ��� �������������

ATtotal = 174 + 1501 = 1675

I = 16751000

= 1.675 A

8.5 Rg = 1 10

4 10 30 10

3

7 4

¥¥ ¥ ¥

-

- -p= 2.65 ¥ 105

R C = ( . )15 01 10

4 10 4500 30 10

2

7 4

- ¥¥ ¥ ¥ ¥

-

- -p= 8.78 ¥ 103

R = 40 10

4 10 4500 30 10

2

7 4

¥¥ ¥ ¥ ¥

-

- -p = 23.58 ¥ 103

f = 4 ¥ 10–3 WbF AB = f (R C + Rg) = 4 ¥ 10–3 (2.65 ¥ 102 ¥ 8.78) ¥ 103

= 1095 AT

f2 = 1095 500

2358 103

¥¥.

= 25.23 mWb

f1 = f + f2 = 25.23 + 4 = 29.23 mWbF1 – R f1 = FAB

or F1 = FAB + R f1

= 10.95 + 23.5 ¥ 103 ¥ 25.23 ¥ 10–3 = 1690 AT

8.6 R g = 1 10

4 10 2 2 5 10

3

7 4

¥¥ ¥ ¥ ¥

-

- -p . = 15.9 ¥ 105

F g = R gf = 15.9 ¥ 105 ¥ 0.25 ¥ 10–3 = 398 ATlAB (right limb) = 2 ¥ 12 + 20 – 0.1 = 43.9 cmlAB (right limb) = 2 ¥ 12 + 20 = 44 cm

Bg = B right limb = 0 25 10

2 2 5 10

3

4

.

.

¥¥ ¥

-

- = 0.5 T

H (right limb) = 220 AT/m (from Fig. Q 8.6)FAB = 398 + 220 ¥ 43.9 ¥ 10–2 = 495 AT

H (left limb) = 49344 10 2¥ - = 1125 AT/m

B (left limb) = 1.38 T (from Fig. Q 8.6)f (left limb) = 1.38 ¥ 2 ¥ 2.5 ¥ 10–4 = 0.69 mWb

f (central limb) = 0.25 + 0.69 = 0.94 mWb

B (central limb) = 0 94 10

4 2 5 10

3

4

.

.

¥¥ ¥

-

- = 0.94 T

f1 f2

fF1

F2 = 500 AT

A

B

R g

R c

R

R

�������������� ��

H (central limb) = 460 AT/m (from Fig. Q8.6)F (care of central limb) = 460 ¥ 20 ¥ 10–2 = 92 AT

Fcoil = FAB + F (care of central limb)= 495 + 92 = 587 AT

Exciting coil current = 5871000

= 0.587 A

200 400 600 800 1000 1200 14000

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

AT/m

BT(

)

Fig. Q.8.6

8.7 R g = 1 10

4 10 12 10

3

7 4

¥¥ ¥ ¥

-

- -p = 6.63 ¥ 105

R c = ( . )20 01 10

4 10 12 10

2

7 4

- ¥¥ ¥ ¥ ¥

-

- -p m r

= 0 132 10 9. ¥ +

m r

R = 80 10

4 10 6 10

2

7 4

¥¥ ¥ ¥ ¥

-

- -p m r

= 106 109. ¥

m r

f = FR R R

g c+ + 2

F1

R g

R cRR

375 AT0.4 mW b

�� ���� ��� ��� �������������

0.6 ¥ 10–3 = 875

6 630 662 10

104

5..+ ¥F

HGIKJ ¥m r

or mr = 7609

8.8 Rg = 2 10

4 10 4 4 10

3

7 4

¥¥ ¥ ¥ ¥

-

- -p = 9.95 ¥ 105

R C = [ ( ) . ]2 25 20 0 2 10

4 10 4000 4 4 10

2

7 4

+ - ¥¥ ¥ ¥ ¥ ¥

-

- -p = 1.11 ¥ 105

Rg + R C = (9.95 + 1.11) ¥ 105 = 11.06 ¥ 105

f = 400 4

11 06 105

¥¥.

= 1.45 mWb

Wf (air-gap) = 12

Rg f2 = 12

¥ 9.95 ¥ 105 ¥ (1.45)2 ¥ 10–6 = 1.046 J

Wf (core) = 12

R cf2 = 1

2 ¥ 1.11 ¥ 105 ¥ (1.45)2 ¥ 10–6 = 0.036 J

Now mr (core) = • \ R c = 0

f = 400 4

9 95 105

¥¥.

= 1.61 mWb

Wf (air-gap) = 12

¥ 9.95 ¥ 105 ¥ (1.61)2 ¥ 10–6 = 1.29 J

Wf (core) = 0(b) i = F/N = f R total/N

= ( . sin ) .0 4 314 10 11 06 10

400

3 5t ¥ ¥ ¥-

= 1.11 sin 314t A

e = wN fmax cos wt

= 314 ¥ 400 ¥ 0.4 ¥ 10–3 cos 314t

= 50.24 cos 314t(c) L = N2/R total

= ( ).400

1106 10

2

5¥ = 144.7 mH

If mr = • R total = 9.95 ¥ 105, then L = ( )

.

400

9 95 10

2

5¥ = 161 mH.

�������������� ��

CHAPTER 9

9.1 (a) I2 = 25 1000

200¥

= 125 A, I1 = 25 1000

600¥

= 41.7 A

(b) (i)( )600

21

2

= 25 ¥ 1000 or Z1 = 14.4 W

Similarity

( )200 2

2Z= 25 ¥ 1000 or Z2 = 1.6 W

(c) fmax = V

f N1

12p = 600

2 50 60¥ ¥ ¥p = 0.045 Wb

The core flux will be the source when the transformer is excited at rated voltage in

secondary side VN

VN

1

1

2

2=

FHG

IKJ

(d)Vf1 =

¢¢

Vf1

or V1¢ = ( f ¢/f ) V1 = 6050

¥ 600 = 720 V

V ¢2 = 7203

= 240 V

(e) fmax = 600

2 40 60¥ ¥ ¥p = 0056 mb

The core flux density has increased. As the core loss is proportional to the square offlux density and directly proportional to frequency, core loss would increase causing thecore to get overheated. Also the magnetizing current would increase which is detrimentalto transformer.

9.2 (a) On 600 – V side:VSC = 600 ¥ 0.052 = 31.2 V

ISC = 25 1000

600¥

= 41.7 A

R1 = 24241 7 2( . )

= 0.139 W

X1 = ( . ) ( . )0 748 0 1392 2- = 0.748 WOn 200 – V side:

R2 = 0 139

9.

= 0.0817 W; X2 = 0 735

9.

= 0.0817 W

PU values:

ZB(HV) = 60041 7.

= 14.4 W

�� ���� ��� ��� �������������

R(pu) = 0 13914 4.

. = 0.0097 or 0.97%

X(pu) = 0 73514 4.

. = 0.051 or 5.1%

(b) Since frequency is same (50 Hz) during SC test

(Core flux) SC(Core flux) Rated Voltage

= 5.2%

(c) Core flux and also core density reduce to 5.2% of rated value. The core losses reduceto (0.052)2 or 0.27% of core loss at rated voltage. Hence core losses during SC are ofnegligible order and the total power input constitutes ohmic losses.

