1
Circuit Variables z
Assessment Problems z
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters
z z z z z z z z z z z z z z z
per second to miles per second:
z z z z z z
3 × 108 100 cm 1 in 1 mile 124,274.24 miles
z z z z
2 z
z z z z
· 1 ft
z z z
=
z z z
· ·
z
z
z m ·
z
3 1 1m z 2.54 cm 12 in z z 5280 feet z 1s z
s z
Now set up a proportion to determine how long it takes this signal to travel
z z z z z z z z z z z z z z
1100 miles:
z z
124,274.24 miles 1100 miles z z z
= z
1s x s z z
Therefore,
1100
x= z = 0.00885 = 8.85 × 10−3 s = 8.85 ms
z z z z z z z z z
124,274.24
AP 1.2 To solve this problem we use a product of ratios to change units from
z z z z z z z z z z z z z z z
dollars/year to dollars/millisecond. We begin by expressing $10 billion in
z z z z z z z z z z
scientific notation:
z z
$100 billion = $100 × 109 z z z z z
Now we determine the number of milliseconds in one year, again using a
z z z z z z z z z z z z
product of ratios:
z z z
1 year 1 day 1 hour 1 1 sec 1 year z
· · · ·
z z z z
= z z
1000 ms 31.5576 × 109 ms
z
365.25 days 24 hours 60 mins min z z z z z z z z
60 secs z
Now we can convert from dollars/year to dollars/millisecond, again with a
z z z z z z z z z z
product of ratios:
z z z
$100 × z 1 year
z 100
109
z
© z2015 zPearson zEducation, zInc., zUpper zSaddle zRiver, zNJ. zAll zrights zreserved. zThis zpublication zis zprotected zby zCopyright zand zwritten zpermission zshould zbe zobtained
from zthe zpublisher zprior zto zany zprohibited zreproduction, zstorage zin za zretrieval zsystem, zor ztransmission zin zany zform zor zby zany zmeans, zelectronic, zmechanical,
zphotocopying, z recording, zor zlikewise. zFor zinformation zregarding zpermission(s), zwrite zto: zRights zand zPermissions zDepartment, zPearson zEducation, zInc., zUpper
zSaddle zRiver, zNJ z07458.
, · z = = $3.17/ms
31.5576 × 109 ms
z z
1 year
z z z
z
z 31.5576 z
1–1
© z2015 zPearson zEducation, zInc., zUpper zSaddle zRiver, zNJ. zAll zrights zreserved. zThis zpublication zis zprotected zby zCopyright zand zwritten zpermission zshould zbe zobtained
from zthe zpublisher zprior zto zany zprohibited zreproduction, zstorage zin za zretrieval zsystem, zor ztransmission zin zany zform zor zby zany zmeans, zelectronic, zmechanical,
zphotocopying, z recording, zor zlikewise. zFor zinformation zregarding zpermission(s), zwrite zto: zRights zand zPermissions zDepartment, zPearson zEducation, zInc., zUpper
zSaddle zRiver, zNJ z07458.
