Chapter 2
Problem Solutions
2.1
R 1
K
I
0
10
0
10
I
0.6 V, 20
f
Vr==Ω
γ
()
()
0
0
For 10 V, 10 0.6
1
9.4
10.02
9.22
I
f
R
vv
Rr
v
⎛⎞
== −
⎜⎟
⎜⎟
+
⎝⎠
⎛⎞
=
⎜⎟
+
⎝⎠
=
0
9.22
0.6
10
2.2
R
D
I
0
D
0
0
0
0
ln and
ln
ID
D
DT D
S
IT
S
vvv
v
i
vV i
I
R
v
vvV
IR
=−
⎛⎞
==
⎜⎟
⎝⎠
⎛⎞
=−
⎜⎟
⎝⎠
2.3
(a)
80sin
13.33sin
6
s
t
vt
ω
ω
==
Peak diode current
13.33
(max)
d
i
R
=
(b) (max) 13.3 V
s
PIV v==
(c)
()
[]
()
()
11
() 13.33sin
2
13.33 13.33 13.33
cos 1 1
22
4.24
o
T
oo
o
oo
o
o
v avg v t dt dt
T
x
vavg V
π
π
π
ππ π
== ×
=− = −−−=⎡⎤
⎣⎦
=
∫∫
(d) 50%
2.4
()()
00
2maxmax2
SS
vv V v v V
γγ
=− ⇒ =+
a. For
() () ()
0
max 25 V max 25 2 0.7 = 26.4 V
S
vv= ⇒ =+
11
22
160
6.06
26.4
NN
NN
= ⇒ =
b. For
() ()
0
max 100 V max 101.4 V
S
vv= ⇒ =
11
22
160
1.58
101.4
NN
NN
= ⇒ =
From part (a)
() ()
2max 226.40.7
S
PIV v V
γ
=−=−
or
52.1 PIV V= or, from part (b)
()
2 101.4 0.7PIV =− or 202.1 PIV V=
2.5
(a)
()
(max)122(0.7)13.4 V
13.4
rms (rms) 9.48 V
2
s
ss
v
vv
=+ =
= ⇒ =
(b)
()( )( )
2 2
12
2222 F
260 0.3 150
MM
r
Cr
VV
VC
fR fVR
CC
μ
= ⇒ =
= ⇒ =
(c)
()
2
, peak 1
212
12
1
150 0.3
, peak 2.33 A
MM
d
r
d
VV
i
RV
i
π
π
⎡⎤
=+
⎢⎥
⎢⎥
⎣⎦
⎡⎤
⎢⎥
=+
⎢⎥
⎣⎦
=
2.6
(a)
()
()
()
()
max 12 0.7 12.7
max
rms rms 8.98
2
S
S
SS
vV
v
vvV
=+ =
= ⇒ =
(b)
()( )( )
12
60 150 0.3
MM
r
r
VV
VC
fRC fRV
= ⇒ ==
or
4444 CF
μ
=
(c) For the half-wave rectifier
()
, max
12 12
14 14
2150 20.3
MM
D
r
VV
i
RV
ππ
⎛⎞
⎛⎞
=+ = +
⎜⎟
⎜⎟
⎜⎟
⎜⎟
⎝⎠
⎝⎠
or
, max
4.58
D
iA=
2.8
(a)
() ()
()
p
eak 15 2 0.7 16.4 V
16.4
rms 11.6 V
2
s
s
v
v
=+ =
==
(b)
()( )( )
15
2857 F
2 2 60 125 0.35
M
r
V
C
fRV
μ
== =
2.9
S
O
O
26
0.6
0.6
26
25.4
25.4
0
0
2.10
V
B
12 V
i
D
x
1
x
2
t
R
i
D
S
S
()
24sin
S
vt t
ω
=
Now
() ()
0
1
T
DD
iavg itdt
T
=
∫
We have for
12
x
tx
ω
≤≤
24sin 12.7
D
x
i
R
−
=