9.3(a) Transformer impedance neglected:

I2 = 2001 48 1 04| . . |+ j

= 110.6 A

I1 = 110 63

. = 36.9 A

(b) Transformer impedance accounted for:

I2 = 2000 0154 1 48 0 0817 1 04|( . . ) ( . . )+ + +j

= 106.8 A

I1 = 106 8

3.

= 35.6 A

% Error caused by neglecting transformer impedance

= 110 6 106 8

110 6. .

.-

= 3.4%

(c) V2 = 106.8 | 1.48 + j1.04 | = 1193.2 V

Voltage reg = 200 193 2

200- .

= 3.4%

9.4 P0 = (91.1)2 ¥ 1.48 = 12.28 kWI 2

2R2 = (91.1)2 ¥ 0.0154 = 127 WPC = 195 WPL = 127 + 195 = 322 W

h = 12 2812 28 0 322

.. .+

¥ 100 = 97.4%

9.5 R(HV) = 0.125 + 0.005 ¥ 1100220

2FH IK = 0.25 W

X(HV) = 0.625 + 0.0025 ¥ 1100220

2FH IK = 1.25 W

�������������� �

Similarly

Z (LV) = (0.01 + j0.05) WNow

Z (pu) = (0.25 + j1.25) ¥ 501000 11 2¥ ( . )

= 0.01 + j0.0529.6 I = 1 pu

Z = (0.01 + j0.052) puV2 = 1 pu

(i) 0.8 pf laggingV1 = V2 + I (R cos q + X sin q)

= 1 + 1 ¥ (0.01 ¥ 1.08 + 0.052 ¥ 0.6) = 1.039 pu

V1(HV) = 1.039 ¥ 1100 = 1142.9 V ; % Req = 3.9(ii) 0.8 pf leading

V1 = 1 + 1 ¥ (0.01 ¥ 0.8 – 0.052 ¥ 0.6) = 0.9768 puV1(HV) = 0.9768 ¥ 1100 = 1074.5 V; % Req = – 2.32

9.7 Pi = 580 W

Pi(3/4th fl ) = 50 1000

110034

2¥ ¥FH

IK ¥ 0.25 = 290 W

PL = 580 + 290 = 870 W

P0 = 50 ¥ 34

¥ 0.8 = 30 kW

h = 3030 0 87+ .

= 97.2 %

9.8 Load current = 1 pu, 0.8 lag/lead pf

V2(pu) = V1(pu) – I(pu) (R/pu) cos q ± X (pu) sin q)= 1.0 – 1 ¥ (0.01 ¥ 0.8 ± 0.052 ¥ 0.6)= 0.9688 pu, 1.0232 pu

V2 = 0.9688 ¥ 220 = 213.1 V (0.8 lag pt)= 1.0232 ¥ 220 = 225.1 V (0.8 lead pf )

Note: Full load is assumed to mean full load current.9.10

(a) Pin = Pi = ( )2001000

2

= 40 W

I0 = 200400

2001000

200500

-FH IKj = (0.1 – j0.2) A

I0 = 0.233 A, pf = 0.45 lag

� ���� ��� ��� �������������

(b) ILV ( f l ) = 5 1000

200¥

= 25 A

VSC = 25 ¥ | 0.18 + j0.5 | = 13.3 VPi = Pc = I2R = (25)2 ¥ 0.18 = 112.5 W

pf = cos tan–1 0 50 18

..

= 0.34 lag

9.11

Z(HV) = 52 78 4

..

= 6.27 W

R(HV) = 287

4 2(8. ) = 4.07 W

X(HV) = 4.77 W(a) For max efficiency

I2(HV) ¥ R(HV) = Pi

I(HV) = 2874 07.

= 8.33 A

P0 (for hmax) = 2200 ¥ 8.33 = 18.33 kW

hmax = 18 3318 33 2 0 287

.. .+ ¥

= 96.96%

(b) pf = cos tan–1 RX

= 0.76 leading

9.12

I2 = 25 1000

200¥

= 125 A

(kVA) Auto = 1200 125

1000¥

= 150

I1 – I2 = 25 1000

1000¥

= 25 A

9.13

I1 = SV

I1 + I2 = SaV

I2 = SaV

SV

-FH IK = SV a

11-FH IK

II

1

2= 1

1 1a

- = a

a1 -

I1

I1

I2

I I1 2+V

av

�������������� ��

(a) a = 0.5

II

1

2=

0 51 0 5

..-

= 1

(b) a = 0.1

II

1

2= 0 1

1 0 1.

.- = 1

9

9.14

I2 = 20 1000

200¥

= 100 A

Auto transformer output = 2640 100

1000¥

= 264 kVA

kVA transformed = 20kVA conducted = 264 – 20 = 244Full-load output = 264 ¥ 1 = 264 kW

As two-winding transformer0.975 (20 ¥ 0.8 + PL) = 20 ¥ 0.8

or PL = 0.41 kW

hauto = 264 1

264 0 41¥

+ . = 99.84%

9.15

Load

A

B

N

C

a

c

bI a:

22/345 kV

Transformation ratio of each transformer unit

= 22/345/ 3 = 1a

or a = 9.054

Let us take VA (Y side voltage of phase A to neutral N) as the reference phasor

Vphase = 345/ 3 = 199.2 kV

VA = 199.2 –0°, VB = 199.2 –– 120°, VC = 199.2 –– 240° kV

Rating of each transformer unit = 13

| 500 + j100 | = 170 MVA

VAB = 345 –30°, VBC = 345 –– 90°, VCA =345 –– 210° kV

�� ���� ��� ��� �������������

V IA A* = 500

3 + j 100

3

IA* =

500 1003 199 2

+¥

j.

= 0.837 + j0.167 kA

I A = 0.837 – j0.167 = 0.853 –– 11.3° kA

IB = 0.853 –– 131.3° kA

IC = 0.853 –– 251.3° kA

On D-side:

Vab = 199 29 054

..

–0° = 22 –0° kV

Vbc = 22 –– 120° kV

Vca = 22 –– 240° kV

Iab = 9.054 ¥ 0.853 –– 11.3° = 7.723 –– 11.3° kA

Ibc = 7.723 –– 131.3° kA

Ica = 7.723 –– 251.3° kA

Ia = I Iab ca- = 3 ¥ 7.723 –– 41.3° = 13.376 –– 41.3° kA

Similarly

Ib = 13.376 –– 161.3° kA

Ic = 13.376 –– 281.3° kA

Note: It is easily observed from above that the line voltages and currents on star side leadthose on delta side by 30° the same holds for phase voltages and current.

9.16

22 3

kV

6.6

3 k

V Z Z

Z

I1 I2

IZ

6.6

kV

22 k

V

( )22 3 2

Z= 3 MVA

or Z = 484 W

3 ¥ 22 3 ¥ I2 = 9 ¥ 1600 or I2 = 136.4 A

(transformer current secondary)

�������������� ��

I1 = 136.4 ¥ 226 6.

= 454.5 A (transformer primary current)

IZ (current in leg of delta) = I2

3 = 136 4

3. = 78.75 A

9.17

22 kV

22 3

kV

I1/ 3Z Z

Z

I1

6.6

kV

IZ

As before (Problem 9.16)I2 = 136.4 A (current in transformer secondary)IZ = 78.75 A (current in leg of a connected load)

I1/ 3 = current in transformer primary

= 136.4 ¥ 226 6.

