1
Circuit Variables
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 day 1 hour 1 min 1 sec 1 year
· · · · =
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 a product
of ratios:
$100 × 109 1 year 100
· = = $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 of light. 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:
2.4 × 108 m 1s 100 cm 1 in (2.4 × 108)(100) 9.45 in
· · · = =
1s 109 ns 1m 2.54 cm (109)(2.54) 1 ns
Thus, a signal traveling at 80% of the speed of light will travel 9.45JJ in a
nanosecond.
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
∫∞
—5000x 20 —5000x ∞ 20 ∞ 0
qtotal = 20e dx = e = (e — e )
0 −5000 0 −5000
20 (0 − 1) = 20
= = 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
t 11
i= = + −
e—αt
dt dt α2 α α2
d 1 d t —αt d 1 —αt
= − e − e
dt α2 dt α dt α2
1 — αt t 1
=0− e — α e —αt − −α e—αt
α α α2
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 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
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= e—1 ∼= 10 A
0.03679
, Problems 1–3
AP 1.5 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 of current entering Terminal 2
is the same as 4A of current leaving Terminal 1. We get
(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4A
(c) v = 20 V, i = −4 A; (d) v = 20 V, 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 = dwdt. If we know the power, we can find the
energy by integrating Eq. (1.3). To begin, find the expression for power:
p = vi = (10,000e—5000t)(20e—5000t) = 200,000e—10,000t = 2 × 105e—10,000t W
Now find the expression for energy by integrating Eq. (1.3):
∫ t
w(t) = p(x) dx
0
, 1–4 CHAPTER 1. Circuit Variables
Substitute the expression for power, p, above. Note that to find the total energy, we
let t → ∞ in the integral. Thus we have
∫ ∞
2 × 105 —10,000x
∞
w= 2 × 105e—10,000x dx = e
0 −10,000 0
2 × 105 —∞ 0 2 × 105 2 × 105
= (e — e ) = (0 — 1) = = 20 J
−10,000 −10,000 10,000
AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and thus
entering the lower terminal where the polarity marking of the voltage is negative.
Thus, using the passive sign convention, p = −vi. Substituting the values of voltage
and current given in the figure,
p = −(800 × 103)(1.8 × 103) = −1440 × 106 = −1440 MW
Thus, because the power associated with the Oregon end of the line is negative,
power is being generated at the Oregon end of the line and transmitted by the line to
be delivered to the California end of the line.
Chapter Problems
P 1.1 To begin, we calculate the number of pixels that make up the display:
npixels = (1280)(1024) = 1,310,720 pixels
Each pixel requires 24 bits of information. Since 8 bits comprise a byte, each pixel
requires 3 bytes of information. We can calculate the number of bytes of
information required for the display by multiplying the number of pixels in the
display by 3 bytes per pixel:
1,310,720 pixels 3 bytes
nbytes = · = 3,932,160 bytes/display
1 display 1 pixel
Finally, we use the fact that there are 106 bytes per MB:
3,932,160 bytes 1 MB
· = 3.93 MB/display
1 display 106bytes
Circuit Variables
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 day 1 hour 1 min 1 sec 1 year
· · · · =
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 a product
of ratios:
$100 × 109 1 year 100
· = = $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 of light. 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:
2.4 × 108 m 1s 100 cm 1 in (2.4 × 108)(100) 9.45 in
· · · = =
1s 109 ns 1m 2.54 cm (109)(2.54) 1 ns
Thus, a signal traveling at 80% of the speed of light will travel 9.45JJ in a
nanosecond.
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
∫∞
—5000x 20 —5000x ∞ 20 ∞ 0
qtotal = 20e dx = e = (e — e )
0 −5000 0 −5000
20 (0 − 1) = 20
= = 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
t 11
i= = + −
e—αt
dt dt α2 α α2
d 1 d t —αt d 1 —αt
= − e − e
dt α2 dt α dt α2
1 — αt t 1
=0− e — α e —αt − −α e—αt
α α α2
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 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
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= e—1 ∼= 10 A
0.03679
, Problems 1–3
AP 1.5 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 of current entering Terminal 2
is the same as 4A of current leaving Terminal 1. We get
(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4A
(c) v = 20 V, i = −4 A; (d) v = 20 V, 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 = dwdt. If we know the power, we can find the
energy by integrating Eq. (1.3). To begin, find the expression for power:
p = vi = (10,000e—5000t)(20e—5000t) = 200,000e—10,000t = 2 × 105e—10,000t W
Now find the expression for energy by integrating Eq. (1.3):
∫ t
w(t) = p(x) dx
0
, 1–4 CHAPTER 1. Circuit Variables
Substitute the expression for power, p, above. Note that to find the total energy, we
let t → ∞ in the integral. Thus we have
∫ ∞
2 × 105 —10,000x
∞
w= 2 × 105e—10,000x dx = e
0 −10,000 0
2 × 105 —∞ 0 2 × 105 2 × 105
= (e — e ) = (0 — 1) = = 20 J
−10,000 −10,000 10,000
AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and thus
entering the lower terminal where the polarity marking of the voltage is negative.
Thus, using the passive sign convention, p = −vi. Substituting the values of voltage
and current given in the figure,
p = −(800 × 103)(1.8 × 103) = −1440 × 106 = −1440 MW
Thus, because the power associated with the Oregon end of the line is negative,
power is being generated at the Oregon end of the line and transmitted by the line to
be delivered to the California end of the line.
Chapter Problems
P 1.1 To begin, we calculate the number of pixels that make up the display:
npixels = (1280)(1024) = 1,310,720 pixels
Each pixel requires 24 bits of information. Since 8 bits comprise a byte, each pixel
requires 3 bytes of information. We can calculate the number of bytes of
information required for the display by multiplying the number of pixels in the
display by 3 bytes per pixel:
1,310,720 pixels 3 bytes
nbytes = · = 3,932,160 bytes/display
1 display 1 pixel
Finally, we use the fact that there are 106 bytes per MB:
3,932,160 bytes 1 MB
· = 3.93 MB/display
1 display 106bytes
