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1
Circuit Variables
Assessment Problems
AP 1.1
To solve this problem we use a product of ratios to change units from dollars/year to
dollars/millisecond. We begin by expressing $10 billion in scientiFc notation:
$
100
billion
=
$
100
×
10
9
Now we determine the number of milliseconds in one year, again using a product of
ratios:
1
year
365
.
25
days
·
1
day
24
hours
·
1
hour
60
mins
·
1
min
60
secs
·
1
sec
1000
ms
=
1
year
31
.
5576
×
10
9
ms
Now we can convert from dollars/year to dollars/millisecond, again with a product
of ratios:
$
100
×
10
9
1
year
·
1
year
31
.
5576
×
10
9
ms
=
100
31
.
5576
=
$
3
.
17
/
ms
AP 1.2
±irst, we recognize that 1 ns =
10
−
9
s. The question then asks how far a signal will
travel in
10
−
9
s if it is traveling at 80% of the speed of light. Remember that the
speed of light
c
=3
×
10
8
m/s. Therefore, 80% of
c
is
(0
.
8)(3
×
10
8
)=2
.
4
×
10
8
m/s. Now, we use a product of ratios to convert from meters/second to
inches/nanosecond:
2
.
4
×
10
8
m
1
s
·
1
s
10
9
ns
·
100
cm
1
m
·
1
in
2
.
54
cm
=
(2
.
4
×
10
8
)(100)
(10
9
)(2
.
54)
=
9
.
45
in
1
ns
Thus, a signal traveling at 80% of the speed of light will travel
9
.
45
00
in a
nanosecond.
1–1
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CHAPTER 1. Circuit Variables
AP 1.3
Remember from Eq. (1.2), current is the time rate of change of charge, or
i
=
dq
dt
In
this problem, we are given the current and asked to Fnd the total charge. To do this,
we must integrate Eq. (1.2) to Fnd an expression for charge in terms of current:
q
(
t
)=
Z
t
0
i
(
x
)
dx
We are given the expression for current,
i
, which can be substituted into the above
expression. To Fnd the total charge, we let
t
→∞
in the integral. Thus we have
q
total
=
Z
∞
0
20
e
−
5000
x
dx
=
20
−
5000
e
−
5000
x
±
±
±
±
∞
0
=
20
−
5000
(
e
∞
−
e
0
)
=
20
−
5000
(0
−
1) =
20
5000
=0
.
004
C
= 4000
µ
C
AP 1.4
Recall from Eq. (1.2) that current is the time rate of change of charge, or
i
=
dq
dt
.In
this problem we are given an expression for the charge, and asked to Fnd the
maximum current. ±irst we will Fnd an expression for the current using Eq. (1.2):
i
=
dq
dt
=
d
dt
²
1
α
2
−
³
t
α
+
1
α
2
´
e
−
αt
µ
=
d
dt
³
1
α
2
´
−
d
dt
³
t
α
e
−
αt
´
−
d
dt
³
1
α
2
e
−
αt
´
−
³
1
α
e
−
αt
−
α
t
α
e
−
αt
´
−
³
−
α
1
α
2
e
−
αt
´
=
³
−
1
α
+
t
+
1
α
´
e
−
αt
=
te
−
αt
Now that we have an expression for the current, we can Fnd the maximum value of
the current by setting the Frst derivative of the current to zero and solving for
t
:
di
dt
=
d
dt
(
te
−
αt
e
−
αt
+
t
(
−
α
)
e
αt
=(1
−
αt
)
e
−
αt
Since
e
−
αt
never equals
0
for a Fnite value of
t
, the expression equals
0
only when
(1
−
αt
)=0
. Thus,
t
=1
/α
will cause the current to be maximum. ±or this value
of
t
, the current is
i
=
1
α
e
−
α/α
=
1
α
e
−
1
Remember in the problem statement,
α
.
03679
. Using this value for
α
,
i
=
1
0
.
03679
e
−
1
∼
= 10
A
Problems
1–3
AP 1.5
Start by drawing a picture of the circuit described in the problem statement:
Also sketch the four Fgures from ±ig. 1.6:
[a]
Now we have to match the voltage and current shown in the Frst Fgure with the
polarities shown in ±ig. 1.6. Remember that 4A of current entering Terminal 2
is the same as 4A of current leaving Terminal 1. We get
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This note was uploaded on 11/26/2009 for the course ECE e circuit taught by Professor Nelson during the Spring '09 term at Adelphi.
 Spring '09
 nelson

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