= 454.6 A

I1 (primary line current) = 454.6 3 = 787.4 A

CHAPTER 10

10.1 slots/pole = 3 ¥ 3 = 9

g = 180

9∞

= 20°

Short pitching angle, b = 20°

Kb = sin

sinm

m

g

g2

2

= sin /

sin /3 20 2

3 20 2¥ ∞

∞ = 0.96

Kp = cos 20°/2 = 0.985

10.2 f = 300 20

120¥

= 50 Hz

Total turns = 180 8

2¥

= 720

Nph (series) = 7203

= 240

m = 1803 20¥

= 3

g = 180 20

180∞ ¥

= 20°

�� ���� ��� ��� �������������

Kb (as calculated in Problem 10.1) = 0.96Ep = 4.44 ¥ 0.96 ¥ 50 ¥ 240 ¥ 0.05 = 426 V

El = 426 3 = 738 V

10.3 P = 120 f

ns =

120 50750

¥ = 8

m = 4S = 4 ¥ 3 ¥ 8 = 96

g = 180 8

96∞ ¥

= 15°

Full pitch = 968

= 12 slots

Coil pitch = 11 slotsShort pitching angle, b = 15°

Total armature turns = 2 8 96

2¥ ¥

= 768

Nph (series) = 7683 2¥

= 128

Kb = sin /

sin /4 15 2

4 15 2¥ ∞

∞ = 0.96

Kp = cos 15°/2 = 0.99Ep = 4.44 Kw f Nph (series) f

= 4.44 ¥ 0.96 ¥ 0.99 ¥ 50 ¥ 128 ¥ 0.05= 1350 V

El = 1350 3 = 2338 VIconductor = 10 A

IP = 2 ¥ 10 = 20 A = Il

kVA = 3 ¥ 2338 ¥ 20 = 81

10.4

P = 120 50

250¥

= 24

g = 180 24

288∞ ¥

= 15°

f = Bav ¥ Polc aue

= 2 123 2

240 8

pp¥FH IK ¥ ¥ ¥F

HIK.

.. = 0.256 Wb

Total turns = 288 8

2¥

= 11.52

�������������� ��

(a) Single-phase case

m = 28824

= 12

Kb = sin /

sin /12 15 2

12 15 2¥ ∞

∞ = 0.638

N(single-phase) = 1152E = 4.44 ¥ 0.638 ¥ 50 ¥ 1152 ¥ 0.256

= 41.77 kVkVA = 41.77 ¥ 10 = 417.7

(b) 3-phase case

m = 2883 24¥

= 4

Kb = 0.96 (see Prob. 10.3)

Nph (series) = 1152

3 = 384

Ep = 4.44 ¥ 0.96 ¥ 50 ¥ 384 ¥ 0.256= 20.96 kV

kVA = 3 ¥ 20.96 ¥ 10 = 6.28.5Observation: For the same winding/same amount of copper) 3-phase winding has 33.5%higher kVA capacity.

10.5S = 24

g = 180 2

24∞ ¥

= 15°

f = 2.2 Wb(a) Single-phase winding

m = 242

= 12

Kb = sin /

sin /12 15 2

12 15 2¥ ∞

∞ = 0.638

Nph(series) = 24 2

2¥

= 24

E = 4.44 ¥ 0.638 ¥ 50 ¥ 24 ¥ 2.2= 7478 V

kVA = 7.478 ¥ 8 = 59.8(b) Two-phase winding

m = 242 2¥

= 6

�� ���� ��� ��� �������������

Kb = sin /

sin /6 15 2

6 15 2¥ ∞

∞ = 0.9

Nph (series) = 242

= 12

Ep = 4.44 ¥ 0.9 ¥ 50 ¥ 12 ¥ 2.2= 5272 V

El = 5.272 2 = 7458 VkVA = 2 ¥ 5.272 ¥ 8 = 84.4

(c) Three-phase winding

m = 243 2¥

= 4

Kb = sin /

sin /4 15 2

4 15 2¥ ∞

∞ = 0.96

Nph (series) = 243

= 8

Ep = 4.44 ¥ 0.96 ¥ 50 ¥ 8 ¥ 2.2= 3751 V

El = 3751 3 = 6497 V

kVA = 3 ¥ 6.497 ¥ 8 = 90

10.6 Br = m 0Fr

g

Bav = 2p

Br

fr = p Dl

PFH

IK Bav =

2 0DlP g

¥FHG

IKJ

m Fr

P = 2 0DlP g

¥m

Wb/AT

10.7 Set speed = 120 50

2¥

= 3000 rpm

P (gen) = 120 400

3000¥

= 16

10.8 n(no-load) = 990 rpmn(full-load) = 950 rpm

ns = 1000 rpm

(a) P = 120 50

1000¥

= 6

�������������� ��

(b) s(no-load) = 101000

¥ 100 = 1%

s(full-load) = 50

1000 ¥ 100 = 5%

(c) No-load case

Speed of rator field wrt rotor surface = 120 0 5

6¥ .

= 10 rpmSpeed of rator field wrt stator = 990 + 10 = 1000 rpmSpeed of rator field wrt (rotor field = 0 rpmFull-load case:

f = 0.05 ¥ 50 = 2.5 HzSpeed of rotor field wrt rotor surface = (120 ¥ 0.5)/6 = 50 rpmSpeed of rotor field wrt stator surface = 950 + 50 = 1000 rpm

10.9 ns = 1000 rpm

(a) s = 1000 960

1000-

¥ 100 = 4%

(b) P = 6(c) T = kI2 cos q2 (Eq. (10.45))

q2 = 0\ T = kI2

For same torque I2 should remain the same. If R2 is doubled, the slip will became doublei.e., s = 2 ¥ 0.04 = 0.08 or n = 1000 (1 – 0.8) = 920 rpm. Thus adding external resistancein rotor circuit gives control over motor speed. This is only possible in slip ring inductionmotor and not in squirrel cage motor where rotor bars are premanentaly short circuited.

CHAPTER 11

11.1 VOC for the OCC is now reduced in the ratio of 1600/1000. The revised data are as under:

VOC (V) 35.6 66 100 135 189 208 220 236 247If (A) 0.5 1.0 1.5 2.0 3.0 3.5 4.0 5.0 6.0

The OCC is drawn in Fig. Q–11.1

(a) V0(Rf = 50 W) = 220 V, If = 4 A

(b) R (critical) = 130

2 = 65 W

n(critical) = 1600 ¥ 110130

= 1354 rpm

I2

R2

sV2

�� ���� ��� ��� �������������

300

250

200

150

100

50

0 1 2 3 4 5 6

Rcritical

Rf = 50 W

V (V)OC

I Af ( )

Fig. Q.11.1

(c) V will come down from 220 V as in part (a) because of IaRa drop.Let V = 200 V

If = 22055

= 3.6 A

Ea = 210

IL = 200

8 = 25 A

Ia = 25 + 3.6 = 28.6 AV = 210 – 28.6 ¥ 0.5 = 195.7 V

Let V = 190 V

If = 19055

= 3.45 A, IL = 1908

= 23.75 A, Ia = 27.2 A

Ea = 205V = 205 – 27.2 ¥ 0.5 = 191.4 V (almost converged)

Ea Ra V RL = 8 WRf = 55 W

If IL

Ia+

–

�������������� �

11.2

250 V50 W

0.02 W

If IL

Ia +

–

400 rpm

(a) Genrator

250 V50 W

0.02 W

If IL

Ia +

–

(b) Motor

As generator:

IL = 50 1000

250¥

= 200 A

If = 25050

= 5 A \ Ia = IL + If = 205 A

Eag = 250 + 205 ¥ 0.02 = 254.1 V254.1 = k ¢E If ng = k¢E ¥ 5 ¥ 400 (i)

As motor:

IL = 50 1000

250¥

= 200 A

If = 5 AIa = 200 – 5 = 195 A

Eam = 250 – 195 ¥ 0.2 = 246.1 V246.1 = k ¢E ¥ 5 ¥ nm (ii)

Dividing Eq. (i) by Eq. (ii)

nm

400=

24612541

.

