P 7.2-1
Solution:
v t   v  0 
1 t
i   d
C 0
and q  Cv
In our case, the current is constant so
 Cv  t   Cv  0  i t

 i   d .
t
0

6
6
q Cv 0 18010  2010  5
 t

 2.7 ms
i
30103
P 7.2-2
Solution:
i t   C
d
1d
1
v t  
12cos  2t 30   12  2  sin  2t  30   3cos  2t  120  A
dt
8 dt
8
P 7.2-3
Solution:
v t  
0  t  2 109
2 109  t  3 109
1 t
1
i   d  v  0  

C 0
2 1012
is  t   0  v  t  
1
2 1012
t
 i   d 10
t
 0 d 10
0
3
0
3
 103
is  t   4 106 A
t
1
4 106  d  103  5 103   2 106  t
12 2ns 
2 10
In particular, v  3 109   5 103   2 106   3 109   103
 v t  
3 109  t  5 109
is  t   2 106 A
t
1
2 106  d  103  4 103  106  t
12 3ns 
2 10
In particular, v  5 109   4 103  106   5 109   103 V
 v t  
5 109  t
P 7.2-5
Solution:
is  t   0  v  t  
1
2 1012

t
5ns
0 d  103  103 V
v  t   v  0 


1 t
4 t
i

d


30

2.5

10
6103 e 6 d 




0
0
C
 30  150 0 e 6 d 
t
t
 1

 30  150   e 6   55 25e 6t V
6

0
P 7.2-6
Solution:
v
1
 1 2e2t  103  25 1 2e2t   A
3
20010
40
dv
iC  C
 10106  2   10 e 2t   200 e 2 t  A
dt
i  iR i C  200 e 2t  25  50 e 2t
iR 


 25  150e 2t
A
P 7.2-7
Solution:
t
v  t   1 i  t  dt  2
0
2
0t 2
for
 1 4 dt  2  4  t  2   2  4t  6
t
2
for
2t 3
 1 4 dt  1 4 dt  2  4  4  t  3  2  4t  18
3
t
2
3
3
4
2
3
 1 4 dt  1 4 dt  2  2
In summary
for
for
3t 4
t4
0t 2
 2
 4t  6 2  t  3

v t   
4t  18 3  t  4
 2
4t
P 7.2-11
Solution:
Apply KCL to get
v t 
d
 0.030 v  t   0
90
dt
5t
10  8e
d
1 10
i t   
 0.030 10  8e5t    e5t A for t  0
90
dt
9 9
i t  


P7.2-14
Solution:
Apply KVL to the mesh to get
 1 t

v  t   8i  t   vC  t   8i  t     i    d   v  0 
 0.1 0

That is,


1 t 25
3e d   2
0.1 0
3

e 25t  1  2
0.1 25
v  t   8 3e 25t 
 24e 25t




 24e 25t  1.2 e 25t  1  2
 22.8e 25t  0.8 V for t  0
P7.2-15
Solution:
Apply KVL to the mesh to get
1 t

v  t   R i  t   vC  t   R i  t     i   d  v  0  
C 0

That is
1 t

9.8e25t  0.6  R 5e 25t    5e 25 d   2 
C 0



 5Re25t 

5
1  25t
1
e25t  1  2  5  R 
e 
2

C  25
25 C 
5C



Equating coefficients gives
0.6 
1
 2  C  0.08  80 mF
5C



1 
1
9.8  5  R 

5
R

 5  R  0.5 R  2.46 

25 C 
25  0.08 

and
P7.2-16
Solution:
Apply KCL at either node to get
3  4 e2 t
d
C
3  4 e2t
R
dt
2 t
3 4e
3 4


  2  4 C e2 t     8 C  e2 t
R
R R


0.3  1.6 e2t 

Equating coefficients:
0.3 
3
 R  10 
R
P7.2-17
Solution:
At t = 0.5 s
and
1.6 
4
 8 C  C  0.25 F
10
v  0.5  2  0.5  8.6  9.6 V
For 0.5  t  1.5
v t  
At t = 1.5 s
For t  1.5
1 t
2 d  9.6  8
0.25 0.5
t
0.5
 9.6  8  t  0.5  9.6  8 t  5.6 V
v 1.5  8 1.5  5.6  17.6 V
v t  
1 t
0 d  17.6  17.6
0.25 1.5
Checks:
i t  
At t = 1.0 s
1 d
1 d
1
v t  
8 t  5.6  8  2 A
4 dt
4 dt
4

