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EE2111 Homework #7 A03
Assigned 09 October 03
Due 15 October 03
Problem 1. Hambly 15.29
Solution:
2
N 2 200
L

 80 mH
 5  10 5
Problem 2. Hambly 15.30
Solution: The impedance of an inductance to a sinusoidal current
is
V
120
ZL  L 
90 0  j 90
IL
1
L
ZL
j 90

 0.3183 H
j j 2  60
The reluctance is found by rearranging equation 15.25
N 2 500


 78.54  10 4
L
0.3183
2
Rearranging equation 15.21, we have
l
20  10 2


 509.3  10 6
4
4
A 78.54  10  5  10
 509.3  10 6
r 

 405.3
0
4  10 7
Problem 3. Hambly 15.35
Solution: Because coil 2 is open circuited, i2 t   0 . Thus we have
L1
di1 t 
 12
dt
Integrating and using the fact that i1 0  0 , we have
i1 t   12 / L1 t  120t
Substituting into equation 15.37, we have
v 2 t   M
di1 t 
 120 V
dt
Problem 4. Hambly 15.45
Solution: If residential power was distributed at 12 V rather than
120 V higher currents (by an order of magnitude) would be
required to deliver the same amounts of power. This would require
much larger wire sizes to avoid excessive power losses in the
resistances of the conductors.
Problem 5. Hambly 15.48
Solution: If we tried to make the 25Ω load look like 100 Ω by
adding 75 Ω in series, 75% of the power delivered would be
dissipated in the 75 Ω resistor. On the other hand, when using the
transformer, virtually all of the power taken from the source is
delivered to the load. Thus, from the standpoint of efficiency, the
transformer is a much better choice.
Problem 6. Hambly 15.49
Solution:
I L  Vs / Rline  RL   5 A rms
PL  Rline I L  250 W
2
(a) PS  VL I L  500 W
PL  RL I L  250 W
2
Efficiency  PL / PS   100%  50%
(b) The step down transformer has a turns ratio of N=10:1. The
load as seen on the primary side of the transformer is RL '  1000  .
Reflecting both the line resistance and the load to the primary of
the step-up transformer, we have
Rline '  Rline / N 2  0.1 
RL "  RL ' / N 2  10 
Then the source current is I S  VS / Rline ' RL "  0.9901 A rms .
The power dissipated in the line is
2
Pline  Rline I S  9.803 W
The power delivered to the load is
2
PL  RL " I S  980.3 W
The power delivered by the source is
PS  VS I S  990.1 W
The efficiency is
E  PL / PS   100%  99.01%
Problem 7. Hambly 15.52
Solution:
(a) Let N1  200 turns and N 2  300 turns (i.e. the total number
of turns). Then we have
d
dt
d
v2  N 2
dt
and thus
v1 N1 2


v2 N 2 3
v1  N 1
(b) The net mmf is:
F  N1 i1  i2   N 2  N1 i2  0
Simplifying, we have
i1 N 2 3


i2 N1 2
Problem 8. A lossless coaxial cable has characteristic impedance
Z C  100  and a propagation constant of   j10 radians / m at a
frequency of   108 radians/sec. Calculate its inductance and
capacitance per unit length.
Solution:
  j LC  j10
108 LC  10
LC  10 7
Z c  100 
L
C
L  10 4 C
10 4 C 2  10 7
C  10 11  3.16  10 6 F
L  3.16  10  2 H
Problem 9. A lossless transmission line has a characteristic
impedance of 50 Ω. It is terminated in a short circuit. It is 1 meter
long and β is 3 radians per meter. If the short circuit is 10 amperes
rms, calculate the sending end voltage, current, and impedance.
Solution: Let the receiving end be at x=0 and the sending end at
x=1 meter.
VS cosx 
V cos3x 
j S
Z c sin l 
50 sin 3
Since I 0  10 , we find
VS  j10  50  sin 3  j 70.56 V
I x    j
 j 70.56 cos3  -9.8999  9.89991800
50 sin 3
 Z 1  jZ c tan 3   j 7.1273  7.1273 - 900
I S  I 1   j
ZS
Problem 10. A lossless line with characteristic impedance
Z C  300  is terminated in an open circuit. The line is 6.375
wavelengths long. If the input voltage is 100 volts, determine :
(a) The input current.
(b) The output voltage.
(c) The output current.
(d) The input impedance.
Solution:
(a)
2


l  6.375 
6.375  2

l  6.375  2  40.06 radians
For the open ended line
V sin x 
100 sin x 
I x   j S
 j
Z C cosl 
300 cos40.06
I l   j
100 sin 40.06
  j 0.333 A
300 cos40.06
(b)
cosx 
cosx 
 100
cosl 
cos40.06
100
VR  V 0 
 141.4 V
cos40.06
(c)
V  x   VS
IR  0
(d)
Z  x    jZ c tan x 
Z l    j 300 tan 40.06  j 300 