.

or nm = 387.4 rpm11.3

At rated output and speed:

Input = 1000 9.

= 111.1 kW

(a) Field current = 600500

= 1.2 A

(b) IL = 1111 1000

600. ¥

= 185.2 A

Ia = 185.2 – 1.2 = 184 A(c) Ea = 600 – 184 ¥ 0.13 = 576 V(d) Mechanical power developed

= Ea Ia

= 576 ¥ 184 = 106 kW

� ���� ��� ��� �������������

(e) T ¥ 2 1200

60p ¥

= 106 ¥ 1000

or T = 843.5 Nm11.4

Running light:Ia1 = 6.32 – 0.92 = 5.4 A

Ea1 = 250 – 5.4 ¥ 0.252 = 248.6 V248.6 = k ¢E ¥ 1280 (i)

Motor current 85 A:No change in field current

Ea2 = 250 – (85 – 0.92) ¥ 0.252 = 228.8228.8 = k ¢E ¥ n2 (ii)

From Eqs (i) and (ii)

n2

1280= 228 8

248 6..

or n2 = 1178 rpmMotor current 60 A:No change in field current

Ea2 = 250 – (60 – 0.92) ¥ 0.252 = 235 V235 = k ¢E ¥ n2 (iii)

From Eqs (i) and (iii)

n2 = 1280 ¥ 235248 8.

= 1209 rpm

11.5 Speed 750 rpm:IL = 28 A, If = 1.67 AIa = 28 – 167 = 26.33 A

Ea = 500 – 0.8 ¥ 26.33 = 479 V

479 = (Kaf) ¥ 750 2

60¥F

HIK

p

or Kaf = 6.1T = (Kaf) Ia = 6.1 ¥ 26.63 = 160.6 Nm

Speed 1000 rpm:

500 – 0.8Ia = Kaf¢ ¥ 1000 ¥ 260p

(i)

160.6 = Kaf¢ Ia (ii)Dividing

500 0 8160 6

- ..

Ia = 100 2 60¥ p /

Ia

500 V300 W

0.8 W

IL

Ia +

–

1.67 A

�������������� ��

or 0.8 Ia2 – 500 Ia + 160.6 ¥ 104.72 = 0

or I 2a – 625 Ia + 210225 = 0

Ia = 589.33 or 35.67 A (Note: the first solution is not feasible)Ea = 500 – 0.8 ¥ 35.67 = 471.5 V

1.67 ¥ 750 a 479If ¥ 1000 a 471.5

or If = 4715749

750 1671000

. .¥ ¥ = 1.23 A

Rf (total) = 5001 23.

= 406.5 W

Rf (Ext) = 406.5 – 300 = 106.5 WIL = 35.67 + 1.23 = 36.9 A

11.6 At no-load (300 V mains)

Ea ª V = 300 V = Kaf ¥ 1200 2

60¥F

HIK

p

or Kaf = 300 60

1200 2¥¥ p

= 2.387

On load (600 V mains)Ea = 600 – 150 ¥ 0.2 = 570 V

570 = 2.387 ¥ n ¥ 2

60p

or n = 2280 rpm11.7

At no load:Ea ª 230 V

230 = Kaf ¥ 2 1000

60p ¥

or Kaf = 1.464On load:

Ea = 230 – 200 ¥ 0.1 = 210 V

n = 1500 ¥ 210230

10 96

¥.

= 1427 rpm

T = 1.464 ¥ 0.96 ¥ 200 = 281 Nm11.8

Ea = kafw = k ¢a Ia w

12 = k ¢a ¥ 1 ¥ 2 1200

60p ¥

�� ���� ��� ��� �������������

or k ¢a = 0.0955T = Kaf Ia = K¢a Ia

2

44 = 0.0955 ¥ Ia2

or Ia = 20.5 A

250 – 20.5 ¥ 0.6 = 0.0955 ¥ 20.5 ¥ 2

60p ¥ n

or n = 1159 rpm11.9 Ea1 = 250 – 85 ¥ (0.12 + 0.1) = 231.3 V

231.3 = K ¢a Iaw = k ¢a ¥ 85 ¥ 2 600

60p ¥

or K ¢a = 0.0433(a) Ia = 100 A

Ea2 = 250 – 100 ¥ (0.12 + 0.1) = 228 V

228 = 0.0433 ¥ 100 ¥ 2

602p ¥ n

or n2 = 503 rpm(b) Ia = 40 A

Ea2 = 250 – 40 ¥ (0.12 + 0.1) = 241.2 V

241.2 = 0.0433 ¥ 40 ¥ 2

602p n

or n2 = 1330 rpm(c) n2 = 800 rpm or 83.78 rad/s

Ea2 = 0.0433 ¥ Ia2 ¥ 83.78 (i)

Ia2 = 250

0 222- Ea

.(ii)

Eliminating Ia2 in Eqs. (i) and (ii)

Ea2 = 0 0433 83 78

0 22. .

.¥

¥ (250 – Ea2)

or Ea2 = 235.6

Ia2 = 250 235 6

0 22- ..

= 65.6 A

11.10At 600 V1, 80 A, 70 rpm

Ea1 = 600 – 80 ¥ (0.215 + 0.08) = 576.4 V

576.4 = K¢a ¥ 80 ¥ 2 750

60p ¥

or K¢a = 0.092

�������������� ��

At 95A current:Ea2 = 600 – 95 ¥ (0.215 + 0.08) = 572 V

572 = 0.092 ¥ 95 ¥ 2

602p ¥ n

or n2 = 625 rpmT = K ¢a Ia

2

= 0.092 ¥ (95)2 = 830.3 Nm11.11

(a) Ia = 220 AEa = 600 – 220 ¥ 0.15 = 567 V

n = 600 ¥ 507480

= 709 rpm

480 = Kaf ¥ 2 600

60p ¥F

HIK = K ¢a ¥ 220 ¥

2 60060

p ¥FH

IK

or K ¢a = 0.037T = K ¢aIa

2 = 0.0347 ¥ (220)2 = 1681 Nm

(b) Ia = 300 = 600R( )total

or R(total) = 2WR(ext) = 2 – 0.15 = 1.85 W

T(start) = 0.0347 ¥ (300)2 = 3132 NmObservation: With an armature current which is 36% more than the full-load current,the motor give 186% full-load torque. The series motor has therefore excellent startingcharacteristic and is capable of starting high torque loads.