,
, 1–2 CHAPTER 1. Circuit z z z
zVariables
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or
z z z z z z z z z z z z z z z
i = dqdt In this problem, we are given the current and asked to find the total
z z z z z z z z z z z z z z z
charge. To do this, we must integrate Eq. (1.2) to find an expression for
z z z z z z z z z z z z z
charge in terms of current:
z z z z z
∫ t z
q(t) = z i(x) dx z
0
We are given the expression for current, i, which can be substituted into the
z z z z z z z z z z z z z
above expression. To find the total charge, we let t → ∞ in the integral. Thus
z z z z z z z z z z z z z z z z
we have
z z
∫ ∞ z z 20 ∞ 20 −∞
z z
qtotal = 20e−5000x dx = e−5000x = (e — e0)
−5000
z z
−5000
z
z
0 0 z
z
z
20 20
= (0 − 1) = = 0.004 C = 4000 µC z
−5000
z z z z z z z
5000
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or
z z z z z z z z z z z z z z z z z
i = dqdt. In this problem we are given an expression for the charge, and asked to
z z z z z z z z z z z z z z z z z
find the maximum current. First we will find an expression for the current
z z z z z z z z z z z z z
using Eq. (1.2):
z z z
dq
z z z z z
1 1
i= − e−αt
z z z z z z
= z z + z z
d dt dt t z
z z
α2 α α2
d t −αt
z z z z z z z z z z z
1 1 −αt
− e −
z z z
e
z z z z
= z
d dt α 2
z
dt d z
α dt
z z
z α2
1 t z z z z
1
z
= 0− e−αt — α −αt
z z
− −α e−αt
z z
z z
z z
α e z
α z
α2
z z z z
1 1
− +t+ e−αt
z z z
= z z z
α α
= te−αt z
Now that we have an expression for the current, we can find the maximum
z z z z z z z z z z z z z
value of the current by setting the first derivative of the current to zero and
z z z z z z z z z z z z z z z
solving for t:
z z z
di z d z
= z (te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0 z z z z z z z z z z
dt dt
Since e−αt never equals 0 for a finite value of t, the expression equals 0 only
z z z z z z z z z z z z z z z
when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For
z z z z z z z z z z z z z z z z z z
this value of t, the current is
z z z z z z z
1 1
i= e−α/α = e−1
z z
z
z z z
© z2015 zPearson zEducation, zInc., zUpper zSaddle zRiver, zNJ. zAll zrights zreserved. zThis zpublication zis zprotected zby zCopyright zand zwritten zpermission zshould zbe zobtained
from zthe zpublisher zprior zto zany zprohibited zreproduction, zstorage zin za zretrieval zsystem, zor ztransmission zin zany zform zor zby zany zmeans, zelectronic, zmechanical,
zphotocopying, z recording, zor zlikewise. zFor zinformation zregarding zpermission(s), zwrite zto: zRights zand zPermissions zDepartment, zPearson zEducation, zInc., zUpper
zSaddle zRiver, zNJ z07458.
Circuit Variables z
Assessment Problems z
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters
z z z z z z z z z z z z z z z
per second to miles per second:
z z z z z z
3 × 108 100 cm 1 in 1 mile 124,274.24 miles
z z z z
2 z
z z z z
· 1 ft
z z z
=
z z z
· ·
z
z
z m ·
z
3 1 1m z 2.54 cm 12 in z z 5280 feet z 1s z
s z
Now set up a proportion to determine how long it takes this signal to travel
z z z z z z z z z z z z z z
1100 miles:
z z
124,274.24 miles 1100 miles z z z
= z
1s x s z z
Therefore,
1100
x= z = 0.00885 = 8.85 × 10−3 s = 8.85 ms
z z z z z z z z z
124,274.24
AP 1.2 To solve this problem we use a product of ratios to change units from
z z z z z z z z z z z z z z z
dollars/year to dollars/millisecond. We begin by expressing $10 billion in
z z z z z z z z z z
scientific notation:
z z
$100 billion = $100 × 109 z z z z z
Now we determine the number of milliseconds in one year, again using a
z z z z z z z z z z z z
product of ratios:
z z z
1 year 1 day 1 hour 1 1 sec 1 year z
· · · ·
z z z z
= z z
1000 ms 31.5576 × 109 ms
z
365.25 days 24 hours 60 mins min z z z z z z z z
60 secs z
Now we can convert from dollars/year to dollars/millisecond, again with a
z z z z z z z z z z
product of ratios:
z z z
$100 × z 1 year
z 100
109
z
© z2015 zPearson zEducation, zInc., zUpper zSaddle zRiver, zNJ. zAll zrights zreserved. zThis zpublication zis zprotected zby zCopyright zand zwritten zpermission zshould zbe zobtained
from zthe zpublisher zprior zto zany zprohibited zreproduction, zstorage zin za zretrieval zsystem, zor ztransmission zin zany zform zor zby zany zmeans, zelectronic, zmechanical,
zphotocopying, z recording, zor zlikewise. zFor zinformation zregarding zpermission(s), zwrite zto: zRights zand zPermissions zDepartment, zPearson zEducation, zInc., zUpper
zSaddle zRiver, zNJ z07458.
