Section 6 5 TRƯỜNG ĐIỆN TỪ

No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]

Introduction to Electromagnetic Fields,

to supplement “Elements of Engineering

Electromagnetics, Sixth Edition”

by

Nannapaneni Narayana Rao

Edward C Jordan Professor of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign, Urbana, Illinois, USA

Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India

6.5 Lines with Initial Conditions

C D

V, V

t  l

2v p

50 0

p l/vp 3l/2vp

z

(–) (+)

V, V

50 50

(+) (–)

I, A

1 0 –1

V, V

l z

(–) (+)

(–)(+)

aa

z l

(–)

I, A

1 0 –1

(+) (+)

I, A

0.5 0 –0.5

[V]RL, V

75

18.75 37.5

[V] RL

2 4 0

z = 0

1 3 5

75 37.5 18.75

–25 –12.5

–25 –12.5 –6.25

100 50

Energy Storage in Transmission Lines

w e, Electric stored energy density =

W e, Electric stored energy =

V 20

Z0 T

w m, Magnetic stored energy density =

W m, Magnetic stored energy =

Check of Energy Balance

Initial stored energy

W e  W m

12

V 20

Z0 T  12 I 20Z0T

12

(100)2

50 10–3  0

0.1 J

+ –

t = 0+:

60  V – 60  V 0.6  I – = 0.6  I+  60 + V +

Solving, we obtain

V – V   – 15

I – 0.15

I  – 0.3

aa

0 1 2 3

60 –15

45 40

z = 0

60 45 40

–15 45

0 1 2 3

0.6 –0.3

0.3 0.4

0.1 0.4

z = l+ z = 2l

0.6 A

t, s

0 1 2 3

z = 0

0.6 0.75 0.8

0.15 0.75 0.05 0.8

z = l

0.6 A

Current

 = 0  = 0, Ceff = 0.5C = 1

t = 3 s + : New steady state

100 

+ –

+ –