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nilsson electric circuits 10e solutions
manual
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 cm 12 in 5280 feet 1s
Now set up a proportion to determine how long it takes this signal to travel
1100 miles:
124,274.24 miles 1100 miles
=
1s xs
Therefore,
1100
x= = 0.00885 = 8.85 × 10−3 s = 8.85 ms
124,274.24
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 3 6 5 . 25 days
, 1 day 1 hour 1 min 1 sec 1
· · · · y
24 hours 60 mins 60 secs 1000 e
ms a
r
=
31.5576
× 109 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
1–1
,
, 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 = dtdq 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:
∫ 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 ∞ 20
qtotal = 20e−5000x dx = e−5000x = (e−∞ — e0)
0 −5000 0 −5000
20 20
= (0 − 1) = = 0.004 C = 4000 µC
−5000 5000
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or
i = dqdt. 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 d 1 t 1
i= = − + e−αt
dt dt α 2 α α2
d 1 d t −αt d 1 −αt
= − e − e
dt α2 dt α dt α2
1 −αt t
= 0− e — α e −αt − −α 1 e−αt
α α 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 −1
i= e = e
α α
nilsson electric circuits 10e solutions
manual
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 cm 12 in 5280 feet 1s
Now set up a proportion to determine how long it takes this signal to travel
1100 miles:
124,274.24 miles 1100 miles
=
1s xs
Therefore,
1100
x= = 0.00885 = 8.85 × 10−3 s = 8.85 ms
124,274.24
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 3 6 5 . 25 days
, 1 day 1 hour 1 min 1 sec 1
· · · · y
24 hours 60 mins 60 secs 1000 e
ms a
r
=
31.5576
× 109 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
1–1
,
, 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 = dtdq 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:
∫ 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 ∞ 20
qtotal = 20e−5000x dx = e−5000x = (e−∞ — e0)
0 −5000 0 −5000
20 20
= (0 − 1) = = 0.004 C = 4000 µC
−5000 5000
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or
i = dqdt. 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 d 1 t 1
i= = − + e−αt
dt dt α 2 α α2
d 1 d t −αt d 1 −αt
= − e − e
dt α2 dt α dt α2
1 −αt t
= 0− e — α e −αt − −α 1 e−αt
α α 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 −1
i= e = e
α α