11.12(a) Ia = 220 A

Ea = 600 – 220 ¥ 0.15 = 567 V

n = 600 ¥ 567450

= 756 rpm

450 = Kaf ¥ 2 600

60p ¥

or Kaf = 7.162T = Kaf Ia = 7.162 ¥ 220 = 1576 Nm

(b) R (ext) = 1.85 W (as before)

518 = Kaf ¥ 2 600

60p ¥

Kaf = 8.244T(start) = 8.244 ¥ 300 = 2473 Nm

Remark: Because of saturation torques have reduced

�� ���� ��� ��� �������������

11.13From the solution of Example 11.13

KaKf Nsc = 0.0955T = 59.7 Nm (field halves in series)

Field halves connected in parallels:

59.7 = (Ka Kf Nsc/2) Ia2 =

0 09552

. ¥ Ia

2

or Ia = 35.36 AEa = (Ka Kf Nsc/2) Ia

w

250 = 0 0955.x

¥ 35.36 ¥ w

or w = 148 rad/s

or n = 148 60

2¥p

= 1413 rpm

11.14When cold (20°C ):

Ra = 0.15 W, Rf = 200 WAt no load; Ea1 ª V = 500 V

n1 = 1000 rpm

If 1 = 500200

= 2.5 A

Ea = kaf n = k ¢e If n

Substituting values500 = K ¢E ¥ 2.5 ¥ 1000

or K ¢E = L/5when had (20° + 40° = 60° C):

R a = 0.5 ¥ 234 5 60234 5 20

.

.++

= 0.579 W

Rf = 200 ¥ 234 5 60234 5 20

.

.++

= 232 W

If 2 = 500232

= 2.18 W

Ea2 = 500 – 70 ¥ 0.579 = 459.5 V

459.5 = 15

¥ 2.16 ¥ n2

or n2 = 1063.6 rpm11.15 Neglecting windage, friction and iron losses the load torque equals developed torque.

With linear magnetisation characteristicEa = K ¢e ¥ If ¥ n (i)T = k ¢T ¥ If ¥ Ia (ii)

�������������� ��

At n1 = 800 rpmEa1 = 220 – 40 ¥ 0.3 = 208 V

If 1 = 220200

= 1.1 A

Substituting in Eqs (i) and (ii)208 = K ¢e ¥ 1.1 ¥ 800 (iii)

T = K ¢T ¥ 1.1 ¥ 40 (iv)At n2 = 1050 rpm

Ea2 = 220 – 0.3 Ia2 = K ¢e ¥ If 2 ¥ 1050 (v)T = K ¢T ¥ If 2 ¥ Ia2 (vi)

Dividing Eq. (v) by (iii) and (vi) by (iv)

1050

2082I f

= 220 0 3

11 8002-

¥.

.Ia (vii)

I If a2 2

11 40. ¥= 1 (viii)

Solving Eqs (vii) and (viii)Ia2

2 – 733.3 Ia2 + 36400 = 0or Ia2 = 53.65 A, 679.7 AThe higher value of current i not an acceptable solution as the motor would operate at toopoor on efficiencyThus Ia2 = 53.65 A

If 2 = 11 4053 65.

.¥

= 0.82 A

Hence

Rf2 = 2200 82.

= 268 W

Rf (ext) = 268 – 200 = 68 W11.16 Since field excitation remains constant

Ea = K¢E n (i)T = K ¢T Ia (ii)

TL = KLn2 (iii)At n1 = 1500 rpm

Ea1 = 250 – 35 ¥ 0.2 = 243 V (iv)T1 = K ¢T ¥ 35 (v)

At n2 = 1200 rpm

(i) Ea2 = 243 ¥ 12001500

= 195 V

T2 = K ¢T ¥ Ia2

�� ���� ��� ��� �������������

But T = TL

Thus

TT

2

1

FHG

IKJ =

Ia2

35 = 1200

1500

2FH IKor Ia2 = 22.4 ANow 195 = 250 – 22.4 ¥ (Ra (1 ¥ t) + 0.12)or Ra(1 ¥ t) = 2.25 WLoss in external armature resistance

= (22.4)2 ¥ 2.25 = 1129 W(ii) Ea2 = 195 V as calculated in part (i)

K ¢T ¥ 35 = K ¢T ¥ Ia2

or Ia2 = 35 A195 = 250 – 35 ¥ (Ra(ext) + 0.2)

or Ra(ext) = 1.37 WLoss in external armature resistance

= (35)2 ¥ 1.37 = 1678 W11.17 Assume linear magnetisation characteristic

Ea = K ¢E If n ; T = K ¢T If Ia

(1) Motor operating at Tfe at 1500 rpmEa1 = 115 – 25 ¥ 0.3 = 107.5 V

107.5 = K ¢E ¥ If 1 ¥ 1500 (i)Tf e = K ¢T ¥ If 1 ¥ 25 (ii)

(2) n2 = 1400 rpmEa2 = K¢e ¥ If 2 ¥ 1400 = 115 – Ia2 (0.3 + 0.6)

or 115 – 0.9 Ia2 = K ¢e ¥ If 2 ¥ 1400 (iii)Dividing Eq. (iii) by Eq. (i)

115 0 9107 5

2- ..

Ia = I

If

f

2

1

FHG

IKJ =

14001500

LetI

If

f

2

1

= k

\ 115 – 0.9 Ia2 = 100.33 k (iv)

T2 = 12

Tf l = 12

¥ K ¢T If 1 ¥ 25

K ¢T If2 Ia2 = 12

K ¢T If 1 ¥ 25

or Ia2 = 12.5/k (v)

�������������� ��

Substituting Eq. (v) in Eq. (iii)

115 – 0.9 ¥ 12 5.

k= 100.33 k

100.33 k2 – 115 k + 11.25 = 0or k = 1.037, 0.108Smaller value of k is rejected because it would mean a very large reduction in fieldcurrent resulting in heavy current in the armature circuit which is practically notacceptable (efficiency would be too low)

k = I

If

f

2

1 = 1.037 =

R

Rf

f

1

2

orR

Rf

f

2

1= 0.965

Thus a reduction of 3.5% is needed in the shunt field resistance11.18 For maximum efficiency of 215 A input current

(2/5)2 ¥ Ra = 3300or Ra = 0.071 WNote: Less resistance would mean larger copper conductor cross-section and therefore morevolume of copper to be used in the machineNow PL = 2 ¥ 3300 = 6600 W

Pn = 230 ¥ 215 = 49.45 kW

h = 49 45 6 6

49 45. .

.-

= 86.65%

11.19IL = 79.8 A, If = 2.6 A

Ia = 79.8 – 2.6 = 77.2 A

Ea = 220 – 77.2 ¥ 0.18 = 206 V

206 = f ¥ ¥ ¥ FH IK

1200 62060

42

(a) or f = 8.31 ¥ 10–3 or 8.31 mWb(b) EaIa = Tw

or T = 206 77 22 1200

60

¥¥

.p = 126.6 Nm

(c) Net output = 15000 WMech power developed = 206 ¥ 77.2 = 15903 W

Rotational loss = 15903 – 15000 = 903 W

�� ���� ��� ��� �������������

(d) Armature copper loss= (77.2)2 ¥ 0.18 = 1073 W

Field copper loss = 220 ¥ 2.6 = 572 WTotal copper loss = 1073 + 572 = 1645 W

Total loss = 1645 + 903 = 2548 W

h = 1515 2 548+ .

