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Electrical EngineeringElectromagnetic Theory
WORKBOOKWORKBOOKWORKBOOKWORKBOOKWORKBOOK
2016
Detailed Explanations ofTry Yourself Questions
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Static Electromagnetic Fields1T1 : Solution
Consider the coaxial cone of figure, where the gap serves as an insulator between the two conductingcones. Here V depends only on , so Laplaces equation coordinates becomes
2 V =
=
GapPotential ( ) due to conducting cones.V
1
Vo
z
2
Since r = 0 and = 0, are excluded, we can multiply by r2 sin to get
= 0Integrating once gives
= A
or
=
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Integrating this results in
V =
=
=
=
= Aln(tan /2) + B
We now apply the boundary conditions to determine the integration constants A and B.V( = 1) = 0 0 = A ln(tan 1/2) + B
or B = A ln(tan 1/2)
Hence V =
l
Also V( = 2) =
=
l
or A =
l
Thus, V =
l
l
E =
= = = l
Taking 1 = /10, 2 = /6, and Vo = 50 gives
V =
=
ll
l
E =
T2 : Solution
1
01
32
90
y
x
l =
l =
l =
l =
= =
+
=
+ = 4
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4 Electrical Engineering Electromagnetic Theory
T3 : Solution
at (3, 2, 1) due to 10 C/m2 is
=
=
xx
due to 20 C/m2 is
= ( )
=
Total,
=
x =
x
T4 : Solution
= + xTangential components are continuous across the boundary ie
=
= + +xNormal components are continuous
D1z = D2z =
2.(2) = 4 E2z
= 1
= + x V/m
T5 : Solution
= I
= ( ) x
= ( ) x = ( ) ( ) x
= ( ) +x N/mT6 : Solution
Let,
r = 3, , E D2 2
r = 2, , E D1 1
V2
V1
+
+
20 V
d = 1 cm
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A = 100 cm2
From boundary conditions,D1 = D2
2 0E1 = 3 0E2
E2 =
Also V2 + V1 = 20 V E2 d + E1 d = 20 V
+ = 20
E1 =
=
= 1200 V/m
D1 = D2 = 2 0 E1 = 21.2 nC/m2
T7 : Solution
=
=
0r =
r =
r =
=
= 1.998 2
T8 : Solution
= tan2 = tan 60 = 1.732
= 240 =
+ =
( )
+
r =
= 1.234
Complex permittivity c =
=
= (10.91 j18.9) 1012 F/m x = 10.91
y = 18.9
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Electromagnetic WavesPropagation3
T1 : Solution
We know thatfor wave number K = is propagation constant.
= +
= ( ) +but for metals
>> 1
=
=
=
=
+
=
+
K = =
Intrinsic impedance =
= =
+
In the case of a good conductor (metal)
metal =
= = +
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T2 : Solution
Loss tangent = tan (2 )Given = 130 + j 75 =
= tan (60) =
T3 : Solution
= x
=
=
x
x
=
=
=
= =
=
= +
T4 : Solution
i=
+ =
+
=
HT = ( )
i
= ( ) ( )
=
T5 : Solution
= 0.01, = 2 , are always positive in lossy medium. = + j = 0.01 + j2 (m1)
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8 Electrical Engineering Electromagnetic Theory
T6 : Solution
Given,
=
=
+ x
Let,
=
+ + xx
From boundary conditions, normal components of B are continuous i.e.B1n = B2n
B1z = 4
Now,
=
= [ ]
+ x
=
=
+ + x x
Again from boundary conditions,
( ) =
= +
+ x x = [ ]
+
( ) =equating terms on both sides
x =
+
B1x = 9 12 = 3
= +x Wb/m2
T7 : Solution
Medium (1) :Medium (1) :Medium (1) :Medium (1) :Medium (1) : Dielectric (c = 4), 1Medium (2) :Medium (2) :Medium (2) :Medium (2) :Medium (2) : Air, 2wave from the dielectric enters free space
So, =
+ =
+ =
+ =
Power density for z > 0 (free space) is
Pt = ( ) i=
i(Pi = incident power density (in dielectric))
=
= 235.8 mW/m2
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Transmission Lines4T1 : Solution
(i) Zin =
+
+
=
+
+ =
+
+ = 33.74 + j24.07
(ii)Reflection coefficient at input end
i =
+
but here ZL = 100 Z0 = 50
i =
+ =
=
(iii) Net input impedance seen from the transmission end of line isZin = 33.74 + j 24.07
Net equivalent diagram of line is
Zin0.5 A 0 100
Current through 100 resister is
=
+
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10 Electrical Engineering Electromagnetic Theory
i =
+
i = 0.122 0.02198 j = 0.12410.21
Current through Zin i1 = 0.5 i
= 0.378 + 0.02198 j
= 0.37863.327
So power dissipated in 100 resistance is
P1 = i2 R= (0.124)2 100 = 1.537 Watt
Power dissipated is 25 resistance is equal to power dissipated in Zin ( line is lossless)
P2 = i2 Real part of ZinP2 = (0.3786)
2 33.74
P2 = 4.836 W
T2 : Solution
Zin =
+ +
ll
l =
=
tan l =
= 1
Zin = 75
+
= 75 j 5
=
T3 : Solution
Zin at input of Z02 line is
Zin 2 =
Similarly, Zin1 =
= Z0
Zin2 =
=
Z01 =
=
= 53
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T4 : Solution
For a lossless line
Z0 =
= 100
v =
= 3 108 m/s (Vacuum)
L =
=
= 0.33 106 H/m
T5 : Solution
= sZ0
=
= s2 =
= Where s is the standing wave ratio (VSWR)
