CHAPTER 1
Solutions for Exercises
E1.1
Charge= Current
×
Time =
(2 A)
×
(10 s) = 20 C
E1.2
A
)
2cos(200
)
200cos(200
0.01
0t)
0.01sin(20
(
)
(
)
(
t
t
dt
d
dt
t
dq
t
i
=
×
=
=
=
E1.3
Because
i
2
has a positive value, positive chargemoves in the same
direction as the reference.
Thus
positive charge moves downward in element C.
Because
i
3
has a negative value, positive charge moves in the opposite direction to the reference.
Thus positive chargemoves upward in element E.
E1.4
Energy = Charge
×
Voltage = (2 C)
×
(20 V) = 40 J
Because
v
ab
is positive, the positive terminal is a and the negative terminal is b.
Thus the charge
moves from the negative terminal to the positive terminal, and energy is removed from the circuit
element.
E1.5
i
ab
enters terminal a.
Furthermore, v
ab
is positive at terminal a.
Thus the current enters the
positive reference, and we have the passive reference
configuration.
E1.6
(a)
2
20
)
(
)
(
)
(
t
t
i
t
v
t
p
a
a
a
=
=
J
6667
3
20
3
20
20
)
(
3
10
0
3
10
0
10
0
2
=
=
=
=
=
∫
∫
t
t
dt
t
dt
t
p
w
a
a
(b)
Notice that the references
are opposite to the passive sign convention. Thus we have:
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 Fall '07
 Zatet
 Electrical Engineering, Addition, Energy, Chemical element, dt dt, 2A current sourcein, current sourcein serieswith

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