¥ 100 = 85.5%

CHAPTER 12

12.3No-load voltage = 400 V

If (nl ) = 5 AAt 125% full-load, 0.8 pf lagging

Ia = 62 1000 1 25

3 400

¥ ¥¥

. = 112 A

Ia = 112 –– 36.9° A

Ef = 4003

+ 112 –– 36.9° ¥ j1.08

= 303.65 + j96.76Ef = 318.6 V or 552 V (line)

By linear interpolation

If = 7.5 + 2 560. ¥ 32 = 8.83 A

12.4(i) Unity pf

Ia = 25 –0° A

Ef = 231 – j6.1 ¥ 25 –0° = 231 – j152.5

Ef = 276.8 V or 479.4 V(line)

Pe = 3 ¥ 400 ¥ 23 ¥ 1 = 17.32 kWQe = 0

(ii) 0.8 pf leading

Ia = 25 –36.9°

Ef = 231 – j6.1 ¥ 25 –36.9° = 322.6 – j122

Ef = 345 V or 579 V (line)

Ia

Vt

xs

Ef–

– ++

�������������� �

Pe = 3 ¥ 400 ¥ 25 ¥ 0.8/1000

= 13.86 kW

Qe = + 3 ¥ 400 ¥ 25 ¥ 0.6/1000

= + 10.39 kVAR12.5

(a) Vt = Ef = 113

kV

3 113

4 22

¥ FHG

IKJ

FHG

IKJ

. sin d = 15

or d = 31.4°

(b) Ia xs = 2 ¥ 113

sin 15.7° kV = 3.44 kV

Ia = 34404 2.

= 819 A

q = 15.7°pf = cos q = 0.96 leading

(c) Qe = + 3 ¥ 11 ¥ 819 sin 15.7°= + 4222 kVAR

12.6

(a) 3 ¥ 12.5 ¥ 60 cos q = 1050

cos q = 0.81 pf leading

(b) Ia = 60 –36.1°

Ef = 12 53. – (2 + j48) ¥ 60 –36.1°/1000

= 8.86 – j2.4Ef = 9.18 kV or 15.9 kV(line)

(c) Mechanical power developed

= 1050 – 3 60 2

1000

2¥ ¥( )

= 1028.4 kWns = 1000 rpm, ws = 157 rad/s

T = 1028 4 1000

157. ¥

= 6650 Nm

(d) Vt = 7.22 kV, Ef = 9.18 kV

15.7°

31.4°q

Ia

I xa s

VL =

Ef =

113

113

Ia

VtEf

–

+

–

+

2 W 48 W

���� ��� ��� �������������

Iaxs = 9.18 – 7.22 = 1.96 kV

Ia = 196 1000

48. ¥

= 40.8 A

pf = cos 90° = 0° leading

Ia

VtEf

Ef

Vt

Ia

–

– ++

48 W

jI xa s

jI xa s

12.7

Ef = 13 23. = 7.62 kV (= OC Voltage)

Vt = 11 53. = 6.64 kV

At max load d = 90°

Pe(max) = 3 ¥ 7 62 6 64

120. .¥

= 1.264 MW

ns = 1500 rpm, ws = 157 rad/s

T = 1264 10

157

6. ¥ = 8051 Nm

From phase diagram

Iaxs = ( . ) ( . )7 62 6 602 2+

= 10.09 kV

Ia = 10 09 1000

120. ¥

= 84.1 A

pf = cos q = 6 6210 09

..

= 0.656 lag

12.8

Vt = Ef = 223

= 12.7 kV

Ia = 200 1003 12 7

¥¥ .

= 5249 –0°

Ef = 12.7 + 52491000

¥ j1.5 = 12.7 + j7.87

= 14.94 –31.8° kV

q7.62

6.64

Ia

I xa s

Ia

VtEf–

– ++

1.5 W

Ia

V Lt 0°E Lf –d–

– ++

120 W

�������������� �

(i) Ef is increased toE ¢f = 14.94 ¥ 1.15 = 17.18 kW

3 ¥ 1718 12 7

1 5. .

.¥

sin d = 200

d = 27.3°From the phase diagram

y = 17.18 sin 27.3° = 7.88x = 17.18 cos 27.3° – 12.7

= 2.57

Iaxs = ( . ) ( . )2 57 17 882 2+ = 8.28 kV

Ia = 8 29 1000

15.

.¥

= 5527 A

cos q = 7 888 29..

= 0.95 lag

(ii) At E ¢f = 17.18, turbine power increased to 250 MW

3 ¥ 1718 12 7

1 5. .

.¥

sin d = 250 MW

d = 35°From the phasor diagram

y = 17.18 sin 35° = 9.85x = 17.18 cos 35° – 12.7 = 1.37

Iaxs = ( . ) ( . )9 85 1 372 2+ = 9.94 kV

Ia = 9 94 1000

1 5.

.¥

= 6627 A

pf = cos q = 9 859 94..

= 0.99 lag

12.9 750 kW, 0.8 leading

Ia = 750

3 3 3 0 8¥ ¥. . –36.9° = 164 –36.9° A

Ef = 33003

– 164 –36.9° ¥ j5.5

= 2446 – j721Ef = 2550 V

1000 kW, same excitation

Pe = V E

xt f

s sin d

Ia

V Lt 0°E Lf –d–

+

xs

q

q

Ia

x

y

Vt = 12.7

E ¢f = 17.18

27.3°

q

q

Ia

x

y

12.7

17.18

35°

� ���� ��� ��� �������������

1000 10003¥

= 1905 2550

5 5¥.

sin d or d = 22.2°

From the phasor diagram(Vt leads Ef for motoring)

x = 2550 sin 22.2° = 963y = 2550 cos 22.2° = 456

Iaxs = x y2 2+ = 1065.5 V

Ia = 1065 55 5

..

= 193.7 A

pf = cos tan-FH IK1 456

963 = 0.9 leading

12.10

(a) Vt = 220003

= 12702 V

Ia = 10003 22¥

= 26.24 V

Iaxs = 26.24 ¥ 250 = 6560 VFrom the phasor dirgram

x = Ef sin d = 0.259 Ef

y = 12.702 –Ef cos d = 12.702 – 0.966 Ef

(6560)2 = (0.259 Ef)2 + (12.702 – 0.966 Ef)

2

or Ef2 – 24.54 Ef + 118.3 = 0

or Ef = 17.95 kV; 6.59 kV (rejected; would give lagging pf )= 31.1 kV(line)

pf = cos tan–1 yx

FH

IK ; x = 4.05, y = 4.64

= 0.708 leading

(b)V E

xt f

s= pe

12 702

250

. E f= 800

31

1000¥

or Ef = 5.25 kV= 9.09 kV (line)

Iaxs = ( . ) ( . )12 702 5 252 2+ = 13.744 kV

22.2°

q 1905

2550x

y

Ia

Ia

VtEf–

+

xs

q

q

12.702

x

y

Ia

Ef

6.56

q12.702

Ia

Ef

I xa s

(min)

�������������� �

Ia = 13 744 1000

250. ¥

= 54.98 A

pf = cos q = 5 2313 744

..

= 0.382 lagging

12.11

(a) ns = 120 50

8¥

= 750 rpm

n = (1 – 0.03) ¥ 750 = 727.5 rpm, w = 76.18 rad/sTs ¥ 76.18 = 40 ¥ 1000

Ts = 525 Nm

(b) Motor input = 40

0 896. = 44.64 kW

3 ¥ 440 ¥ 68.9 ¥ pf = 44.64 ¥ 1000

pf = 0.85 lag12.12

As per Eq. (12.46) (stator impedance negligible)

T = 3 12

2w s

s V ar

◊( / )

= KT V 12s

or T = KT V12 ¥ 0.04

1.25 T = KT (0.8V1)2 ¥ s

Dividing

0 640 64..

s= 1.25

or s = 0.078ns = 1000 rpmn = (1 – 0.078) ¥ 1000

= 922 rpm12.13 Windage and friction losses are included in the no-load test data. The torque developed is

therefore assumed to the net torque.(a) ns = 1500 rpm

n = 1500 (1 – 0.05) = 1425 rpm, w = 149.2 rad/sMechanical output = 95.6 ¥ 149.2 = 14.26 kW

(b) Pm = 3I ¢22 r2 =

ss1 -

FHG

IKJ Pm

= 0 050 95..

