SOLUTION MANUAL
All Chapters Included
, 1
Circuit Variables
Assessment Problems
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from
meters per second to miles per second:
2 3 × 108 m 100 cm 1 in 1 ft 1 mile 124,274.24 miles
· · · · =
3 1s 1m 2.54 12 5280 feet 1s
cm in
Now set up a proportion to determine how long it takes this signal to
travel 1100 miles:
124,274.24 miles 1100 miles
=
1s x s
Therefore,
1100
x= = 0.00885 = 8.85 × 10−3 s = 8.85 ms
124,274.2
4
AP 1.2 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 1 day 1 hour 1 min 1 sec 1 year
· · · · =
365.25 days 24 60 mins 60 secs 1000 31.5576 × 109 ms
hours ms
Now we can convert from dollars/year to dollars/millisecond, again
with a product of ratios:
$100 × 109 1 year 100
· = = $3.17/ms
1 year 31.5576 × 109 ms 31.5576
, 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 dtIn 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:
∫ t
q(t) = i(x) dx
0
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
∫ ∞ 20 ∞
qtotal = 20 −5000
x
dx e−5000x = 20 (e−∞ — e0)
e = −5000 0 −5000
0
20
20
= (0 − 1) = 0.004 C = 4000 µC
−500 = 500
0 0
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):
dq 1 1
i= = − + e−αt
d dt dt t
α2 α α2
d 1 t −αt 1 −αt
= − e − e
d dt α2 dt d
α dt α2
1 t
= 0 − e−αt — α e −αt − −α 1 e−αt
α α
, Problems 1–3
α2
1 1
= − +t+ e−αt
α α
= 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 d
= (te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0
dt dt
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
1 −α/α 1
i= e = e−1
α α
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= ∼ 10 A
e−1 =
0.03679
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
Also sketch the four figures from Fig. 1.6:
[a] Now we have to match the voltage and current shown in the first
figure with the polarities shown in Fig. 1.6. Remember that 4A of
current entering Terminal 2 is the same as 4A of current leaving
Terminal 1. We get
(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4A
All Chapters Included
, 1
Circuit Variables
Assessment Problems
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from
meters per second to miles per second:
2 3 × 108 m 100 cm 1 in 1 ft 1 mile 124,274.24 miles
· · · · =
3 1s 1m 2.54 12 5280 feet 1s
cm in
Now set up a proportion to determine how long it takes this signal to
travel 1100 miles:
124,274.24 miles 1100 miles
=
1s x s
Therefore,
1100
x= = 0.00885 = 8.85 × 10−3 s = 8.85 ms
124,274.2
4
AP 1.2 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 1 day 1 hour 1 min 1 sec 1 year
· · · · =
365.25 days 24 60 mins 60 secs 1000 31.5576 × 109 ms
hours ms
Now we can convert from dollars/year to dollars/millisecond, again
with a product of ratios:
$100 × 109 1 year 100
· = = $3.17/ms
1 year 31.5576 × 109 ms 31.5576
, 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 dtIn 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:
∫ t
q(t) = i(x) dx
0
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
∫ ∞ 20 ∞
qtotal = 20 −5000
x
dx e−5000x = 20 (e−∞ — e0)
e = −5000 0 −5000
0
20
20
= (0 − 1) = 0.004 C = 4000 µC
−500 = 500
0 0
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):
dq 1 1
i= = − + e−αt
d dt dt t
α2 α α2
d 1 t −αt 1 −αt
= − e − e
d dt α2 dt d
α dt α2
1 t
= 0 − e−αt — α e −αt − −α 1 e−αt
α α
, Problems 1–3
α2
1 1
= − +t+ e−αt
α α
= 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 d
= (te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0
dt dt
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
1 −α/α 1
i= e = e−1
α α
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= ∼ 10 A
e−1 =
0.03679
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
Also sketch the four figures from Fig. 1.6:
[a] Now we have to match the voltage and current shown in the first
figure with the polarities shown in Fig. 1.6. Remember that 4A of
current entering Terminal 2 is the same as 4A of current leaving
Terminal 1. We get
(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4A