v  0.5  8  0.5  5.6  9.6 V
At t = 0.5 s
P 7.3-4
Solution:
vc  t   vc  0 
1 t
1 t


id


v
0

60 cos10t   d 


c


0
0

4
4
6
3  3 


 vc  0  sin   sin 10t  

2
6 2
6

3

Now since vc  t  ave  0  vc  0  sin  0
2
6
3 

 vc  t   sin 10t   V
2 
6


4 106 1.5
1 2
 Wmax  Cv 
 4.5  J
2 cmax
2
First non-negative t for max energy occurs when :
10t 

6


2
 t 

30
2
 0.1047 s
P 7.3-5
Solution:
Max. charge on capacitor  Cv  15106  9  135 C


q 135106

9 sec to charge
i
15106
1
1
2
stored energy W  Cv 2  15106  9  607.5 J
2
2
t 


P 7.4-2
Solution:
4  F in series with 4  F 
4  F4  F
 2 F
4  F+4  F
2 F 2 F = 4 F
4  F in series with 4  F = 2  F
d
i (t ) (2106 ) (53 e 250t )  (2106 ) (0 3( 250) e 250t ) A  1.5 e 250t mA
dt
P 7.4-3
Solution:
C in series with C 
C C
C C C

C C 2
C
5
 C
2
2
C 5 C
5
5
2
C in series with C 
 C
5
2
7
C C
2
5  d
5 
(25103 ) cos 250t   C  (14sin 250t )   C (14)(250) cos 250t
 7  dt
7 
3
6
so 2510  2500 C  C  1010  10 μF
P 7.4-8
Solution:
v1  v2 
dv1 dv2
i
i
C

 1  2  i1  1 i2
dt
dt
C1 C2
C2
C

KCL: i  i1  i2   1  1 i2
 C2


i2 
C2
i
C1 C2
P 7.5-3
Solution:
v(t )  (300  103 )
d
(150  103 )sin(500t  30)  (0.3)(0.15)(500) cos(500t  30)
dt
 22.5 cos(500t  30)
P 7.5-6
Solution:
(a)
(b)
 0 0t 2
d

v(t )  L i (t )  0.1 2  t  6
dt
0
6t

t
1 t
i  t    v   d  i  0   2 v   d
0
0
L
For 0 < t < 2, v(t) = 0 V so
i  t   2 0 d  0  0 A
t
0
For 2 < t < 6, v(t) = 0.2 t  0.4 V so

i  t   2  0.2  0.4  d  0  0.2 2  0.8
t
2
 
i(6)  0.2 62  0.8  6   0.8  3.2 A .
For 6 < t, v(t) = 0.8 V so
 2 =0.2 t
t
2
 0.8 t  0.8 A
i  t   2 0.8 d  3.2  1.6 t  6.4  A
t
6
P 7.5-10
Solution:
d
i t 
dt
 9.6e8t V for t  0
vL  t   0.3
Use KVL to get
+
(t)
–




v  t   18  9.6e8t  18  9.6e8t V for t  0
P 7.5-11
Solution:
Apply KVL to get
v  t   9i  t   7.5




d
d
i  t   9 3  2e3t  7.5
3  2e3t  27 1  e3t V for t  0
dt
dt
P7.5-13
Solution:
The inductor current is related to the inductor voltage by
i t  
1 t
v   d  i  0 
L 0
That is
i t  


1 t
15 4 t
15e4 d   4 
e 0  4  1.25 e4t  1  4  5.25  1.25e4t A for t  0

0
3
3  4
P7.5-14
Solution:
Apply KCL at either node to get
i t  
v t 
v t   1 t

 i L t  
   v   d  i  0  
0
R
R L

That is
1.2 e20 t  1.5 
4 e20 t 1 t 20
4 e20 t
4
  4 e d  3.5 

e20 t  1  3.5

0
R
L
R
L  20 
 4 1  20 t 1
 
 3.5
e 
5L
 R 5L 
Equating coefficients gives
1.5 
1
 3.5  L  0.1 H
5L
and
1.2 
4 1
4
1
4

