Solution Manual for Electric Circuits 11th Edition by James Nilsson, Susan Reidel

Instructor Solution Manual
Electric Circuits
AUTHOR: James Nilsson, Susan Reidel

11th Edition




mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.

, 1
Circuit Variables



Assessment Problems

AP 1.1 Use A Product Of Ratios To Convert Two-Thirds The Speed Of Light From
MetersPer 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
Travel1100 Miles:
124,274.24 Miles 1100 Miles
=
1S XS
Therefore,
1100 −3
S = 8.85 Ms
X
10= 124,274.24 = 0.00885 = 8.85 ×
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 1 1 Min 1 Sec 1 Year
Day Hour
· · · · =
365.25 Days 24 Hours 60 Mins 60 Secs 1000 Ms 31.5576 × 109 Ms
Now We Can Convert From Dollars/Year To Dollars/Millisecond, Again With
AProduct Of Ratios:
$100 × 109 1 Year 100
· = = $3.17/Ms
1 Year 31.5576 × 109 Ms 31.5576

mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.

, 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
∞T In The Integral.
The Above Expression. To find The Total Charge, We→Let
ThusWe Have
∫ ∞ .∞
20 .
−5000x.
Qtotal = 20e−5000x Dx E . = — E0 )
0
= −5000 0 (E−
∞ 20
−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
Tofind The Maximum Current. First We Will find An Expression For The
Current Using Eq. (1.2):
Dq D 1 T 1 −Αt
I= =
DtDt Α2 − Α + Α2 E
D T D
D 1— — 1
=
Dt Α2 E −Αt Α2 −Αt
Α E
Dt Dt
1 −Α T 1
= 0− E
t —Α e −Αt −
−Α 2 E −Αt
Α α α
1 1 −Αt
= − +T+ E
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 AndSolving 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