120
PROBLEM SOLUTIONS: Chapter 10
Problem 10.1
part (a):
part (b):
V
rms
=
1
T
_
T
0
v
2
R
(t) dt =
2
T
_
T/4
0
_
4V
0
t
T
_
2
dt =
V
0
6
= 3.67 V
part (c): < p
diss
>= V
2
rms
/R = 9 mW.
Problem 10.2
part (a): The diode does not turn on until the source voltage reaches 0.6 V,
which occurs at time t = (0.6/4V
0
)T = T/60.
121
part (b):
V
rms
=
2
T
_
T/4
T/60
_
4 8.4t
T
_
2
dt = 3.43 V
part (c): < p
diss
>= V
2
rms
/R = 7.8 mW.
Problem 10.3
V
rms
=
2
T
_
T/4
t
d
_
4V
0
t
T
_
2
dt =
32V
2
0
3
_
1
4
3
_
t
d
T
_
3
_
Problem 10.4
Problem 10.5
part (a): Peak V
R
= 310 V.
part (b): Ripple voltage = 25.7 V.
part (c): Time-averaged power dissipated in the load resistor = 177 W.
part (d): Time-averaged power dissipated in the diode bridge = 0.41 W.
Problem 10.6
If v
s
(t) 0, diode D1 is ON, diode D2 is o and the inductor current is
governed by the following dierential equation:
L
di
dt
+Ri = v
s
(t)
of v
s
(t) < 0, diode D1 is o and diode D2 is on and the inductor current is
governed by the dierential equaion:
L
di
dt
+Ri = 0
A simple integration implemented in MATLAB produces the following plot:
122
Problem 10.7
Problem 10.8
part (a): Letting T = 2/
V
dc
=
1
T
_
T/2
0
V
0
sin t =
V
0
= 14.3 V
part (b):
I
dc
=
V
dc
R
= 2.9 A
123
part (c):
part (d):
Problem 10.9
part (a): Letting T = 2/
V
dc
=
1
T
_
T/2
t
d
V
0
sin t =
V
0
2
(1 + cos t
d
)
part (b):
I
dc
=
V
dc
R
=
V
0
2R
(1 + cos t
d
)
124
part (c):
Problem 10.10
part (a): Letting T = 2/
V
dc
=
2
T
_
T/2
t
d
V
0
sin t =
V
0
(1 + cos t
d
)
part (b):
I
dc
=
V
dc
R
=
V
0
R
(1 + cos t
d
)
part (c):
125
part (d):
Problem 10.11
part (a):
(i)
(ii)
V
dc
=
1
_
5/4
/4
V
0
sin d =
V
0
cos
2
=
V
0
2
(iii)
P
load
= V
dc
I
dc
=
V
0
I
dc
2
126
part (b):
(i)
(ii)
V
dc
=
1
_
7/4
3/4
V
0
sin d =
V
0
cos
2
=
V
0
2
(iii)
P
load
= V
dc
I
dc
=
V
0
I
dc
2
The power is negative, hence energy is being extracted from the load.
Problem 10.12
part (a) From Eq. 10.11
I
dc
=
2V
0
R + 2L
s
= 18.3 A
and from Eq. 10.8
t
c
=
1
cos
1
_
1
2I
dc
L
s
V
0
_
= 3.12 msec
part (b): For L
s
= 0
I
dc
=
2V
0
R
= 23.6 A
Problem 10.13
part (a): At 1650 r/min, the generated voltage of the dc motor is equal to
E
a
= 85
_
1650
1725
_
= 81.3 V
127
The motor input power will then be
P
in
= I
a
(E
a
+I
a
R
a
) = 665 W
part (b):
V
dc
=
_
2V
0
_
cos
d
=
_
2
2 115
_
cos
d
= 103.5 cos
d
V
Thus for V
dc
= E
a
+I
a
(R
a
+R
L
) = 90.5 V,
d
= 29.1
.
Problem 10.14
The rated current of this motor is
I
rated
=
P
rated
V
rated
=
1000
85
= 11.8 A
The controller must limit I
dc
to twice I
rated
or 23.6 A. Under this condition,
V
dc
= I
a
(R
a
+R
L
) = 28.5 V.
From part (b) of the solution to Problem 10.13,
V
dc
= 103.5 cos
d
V
and thus the controller must set
d
= 74.0
.
Problem 10.15
The required dc voltage is V
f
= I
f
R
f
= 277 V. From Eq. 10.19,
V
ll,rms
=
V
f
3
2
= 204 V, rms
Problem 10.16
The required dc voltage is V
f
= I
f
R
f
= 231 V. From Eq. 10.20,
d
= cos
1
_
V
f
3
2 V
ll,rms
_
= 39.0
Problem 10.17
part (a): The magnet resistance is suciently small that its voltage drop
can be ignored while the magnet is being charged. The desired charge rate is
di
dt
= 80 A25 sec = 3.2 A/sec
Thus the required dc voltage will be
V
dc
= L
di
dt
= 15.7 V
128
Thus
d
= cos
1
_
V
dc
3
2 V
ll,rms
_
= cos
1
_
15.7
3
2 15
_
= 39.3
part (b): Constant current simply requires a dc voltage of V
dc
= RI
dc
=
3.6 10
3
80 = 0.29 V. Thus
d
= cos
1
_
V
dc
3
2 V
ll,rms
_
= cos
1
_
15.7
3
2 0.288
_
= 89.2
Problem 10.18
part (a):
V
1
=
2
T
_
T
0
v(t) cos
_
2t
T
_
dt =
8
T
_
DT/2
0
V
0
cos
_
2t
T
_
dt
=
V
0
4
sinD = 51.5 V
part (b):
Harmonic number Peak amplitude [V]
1 51.5
2 0
3 6.6
4 0
5 12.7
6 0
7 2.8
8 0
9 5.7
10 0
Problem 10.19
part (a):
V
3
=
3
T
_
T
0
v(t) cos
_
6t
T
_
dt =
4V
0
3
sin (3D)
A
3
= 0 for D = 1/3
129
part (b):
Harmonic number Peak amplitude [V]
1 55.1
2 0
3 0.007
4 0
5 11.0
6 0
7 7.9
8 0
9 0.007
10 0
Problem 10.20
part (a):
Time period S1 S2 S3 S4
0 t <
d
+DT/2 ON OFF ON OFF
d
+DT/2 t < +
d
DT/2 ON ON OFF OFF
+
d
DT/2 t < +
d
+DT/2 OFF ON OFF ON
+
d
+DT/2 t < 2 +
d
DT/2 OFF OFF ON ON
2 +
d
DT/2 t < 2 ON OFF ON OFF
part (b): By analogy to the solution of Problem 10.18, part (a)
I
1
=
I
0
4
sin D
and by inspection
1
=
d
.
part (c):
p(t) =< i
1
(t)v
L
(t) >=
V
a
I
1
2
cos
1
=
V
a
I
0
8
cos
d
Problem 10.21
From Eq. 10.34
(i
L
)
avg
=
[2D1]V
0
R
= 17.65 A
From Eq. 10.29
(i
L
)
min
=
_
V
0
R
_
_
1 2e
T(1D)
+e
_
(1 e
)
= 17.45 A
and from Eq. 10.20
(i
L
)
max
=
_
V
0
R
_
_
1 2e
DT
+e
_
(1 e
)
= 17.84 A
Finally
Ripple = (i
L
)
max
(i
L
)
min
= 0.39 A