¥ 14.26 = 751 W

� ���� ��� ��� �������������

(c) Stator iron, windage and friction loss

= 3 ¥ ( / )

.400 3

242 2

2

= 661 W

h = 14 2614 26 0 751 0 601

.. . .+ +

= 91.1%

12.14

Pm = 3I ¢22 r¢2

11

s-FH IK

copper loss = 3I ¢22 (r1 + r ¢2)

Pm = 3I ¢22 r

sr r2 1 2

1 1-FH IK + + ¢LNM

OQP( )

= 3I ¢22 r

rs12+ ¢F

HGIKJ

\ h = ( )1 1

1 2

- ¢+ ¢s r

sr r

Substituting parameter values

h = 0 496 1

0 975 0 496. ( )

. .-

+s

s

h (s = 0.04) = 0 496 1 0 04

0 975 0 04 0 496. ( . )

. . .-

¥ + = 83%

h (s = 0.1) = 0 496 1 0 1

0 975 01 0 496. ( . )

. . .-

¥ + = 75.2%

h (s = 0.5) = 0 496 1 0 5

0 975 0 5 0 496. ( . )

. . .-

¥ + = 25.2%

Remark: At high values of slip h drops off very sharply. The induction motor should beoperated at two steps (2 to 8%)

12.15No-load test

ri = ( )

.400444 5

2

= 360 W (inclusive of windage and friction loss)

cos q0 = 444 53 400 3 5

..¥ ¥

= 0.183

xm = ri / tan q0 = 67 WBlocked rotor test

r1 + r ¢2 = 2220

3 16 7 2¥ ( . ) = 2.65 W

�������������� �

r1 = 1.25 W, r ¢2 = 1.4 W

z = 200 3

16 7/.

= 6.91 W

x1 + x ¢2 = ( . ) ( . )6 91 2 652 2- = 6.38 W

Running

ns = 1000 rpm, ws = 2 1000

60p ¥

= 104.7 rad/s

n = 935 rpm s = 0.065

r1 + r ¢2/s = 1.25 + 1 4

0 065.

. = 22.79 W

¢I2 = 400 3 022 79 6 38

/. .

– ∞+ j

= 9.76 –– 15.6° = 9.40 – j2.62

I0 = 3.5 –– cos–1 0.18.3 = 3.5 –– 79.5°

= 0.64 – j3.44 A

I1 = I I0 2+ ¢ = 10.04 – j6.06 A

= 11.73 –– 31.1° AI1 = 11.73 A, pf = 0.856 lag

Pi(input) = 3 ¥ 400 ¥ 11.73 ¥ 0.856 = 6.956 kW

P0 (output) = 3I ¢22 r ¢2

11

s-FH IK

= 3 ¥ (9.76)2 ¥ 1.4 10 65

1.

-FH IK = 5.75 kW

h = 5 7556 956..

= 82.7%

Breakdown torque

smax, T = ¢

+ + ¢

r

r x x

2

12

1 22( )

; (Eq. (12.46))

= 1 4

125 6 382 2

.

( . ) ( . )+ = 0.215

Now for s = 0.215z = (1.25 + 0.14/0.215) + j6.38

= 10.05 –39.4°

� ���� ��� ��� �������������

I ¢2 = 400 310 05

/.

= 23 A

Tmax = 3 23 1 4 0 215

104 7

2¥ ¥( ) . / ..

= 98.7 NmSpeed = 1000 (1 – 0.215) = 785 rpm

12.16

(a) Ts = 3 12

2

1 22

1 22w s

V r

r r x x◊ ¢

+ ¢ + + ¢( ) ( ); Eq. (12.41)

with s = 1 (i)Parameter values as calculated in Example 12.8 are:

r1 = 0.42 W, r ¢2 = 0.463 W; r1 + r ¢2 = 0.883 Wx1 + x ¢2 = 2.25 W

Substituting the values in Eq. (i)

250 = 378 54

400 3

0 42 2 25

22

22 2.

( / ) ( )

( . ( )) ( . )◊ ¢

+ ¢ +R

R

total

total

= 2037

0 1764 0 84 5 062

2 22

¢+ ¢ + ¢ +

R

R R. . .

or 250 R ¢22 – 2097 R ¢2 + 1265 = 0

or R ¢22 – 8.39 R ¢2 + 5.06 = 0

or R ¢2 = 0.655 W, 7.735 WIt is seen from the figure that with 7.735 W themotor cannot start though it has a Ts of 250 Nm.Thus

R ¢2 = 0.655 WResistance to be added in rotor circuit

= (0.655 – 0.463)/(2.45)2 = 0.032 W(a) (i) Added resistance left in circuit

T = 30 42

12

2

22

1 2w s

V R s

R s x x◊ ¢

+ ¢ + + ¢( ) ( / )

( . / ) ( )

or 250 = 378 54

400 3

0 42

22

22

1 22.

( / ) ( / )

( . / ) ( )◊ ¢

+ ¢ + + ¢R s

R s x x

From the solution of part (a)R ¢2/s = 0.655 or 7.735

But R ¢2 = 0.655 W

0.655 W

7.735 W

250 Nm

�������������� �

\ s = 1 or 0 6557 735..

= 0.085

n = 750 (1 – 0.085) = 686.25 rpm(ii) Added resistance cut and

Again r¢2/s = 0.655 or 7.735But r ¢2 = 0.463 W

s = 0.707 W or 0.06n = 750 (1 – 0.06) = 705 rpm

This illustrates the speed control action of added rotor resistance

(c) T = 3 12

2

1 22

1 22w s

V r s

r r s x x◊ ¢

+ ¢ + + ¢( ) ( / )

( / ) ( ), r ¢2 = 0.463 W

s = 0.085 shift

or 250 = 378 54

3 0 463 0 085

0 42 0 463 0 085 2 25

2 2

2 2.( / ) ( . / . )

( . . / . ) ( . )◊

+ +V

V = 377.3 V(d) Efficiency comparison:

With external resistance in rotor circuit

P0 = 250 ¥ 2 686 25

60p ¥ .

= 17.966 kW

P0 = 3I ¢22 R ¢2 1

1s

-FH IK

17966 = 3I ¢22R ¢2 1

0 0851

.-FH IK

or 3I ¢22 R ¢2 = 1.669 kW

Power lost in iron loss, windage and friction

= Vri

2

= ( )

.400132 2

2

= 1.21 kW

Stator loss cannot be estimated and is ignored in this comparison.

h = 17 96617 966 1 669 1 21

.. . .+ +

= 86.2%

With reduced stator voltagePower lost in iron loss, windage and friction

= ( . )

.377 3132 2

2

= 1.077 kW

P0 = 17.966 kW3I ¢2

2R ¢2 = 1.669 kW

� ���� ��� ��� �������������

h = 17 96617 966 1 669 1 077

.. . .+ +

= 86.74%

Remark: Observe that efficiency is slightly higher for the reduced stator voltage case.However, the equipment to reduce voltage would be more expensive than the resistanceto be added in rotor circuit and would have associated loss not considered above.