, · z = = $3.17/ms
31.5576 × 109 ms
z z
1 year
z z z
z
z 31.5576 z
1–1
© z2015 zPearson zEducation, zInc., zUpper zSaddle zRiver, zNJ. zAll zrights zreserved. zThis zpublication zis zprotected zby zCopyright zand zwritten zpermission zshould zbe zobtained
from zthe zpublisher zprior zto zany zprohibited zreproduction, zstorage zin za zretrieval zsystem, zor ztransmission zin zany zform zor zby zany zmeans, zelectronic, zmechanical,
zphotocopying, z recording, zor zlikewise. zFor zinformation zregarding zpermission(s), zwrite zto: zRights zand zPermissions zDepartment, zPearson zEducation, zInc., zUpper
zSaddle zRiver, zNJ z07458.
,
, 1–2 CHAPTER 1. Circuit z z z
zVariables
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or
z z z z z z z z z z z z z z z
i = dqdt In this problem, we are given the current and asked to find the total
z z z z z z z z z z z z z z z
charge. To do this, we must integrate Eq. (1.2) to find an expression for
z z z z z z z z z z z z z
charge in terms of current:
z z z z z
∫ t z
q(t) = z i(x) dx z
0
We are given the expression for current, i, which can be substituted into the
z z z z z z z z z z z z z
above expression. To find the total charge, we let t → ∞ in the integral. Thus
z z z z z z z z z z z z z z z z
we have
z z
∫ ∞ z z 20 ∞ 20 −∞
z z
qtotal = 20e−5000x dx = e−5000x = (e — e0)
−5000
z z
−5000
z
z
0 0 z
z
z
20 20
= (0 − 1) = = 0.004 C = 4000 µC z
−5000
z z z z z z z
5000
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or
z z z z z z z z z z z z z z z z z
i = dqdt. In this problem we are given an expression for the charge, and asked to
z z z z z z z z z z z z z z z z z
find the maximum current. First we will find an expression for the current
z z z z z z z z z z z z z
using Eq. (1.2):
z z z
dq
z z z z z
1 1
i= − e−αt
z z z z z z
= z z + z z
d dt dt t z
z z
α2 α α2
d t −αt
z z z z z z z z z z z
1 1 −αt
− e −
z z z
e
z z z z
= z
d dt α 2
z
dt d z
α dt
z z
z α2
1 t z z z z
1
z
= 0− e−αt — α −αt
z z
− −α e−αt
z z
z z
z z
α e z
α z
α2
z z z z
1 1
− +t+ e−αt
z z z
= z z z
α α
= te−αt z
Now that we have an expression for the current, we can find the maximum
z z z z z z z z z z z z z
value of the current by setting the first derivative of the current to zero and
z z z z z z z z z z z z z z z
solving for t:
z z z
di z d z
= z (te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0 z z z z z z z z z z
dt dt
Since e−αt never equals 0 for a finite value of t, the expression equals 0 only
z z z z z z z z z z z z z z z
when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For
z z z z z z z z z z z z z z z z z z
this value of t, the current is
z z z z z z z
1 1
i= e−α/α = e−1
z z
z
z z z
© z2015 zPearson zEducation, zInc., zUpper zSaddle zRiver, zNJ. zAll zrights zreserved. zThis zpublication zis zprotected zby zCopyright zand zwritten zpermission zshould zbe zobtained
from zthe zpublisher zprior zto zany zprohibited zreproduction, zstorage zin za zretrieval zsystem, zor ztransmission zin zany zform zor zby zany zmeans, zelectronic, zmechanical,
zphotocopying, z recording, zor zlikewise. zFor zinformation zregarding zpermission(s), zwrite zto: zRights zand zPermissions zDepartment, zPearson zEducation, zInc., zUpper
zSaddle zRiver, zNJ z07458.