To find
x
1
and x
2
,
1
24sin 12.7x =
1
2
0.558
0.558 2.584
xrad
x
rad
π
=
=− =
Then
()
()
2
1
2
2
1
1
1 24sin 12.7
2
2
1 24 1 12.7 6.482 4.095
cos or 2 1.19
22
x
D
x
x
x
x
x
x
iavg dx
R
xx R
RRRR
−
⎡⎤
==
⎢⎥
⎣⎦
⎛⎞ ⎛ ⎞
=−− ⋅=−⇒ =Ω
⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠
∫
π
ππ
21
2.584 0.558
Fraction of time diode is conducting 100% 100%
22
xx
ππ
−
−
=×= ×
or Fraction = 32.2%
Power rating
() ( )() ( )
() ()()( )()
2
1
2
1
2
2
2
1
1
1
2
22
0
22
2
2 2
24sin 12.7
2
1
24 sin 2 12.7 24 sin 12.7
2
1sin2
24 2 12.7 24 cos 12.7
224
x
T
avg rms D
x
x
x
x
x
x
x
x
x
RR x
PRi idt dx
TR
xxdx
R
xx
x
x
R
π
π
π
−
⎡⎤
=⋅ = =
⎢⎥
⎣⎦
⎡⎤
=−+
⎣⎦
⎡⎤
⎡
⎤
=−−−+
⎢⎥
⎦
⎣
⎣⎦
∫∫
∫
For
1.19 ,R =Ω then 17.9
avg
PW=
2.11
(a)
()() ()
15
150
0.1
max max 15 0.7 or max 15.7
So S
R
vvV v V
γ
==Ω
=+=+ =
Then
()
15.7
11.1
2
S
vrms V==
Now
11
22
120
10.8
11.1
NN
NN
= ⇒ =
(b)
()( )( )
15
2 2 2 60 150 0.4
MM
r
r
VV
VC
fRC fRV
= ⇒ ==
or
2083 CF
μ
=
(c)
() ()
2 max 2 15.7 0.7
S
PIV v V
γ
=−=− or 30.7 PIV V=
2.12
R
1
R
2
R
L
D
2
0
i
R
1
R
2
R
L
0
i
For
0
i
v >
0V
γ
=
Voltage across
1
L
i
RRv+=
Voltage Divider
0
1
1
2
L
ii
L
R
vvv
RR
⎛⎞
⇒
==
⎜⎟
+
⎝⎠
0
20
2.13
For
()
0, 0
i
vV
γ
>=
R
1
R
L
R
2
i
0
a.
2
0
21
2
0
||
||
|| 2.2 || 6.8 1.66 kΩ
1.66
0.43
1.66 2.2
L
i
L
L
ii
RR
vv
RR R
RR
vvv
⎛⎞
=
⎜⎟
+
⎝⎠
==
⎛⎞
==
⎜⎟
+
⎝⎠
4.3
0
b.
()
()
()
0
00
max
rms rms 3.04
2
v
vvV=
⇒ =
2.14
()()
2
3.9
0.975 mA
4
20 3.9
1.3417 mA
12
1.3417 0.975 0.367 mA
0.367 3.9 1.43 mW
L
R
ZZ
TZZ T
II
I
II
PIV P
= ⇒ =
−
==
=−⇒ =
=⋅= ⇒ =
2.15
(a)
()()
40 12
0.233 A
120
0.233 12 2.8 W
Z
I
P
−
==
==
(b) I
R
= 0.233 A, I
L
= (0.9)(0.233) = 0.21 A
So
12
0.21 57.1
L
L
R
R
= ⇒ =Ω
(c)
()( )()
0.1 0.233 12 0.28 WPP= ⇒ =
2.16
R
L
I
I
I
Z
I
L
R
i
V
Z
V
I
V
0
6.3 V, 12 , 4.8
IiZ
VRV==Ω=
