M
Circuit Variables
PR
Assessment Problems
ES
AP 1.1 Use a product of ratios to convert 95% of the speed of light from meters per
second to miles per second:
3 × 108 m 100 cm 1 in 1 ft 1 mile 177,090.79 miles
SI
(0.95) · · · · = .
1s 1m 2.54 cm 12 in 5280 feet 1s
Now set up a proportion to determine how long it takes this signal to travel
950 miles:
VE
177,090.79 miles 950 miles
= .
1s xs
Therefore,
950
x= = 0.00536 = 5.36 × 10−3 s = 5.36 ms.
G
177,090.79
AP 1.2 We begin by expressing $1 trillion in scientific notation:
$1 trillion = $1 × 1012 .
R
Divide by 100 = 102 to find the number of $100 bills:
AD
1012
$1 trillion = = 1010 $100 bills.
102
Calculate the height of a stack of 1010 $100 bills:
0.11 mm 1m
1010 bills · · = 1.1 × 106 m.
bill 1000 mm
ES
Now we can convert from meters to miles, again with a product of ratios:
100 cm 1 in 1 ft 1 mi
1.1 × 106 m · · · · = 683.51 miles.
1m 2.54 cm 12 in 5280 ft
1–1
, 1–2 CHAPTER 1. Circuit Variables
AP 1.3 [a] First we use Eq. (1.2) to relate current and charge:
dq
i= = 0.25te−2000t .
M
dt
Therefore, dq = 0.25te−2000t dt.
To find the charge, we can integrate both sides of the last equation. Note
PR
that we substitute x for q on the left side of the integral, and y for t on
the right side of the integral:
Z q(t) Z t
dx = 0.25 ye−2000y dy.
q(0) 0
We solve the integral and make the substitutions for the limits of the
ES
integral:
t
e−2000y
q(t) − q(0) = 0.25 (−2000y − 1)
(−2000)2 0
= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9
SI
= 62.5 × 10−9 (1 − 2000te−2000t − e−2000t ).
But q(0) = 0 by hypothesis, so
q(t) = 62.5(1 − 2000te−2000t − e−2000t ) nC.
VE
[b] q(0.001) = (62.5)[1 − 2000(0.001)e−2000(0.001) − e−2000(0.001) ] = 37.12 nC.
75 × 10−6 C/s
AP 1.4 n = = 4.681 × 1014 elec/s.
1.6022 × 10−19 C/elec
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
G
R
Also sketch the four figures from Fig. 1.6:
AD
ES
, Problems 1–3
[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 250 mA of current
entering Terminal 2 is the same as 250 mA of current leaving Terminal 1.
M
We get
(a) v = 50 V, i = −0.25 A; (b) v = 50 V, i = 0.25 A;
PR
(c) v = −50 V, i = −0.25 A; (d) v = −50 V, i = 0.25 A.
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (50)(−0.25) = −12.5 W.
[c] Since the power is less than 0, the box is delivering power.
Z t
ES
AP 1.6 p = vi; w= p dx.
0
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
SI
VE
Note that in constructing the plot above, we used the fact that 60 hr
= 216,000 s = 216 ks.
G
p(0) = (6)(15 × 10−3 ) = 90 × 10−3 W;
p(216 ks) = (4)(15 × 10−3 ) = 60 × 10−3 W;
R
1
w = (60 × 10−3 )(216 × 103 ) + (90 × 10−3 − 60 × 10−3 )(216 × 103 ) = 16,200 J.
2
AD
AP 1.7 [a] p = vi = (15e−250t )(0.04e−250t ) = 0.6e−500t W;
p(0.01) = 0.6e−500(0.01) = 0.6e−5 = 0.00404 = 4.04 mW.
Z ∞ Z ∞ ∞
0.6 −500x
[b] wtotal = p(x) dx = 0.6e−500x dx = e
0 0 −500 0
ES
= −0.0012(e∞ − e0 ) = 0.0012 = 1.2 mJ.
