Electronica De Nilson

Circuit Variables

1

Assessment Problems
AP 1.1

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 hour
1 min
1 sec
1 year
1 day
·
·
·
=·
31.5576 × 109 ms
365.25 days 24 hours 60 mins 60 secs 1000 ms
Now we can convert from dollars/year to dollars/millisecond, again with a product
of ratios:
1 year
100
$100 × 109
·
=
= $3.17/ms
1 year
31.5576 × 109 ms
31.5576

AP 1.2

First, we recognize that 1 ns = 10−9 s. The question then asks how far a signal will
travel in 10−9 s if it is traveling at 80% of the speed oflight. Remember that the
speed of light c = 3 × 108 m/s. Therefore, 80% of c is (0.8)(3 × 108 ) = 2.4 × 108
m/s. Now, we use a product of ratios to convert from meters/second to
inches/nanosecond:
100 cm
1 in
(2.4 × 108 )(100)
9.45 in
2.4 × 108 m
1s
·
·
=
=
·9
1s
10 ns
1m
2.54 cm
(109 )(2.54)
1 ns
Thus, a signal traveling at 80% of the speed of light will travel 9.45 in ananosecond.

1–1

1–2
AP 1.3

CHAPTER 1. Circuit Variables
Remember from Eq. (1.2), current is the time rate of change of charge, or i = dq In
dt
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) =

0

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 → ∞ in the integral. Thus we have
∞

qtotal =
=
AP 1.4

0

20e−5000x dx =

20 −5000x
e
−5000

∞
0

=

20
(e∞ − e0 )
−5000

20
20
(0 − 1) =
= 0.004 C = 4000 µC
−5000
5000

Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dq . In
dt
this problem we aregiven 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=
=

dq
d1
1
t
=
+ 2 e−αt
−
2
dt
dt α
αα
d1
d t −αt
d 1 −αt
e
−
−
e
dt α2
dt α
dt α2
1 −αt
t
1
− α e−αt − −α 2 e−αt
e
α
α
α

= 0−
=−

1
1 −αt
e
+t+
α
α

= te−αt
Now that we have an expression for the current, wecan find the maximum value of
the current by setting the first derivative of the current to zero and solving for t:
d
di
= (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
i=

1 −α/α
1
= e−1
e
α
αRemember in the problem statement, α = 0.03679. Using this value for α,
i=

1
e−1 ∼ 10 A
=
0.03679

Problems
AP 1.5

1–3

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 ofcurrent entering Terminal 2
is the same as 4A of current leaving Terminal 1. We get
(a) v = −20 V,
(c) v = 20 V,

i = −4 A;
i = −4 A;

(b) v = −20 V,
(d) v = 20 V,

i = 4A
i = 4A

[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (−20)(−4) = 80 W. Since the power is greater than 0, the box is
absorbing power.
[c] From the calculation in part(b), the box is absorbing 80 W.
AP 1.6

Applying the passive sign convention to the power equation using the voltage and
current polarities shown in Fig. 1.5, p = vi. From Eq. (1.3), we know that power is
the time rate of change of energy, or p = dw . If we know the power, we can find the
dt
energy by integrating Eq. (1.3). To begin, find the expression for power:
p = vi = (10,000e−5000t...