 
 2  R 5 
R 5 L R 5  0.1 R
\P 7.7-1
Solution:
12 H 6 H 
i(t ) 
126
 4 H and 4 H  4 H  8 H
12+6
1 t
6
sin100 |t0   0.0075sin100t A  7.5sin100t mA
6cos100

d



8 0
8100 
P 7.7-2
Soluton:
1210 1210   6 mH
12 mH 12 mH 
3
6 mH  6 mH  12 mH ,
3
12103 12103
and 4 mH  4 mH  8 mH
v(t )  (12103 )
d
(5 3e250t )  (12103 )(0 3(250)e250t ) 9e250t V
dt
P 7.7-8
Solution:
(a) The energy stored by the 0.5 H inductor is w1 
1
 0.5  0.82   0.16 J and the energy
2
1
 2   0.82   0.64 J .
2
(b) 200 ms after the switch opens the current in the inductors is 0.8e0.4  0.536 A .
1
1
w1   0.5  0.5362   71.8 mJ and w2   2   0.5352   287.3 mJ.
2
2
Next, Leq  2  0.5  2.5 H .
stored by the 2 H inductor is w 2 
(c)
(d)
1
 2.5  0.82   0.8 J  w1  w2
2
1
weq   2.5  0.5362   359.12 mJ  w1  w 2
2
weq 
P 7.7-9
Then
Solution:
1 t
i1   v dt  i1  t0  ,
L1 t0
i2 
1 t
 v dt  i 2  t0 
L2 t0
but i1  t0   0 and i 2  t0   0
1
1 t
1 t
1 t
t
 t v dt   t v dt   +
 v dt
  t v dt 
L1 0
LP t 0
0
 L1 L2  0
1 t
1
 t v dt
i
L2
L 0
L1
1  1


1
1
1
t
i
L1  L2
+
 t v dt
LP 0
L1 L2
i  i1  i2 
P 7.8-1
Solution:
Then
i L  0   i L  0   0 and v C  0   v C  0   12 V
Next
P 7.8-2
Solution:
Then
 
 
 
 
i L 0  i L 0  1 mA and v C 0  v C 0  9 V
Next
P 7.8-3
Solution:
Then
i L  0   i L  0   0 and v C  0   v C  0   0 V
Next
P7-8.6
Solution:
The capacitor voltage and inductor current
don’t change instantaneously and so are the
keys to solving this problem.
Label the capacitor voltage and inductor
current as shown.
Before t = 0, with the switch closed and the
circuit at steady state, the inductor acts like a
short circuit and the capacitor acts like an open
circuit.
i 3  0   i  0  
18
 1.33 A
13.5
v 4  0   v  0    9i  0    12 V
v1  0    0 V and i 2  0    0 A
The capacitor voltage and inductor current don’t change instantaneously so
v  0    v  0    8 V and
i  0    i  0    1.33 A
After the switch opens the circuit looks like this:
From KCL:
i 3  t   0 A and i 2  t   i  t 
From KVL:
v1  t   9i  t   v  t 
From Ohm’s Law:
v 4  t   9i  t 
At t  0 
i 3  0    0 A and i 2  0    i  0    1.33 A
v1  0    v  0    9i  0    12  9 1.333  0 V
v 4  0    9i  0    12 V
P 7.8-8
Solution:
The capacitor voltage and inductor current don’t
change instantaneously and so are the keys to
solving this problem.
Label the capacitor voltage and inductor current
as shown.
Before t = 0, with the switch open and the circuit
at steady state, the inductor acts like a short circuit
and the capacitor acts like an open circuit.
i 2 0   i 0  
24
 0.4 A
60
v1  0    0 V
v  0    15 i  0    v1  0    v  0    6 V
v 3  0    15 i  0    6 V
The capacitor voltage and inductor current don’t change instantaneously so
v  0    v  0    6 V and
i  0    i  0    0.4 A
After the switch closes the circuit looks like this:
From Ohm’s Law:
i 2 t   
From KVL:
From KCL:
v t 
30
v1  t   v 3  t   24
v1  t 
10

v 3 t 
15
 i t 
At t  0 
i 2 0   
v 0 
 0.2 A
30
v1  0    v 3  0    24 

  v1  0    7.2 V and v 3  0    16.8 V
v1  0   v 3  0  

 i  0  
10
15

P 7.9-1
Solution:
P 7.9-2
Solution:
P 7.9-3
Solution: