Nilsson&Riedel-ElectricCircuits10ESolutionsManual

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rcuit Variables

Assessment Problems

1.1 Use a product of ratios to convert two-thirds the speed of light from meters

per second to miles per second:





3 × 10

1 s

0 cm

1 m

1 i

2.54 cm

12 in

1 m

ile

5280 feet

4,274.24 miles

1 s

w set up a proportion to determine how long it takes this signal to travel

1100 miles:

124,274.24 miles

1 s

00 miles

x s

erefore,

x =

1100

124,274.2

= 0.00885 = 8.85 × 10

−3

s = 8.85 ms

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 × 10

Now we determine the number of milli se conds in one year, again using a

product of ratios:

1 year

365.25 days

day

24 hours

1 h

our

60 m ins

1 m

60 secs

1 s

1000 ms

1 y

ear

31.5576 × 1

Now we can convert from dollars/year to dollars/millisecond, again with a

product of ratios:

$100 × 10

1 year

1 y

ear

31.5576 × 1

100

31.5576

$3.17/ms

1–1

from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 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

P 1.3 Re member from Eq. (1.2), current is the time rate of change of charge, or

i =

this problem, we are given the current and asked to ﬁnd the total

charge. To do this, we must i ntegrate Eq. (1.2) to ﬁnd an expression for

charge in terms of current:

q(t) =

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 → ∞ i n the integral. Thus

we have

total

∞

20e

−5000x

dx =

−5000

−5

000x



∞

−5000

(

−∞

− e

)

−5000

(

0 − 1) =

5000

= 0.0

04 C = 4000 µC

AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or

i =

. I

n 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 =



−





−αt







−



−αt



−



−αt



0 −



−αt

− α

−αt



−



−α

−αt





−

+ t +



−αt

= t

−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:

−αt

) = e

−αt

+ t(− α)e

αt

= (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 maxi mum. For

this value of t, the current is

i =

−α/

−1

from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,

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