, 1–4 CHAPTER 1. Circuit Variables
Chapter Problems
M
P 1.1 [a] Use a product of ratios to convert 40,075 km to inches:
1000 m 100 cm 1 in
40,075 km · · · = 1,577,755,905.5 in.
PR
1 km 1m 2.54 cm
Now calculate the number of $20 bills this distance represents:
1,577,755,905.5 in
Number of bills = = 256,963,502.5 ≈ 256,963,503.
6.14 in/bill
The total dollar amount represented by this number of $20 bills is
ES
Total dollars = (256,963,503)($20) ≈ $5.14 billion.
[b] The weight of the number of bills calculated in part (a) is
(256,963,503)(1 g) = 256,963.5 kg. Use a product of ratios to convert kg
to tons:
2.2 lbs 1 ton
256,963.5 kg · · = 282.66 tons.
SI
1 kg 2000 lbs
1s
P 1.2 [a] 8(65.5 × 106 ) bits · = 10.48 s.
50 × 106 bits
1s
VE
[b] 8(65.5 × 106 ) bits · = 262 ms.
2 × 109 bits
1s
[c] 8(74 × 1012 ) bits · 6
= 11.84 × 106 s
50 × 10 bits
1 min 1 hr 1 day
11.84 × 106 s · · · = 137 days!
60 s 60 min 24 hr
G
1s
[d] 8(74 × 1012 ) bits · = 296 × 103 s.
2 × 109 bits
R
1 min 1 hr 1 day
296 × 103 s · · · = 3.4 days.
60 s 60 min 24 hr
P 1.3 [a] To begin, we calculate the number of pixels that make up the display:
AD
npixels = (1920)(1080) = 2,073,600 pixels.
Each pixel requires 24 bits of information. Since 8 bits equal one 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:
ES
2,073,600 pixels 3 bytes
nbytes = · = 6,220,800 bytes/display.
1 display 1 pixel
Circuit Variables
PR
Assessment Problems
ES
AP 1.1 Use a product of ratios to convert 95% of the speed of light from meters per
second to miles per second:
3 × 108 m 100 cm 1 in 1 ft 1 mile 177,090.79 miles
SI
(0.95) · · · · = .
1s 1m 2.54 cm 12 in 5280 feet 1s
Now set up a proportion to determine how long it takes this signal to travel
950 miles:
VE
177,090.79 miles 950 miles
= .
1s xs
Therefore,
950
x= = 0.00536 = 5.36 × 10−3 s = 5.36 ms.
G
177,090.79
AP 1.2 We begin by expressing $1 trillion in scientific notation:
$1 trillion = $1 × 1012 .
R
Divide by 100 = 102 to find the number of $100 bills:
AD
1012
$1 trillion = = 1010 $100 bills.
102
Calculate the height of a stack of 1010 $100 bills:
0.11 mm 1m
1010 bills · · = 1.1 × 106 m.
bill 1000 mm
ES
Now we can convert from meters to miles, again with a product of ratios:
100 cm 1 in 1 ft 1 mi
1.1 × 106 m · · · · = 683.51 miles.
1m 2.54 cm 12 in 5280 ft
1–1
, 1–2 CHAPTER 1. Circuit Variables
AP 1.3 [a] First we use Eq. (1.2) to relate current and charge:
dq
i= = 0.25te−2000t .
M
dt
Therefore, dq = 0.25te−2000t dt.
To find the charge, we can integrate both sides of the last equation. Note
PR
that we substitute x for q on the left side of the integral, and y for t on
the right side of the integral:
Z q(t) Z t
dx = 0.25 ye−2000y dy.
q(0) 0
We solve the integral and make the substitutions for the limits of the
ES
integral:
t
e−2000y
q(t) − q(0) = 0.25 (−2000y − 1)
(−2000)2 0
= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9
SI
= 62.5 × 10−9 (1 − 2000te−2000t − e−2000t ).
But q(0) = 0 by hypothesis, so
q(t) = 62.5(1 − 2000te−2000t − e−2000t ) nC.
VE
[b] q(0.001) = (62.5)[1 − 2000(0.001)e−2000(0.001) − e−2000(0.001) ] = 37.12 nC.
75 × 10−6 C/s
AP 1.4 n = = 4.681 × 1014 elec/s.
1.6022 × 10−19 C/elec
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
G
R
Also sketch the four figures from Fig. 1.6:
AD
ES
, Problems 1–3
[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 250 mA of current
entering Terminal 2 is the same as 250 mA of current leaving Terminal 1.
M
We get
(a) v = 50 V, i = −0.25 A; (b) v = 50 V, i = 0.25 A;
PR
(c) v = −50 V, i = −0.25 A; (d) v = −50 V, i = 0.25 A.
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (50)(−0.25) = −12.5 W.
[c] Since the power is less than 0, the box is delivering power.
Z t
ES
AP 1.6 p = vi; w= p dx.
0
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
SI
VE
Note that in constructing the plot above, we used the fact that 60 hr
= 216,000 s = 216 ks.
G
p(0) = (6)(15 × 10−3 ) = 90 × 10−3 W;
p(216 ks) = (4)(15 × 10−3 ) = 60 × 10−3 W;
R
1
w = (60 × 10−3 )(216 × 103 ) + (90 × 10−3 − 60 × 10−3 )(216 × 103 ) = 16,200 J.
2
AD
AP 1.7 [a] p = vi = (15e−250t )(0.04e−250t ) = 0.6e−500t W;
p(0.01) = 0.6e−500(0.01) = 0.6e−5 = 0.00404 = 4.04 mW.
Z ∞ Z ∞ ∞
0.6 −500x
[b] wtotal = p(x) dx = 0.6e−500x dx = e
0 0 −500 0
ES
= −0.0012(e∞ − e0 ) = 0.0012 = 1.2 mJ.
, 1–4 CHAPTER 1. Circuit Variables
Chapter Problems
M
P 1.1 [a] Use a product of ratios to convert 40,075 km to inches:
1000 m 100 cm 1 in
40,075 km · · · = 1,577,755,905.5 in.
PR
1 km 1m 2.54 cm
Now calculate the number of $20 bills this distance represents:
1,577,755,905.5 in
Number of bills = = 256,963,502.5 ≈ 256,963,503.
6.14 in/bill
The total dollar amount represented by this number of $20 bills is
ES
Total dollars = (256,963,503)($20) ≈ $5.14 billion.
[b] The weight of the number of bills calculated in part (a) is
(256,963,503)(1 g) = 256,963.5 kg. Use a product of ratios to convert kg
to tons:
2.2 lbs 1 ton
256,963.5 kg · · = 282.66 tons.
SI
1 kg 2000 lbs
1s
P 1.2 [a] 8(65.5 × 106 ) bits · = 10.48 s.
50 × 106 bits
1s
VE
[b] 8(65.5 × 106 ) bits · = 262 ms.
2 × 109 bits
1s
[c] 8(74 × 1012 ) bits · 6
= 11.84 × 106 s
50 × 10 bits
1 min 1 hr 1 day
11.84 × 106 s · · · = 137 days!
60 s 60 min 24 hr
G
1s
[d] 8(74 × 1012 ) bits · = 296 × 103 s.
2 × 109 bits
R
1 min 1 hr 1 day
296 × 103 s · · · = 3.4 days.
60 s 60 min 24 hr
P 1.3 [a] To begin, we calculate the number of pixels that make up the display:
AD
npixels = (1920)(1080) = 2,073,600 pixels.
Each pixel requires 24 bits of information. Since 8 bits equal one 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:
ES
2,073,600 pixels 3 bytes
nbytes = · = 6,220,800 bytes/display.
1 display 1 pixel
