TEST BANK FOR Electric Circuits 8th Edition By Nilsson, J.W. and Riedel, S

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 scientific notation:
$100 billion = $100 × 109
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 × 109 ms
Now we can convert from dollars/year to dollars/millisecond, again with a product
of ratios:
$100 × 109
1 year
· 1 year
31.5576 × 109 ms
=
100
31.5576
= $3.17/ms
AP 1.2 First, 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× 108 m/s. Therefore, 80% of c is (0.8)(3 × 108) = 2.4 × 108
m/s. Now, we use a product of ratios to convert from meters/second to
inches/nanosecond:
2.4 × 108 m
1s
· 1 s
109 ns
· 100 cm
1 m
· 1 in
2.54 cm
=
(2.4 × 108)(100)
(109)(2.54)
=
9.45 in
1 ns
Thus, a signal traveling at 80% of the speed of light will travel 9.45 in a
nanosecond.
1–1
1–2 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 find the total charge. To do this,
we must integrate Eq. (1.2) to find an expression for charge in terms of current:
q(t) =
t
0
i(x) dx
We are given the expression for current, i, which can be substituted into the above
expression. To find the total charge, we let t→∞in the integral. Thus we have
qtotal =
∞
0
20e−5000x dx =
20
−5000e−5000x

∞
0
=
20
−5000
(e∞ − e0)
=
20
−5000
(0 − 1) =
20
5000
= 0.004 C = 4000 μC
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 find the
maximum current. First we will find an expression for the current using Eq. (1.2):
i = dq
dt
= d
dt
1
α2
−

t
α
+
1
α2

e−αt

= d
dt
1
α2

− d
dt

t
α
e−αt

− d
dt
1
α2 e−αt

= 0−
1
α
e−αt − α
t
α
e−αt

−

−α
1
α2 e−αt

=

−1
α
+ t +
1
α

e−αt
= te−αt
Now that we have an expression for the current, we can find the maximum value of
the current by setting the first derivative of the current to zero and solving for t:
di
dt
= d
dt
(te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0
Since e−αt never equals 0 for a finite value of t, the expression equals 0 only when
(1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For this value
of t, the current is
i =
1
α
e−α/α =
1
α
e−1
Remember in the problem statement, α = 0.03679. Using this value for α,
i =
1
0.03679e−1 ∼=
10 A