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CHAPTER 3
Solutions for
Exercises
E3.1
V
)
10
sin(
5
.
0
)
10
2
/(
)
10
sin(
10
/
)
(
)
(
5
6
5
6
t
t
C
t
q
t
v
=
×
=
=


A
)
10
cos(
1
.
0
)
10
cos(
)
10
5
.
0
)(
10
2
(
)
(
5
5
5
6
t
t
dt
dv
C
t
i
=
×
×
=
=

E3.2
Because
the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = 0.
ms
4
ms
2
for
10
10
4
10
10
ms
2
0
for
10
10
0
)
(
)
(
3
6
3
2E
3
3
2E
0
3
3
0
3
0
≤
≤

×
=

+
=
≤
≤
=
=
+
=







∫
∫
∫
∫
t
t
dx
dx
t
t
dx
dx
x
i
t
q
t
t
t
ms
4
ms
2
for
10
40
ms
2
0
for
10
/
)
(
)
(
4
4
≤
≤

=
≤
≤
=
=
t
t
t
t
C
t
q
t
v
ms
4
ms
2
for
10
10
40
ms
2
0
for
10
)
(
)
(
)
(
3
≤
≤
+
×

=
≤
≤
=
=

t
t
t
t
t
v
t
i
t
p
ms
4
ms
2
for
)
10
40
(
10
5
.
0
ms
2
0
for
5
2
/
)
(
)
(
2
4
7
2
2
≤
≤

×
=
≤
≤
=
=

t
t
t
t
t
Cv
t
w
in which the units of charge, electrical potential, power, and energy are coulombs, volts, watts and
joules, respectively.
Plots of these quantities are shown in Figure 3.8 in the book.
E3.3
Refer to Figure 3.10 in the book.
Applying KVL, we have
3
2
1
v
v
v
v
+
+
=
Then using Equation 3.8 to substitute for the voltages we have
1
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0
(
)
(
1
)
0
(
)
(
1
)
0
(
)
(
1
)
(
3
0
3
2
0
2
1
0
1
v
dt
t
i
C
v
dt
t
i
C
v
dt
t
i
C
t
v
t
t
t
+
+
+
+
+
=
∫
∫
∫
This can be written as
)
0
(
)
0
(
)
0
(
)
(
1
1
1
)
(
3
2
1
0
3
2
1
v
v
v
dt
t
i
C
C
C
t
v
t
+
+
+
+
+
=
∫
(1)
Now if we define
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 Fall '07
 Zatet
 Electrical Engineering, Volt

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