12.17(a) From Eq. (12.46)

T = 3 12

2

1 22

22w s

V r s

s r s x x◊ ¢

+ ¢ + + ¢( / )

( / ) ( )

Load torque, TL = 75 ¥ nns

FHG

IKJ = 75 (1 – s)2

At steady speed

3 12

2

1 22

1 22w s

V r s

r r s x x◊ ¢

+ ¢ + + ¢( / )

( / ) ( ) = 75 (1 – s)2

CHAPTER 14

14.1 Limiting error (dA) = guarantee error ¥ full-scale readingdA = 0.01 ¥ 150 = 1.5 V

\ the limiting error is (1.5/85) ¥ 100 = 1.765%

14.2 Resistance in series, R = G2/2 KJ

K = 2.4 ¥ 10–6 NmJ = 1.6 ¥ 10–7 kg m2

G = BldN = 0.12 ¥ 0.02 ¥ 0.025 ¥ 250 = 0.015

\ External resistance R = (0.015)2/2 ( . . )2 4 1 6 10 13¥ ¥ -

= 181.546 W14.3 If R1 and R2 are the shunt resistances, then, for a range of 15 A,

R2 ¥ 14.975 = 5 ¥ 0.025R 2 = 0.0083 W

and for a range of 10 A(R1 + R2) ¥ 9.975 = 5 ¥ 0.025

R1 = 0.0042 W14.4 Error in W1 = 0.5 ¥ 1500/100 = 7.5 W

Error in W2 = 0.5 ¥ 1500/100 = 7.5 W

tan q = 3 (W1 – W2)/(W1 + W2)

q = tan–1 3 (500 ± 15)/(1500 ± 15)

��������������

q = tan–1 3 (500 ± 3%)/(1500 ± 1%) = tan–1 3 (1/3 ± 4%)

Hence, three different values of q are

q = tan–1 3 (1/3 + 4%) = 30.98°

q = tan–1 3 (1/3 – 4%) = 29°

q = tan–1 3 (1/3) = 30°

The three different cosines of q arecos 30.98° = 0.8573 ; cos 29° = 0.8746

cos 30° = 0.866\ maximum error in computing cos q is 0.0087.

14.5 Let f be the frequency of the ac voltage. From the circuitshown in Figure we have

Z1 = 1000 W | | 1/jw (0.16 mF)Z2 = Rx – j/w (0.65 mF )Z3 = 1000 W, Z4 = 500 W, Z1Z4 = Z2Z3

or 1 = 2Z2Y = 2[RX – j/w (0.65 mF )] ¥ [(1/1000) + jw (0.16 mF )]Equating real and imaginary parts

1 = 2RX /1000 + 0.16/0.652Rx w (0.16 mF ) = 2/1000 w (0.65 mF )

\ Rx = (1 – 0.16/0.65) ¥ 500 W= 376.92 W

f = 803.857 MHz.

14.6 From Fig. P14.6 we haveZ1Zx = Z2Z3

or Rx – j/wCx = (R2/( jwC3)) (1/R1 + jwC1)Equating real and imaginary terms

Rx = R2C1/C3 = 200 WCx = C3R1/R2 = 5 ¥ 10–9 F

D = Rx/Xx = wCxRx = 6.28 ¥ 10–3

14.7 (a) From the condition of bridge balance, we get(R + jwL) = R1R2 (1/R3 + jwC)

Equating the real and imaginary terms, we obtainR = R1R2/R3, L = CR1R2

(b) Percentage error for R = ± 0.6%Percentage error for L = ± 0.5%

14.8 Four diodes bridge is shown in Fig. P14.8Now PMMC meter will read the average value of the signal. Hence the range for voltmeter is

E = 2 Erms = 282.8 V

DAE

Z3 Z4

Z2

Z1

C

D

B

DE1

R2

R1

C3C1

C

B

Z R j wCx x x= = /

Fig. P14.6

� ���� ��� ��� �������������

and maximum current through the meter is 30 mA.\ the internal resistance required for the meter

Ri = 282.8/30 ¥ 10–3

= 9426.67 WThus, the external resistance is given by

Rx = Ri – Rm – 2 Rd

where Rm is the internal resistance of ammeter andRd the forward resistance of a diode.\ Rx = 9426.67 – 150 – 600 = 8676.67 W

14.9 (a) f = sin–1 (0/5) = 0°(b) f = sin–1 (3/5) = 36.87°(c) f = sin–1 (3/5) = – 36.87°

14.10

Erms = 1 2/T e dtzFrom Fig. e = 50 t V

\ Erms = 200/ 3 V

Eav = (2/T) e dtT

0z = 100 V

\ k = 2/ 3 = 1.155

Now, the meter scale is calibrated in terms of the rms value of a sine-wave voltage, whereEm = kEav

\ Erms = 1.11 Eav.But for sawtooth wave,

Erms = 1.155 Eav.\ the meter indication for the square-wave voltage is low by a factor of

ksine wave/ksaw tooth = 1.11/155 = 0.96104Thus, % error = (0.96104 – 1) ¥ 100/1 = – 3.896%The meter indication is low.

14.11 For a 10 V peak and 25% duty cycle.

Erms = 1 102

0

4 0 5

Tdt

T / .zLNM

OQP = 1 100

4

0 5

TT¥L

NMOQP

.

= 5 V

For 10 V dc supply

Erms = ( / )1 102T dtz = 10 V

14.12 Sensitivity = 1/Ifsd

\ Ifsd = 10–4 A; Rm = Vs/Ifsd = 2 MW

ac input

To d

c am

plifi

er

Fig. P14.8

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The diode will rectify the signal and the meter will measure the average value.\ Imax = 800/4 ¥ 106 = 2 ¥ 10–4 AThe peak value of the voltage across the meter = 400 Vand the meter will read an average value of 200 V.

14.13 Self capacitance, Cd = (C1 – 4C2)/3C1 = 450 pF and C2 = 100 pF

\ Cd = 50/3 pF.14.14 Cd = (C1 – 4C2)/3 C1 = 500 pF and C2 = 60 pF

\ Cd = 260/3 pF.14.15 Q = 1/wCR = 1/[2p 106 ¥ 70 ¥ 10–12 ¥ 12] = 189.56

Insertion resistance = 0.1 W\ Q2 = 1/wC (R + 0.2) = 186.45Thus, % error = (189.56 – 186.45) ¥ 100/189.56 = 1.65

14.16 Sensitivity = (lowest full scale value) ¥ ResolutionResolution = 10–4

\ Sensitivity = 10 mV ¥ 10–4 = 10–6 V14.17

(a) Reading is 6.00 V\ 0.5 % of reading = 0.030 V

Now 3 12 digits meter can show four digit. Hence, 6.00 will be shown as 06.00 V

\ 2 digit errror is ± 00.02 VThus, possible error is ± 00.05 V

(b) Reading is 0.20 V\ 0.5 % of reading = 0.001 V

Now 3 12 digits meter can show four digits.

Hence, 0.20 will be shown as 00.20 V\ 2 digit error is ± 00.02 VThus,possible error is ± 00.021 V

(c) Error as a percentage of reading in part (b)= 0.041 ¥ 100/0.20 = 20.5 %