1 Circuit Variables Assessment Problems AP 1.1 Use a product of ratios to convert 95% of the speed of light from meters per
second to miles per second:
(0:95)
310 8 m
- s
100 cm
- m
- in
2:54 cm
- ft
12 in
- mile
5280 feet =
177;090:79 miles
- s
:
Now set up a proportion to determine how long it takes this signal to travel
950 miles:
177;090:79 miles
- s
= 950 miles xs
:
Therefore, x= 950
177;090:79
= 0:00536 = 5:3610
3
s = 5:36 ms.
AP 1.2 We begin by expressing $1 trillion in scientic notation:
$1 trillion = $110 12
:
Divide by 100 = 10 2
to nd the number of $100 bills:
$1 trillion = 10 12 10 2 = 10 10
$100 bills:
Calculate the height of a stack of 10 10
$100 bills:
10 10 bills
0:11 mm
bill
- m
1000 mm
= 1:110
6 m.
Now we can convert from meters to miles, again with a product of ratios:
1:110
6 m 100 cm
- m
- in
2:54 cm
- ft
12 in
- mi
5280 ft
= 683:51 miles.
1{1 1 / 3
1{2CHAPTER 1. Circuit Variables
AP 1.3[a]First we use Eq. (1.2) to relate current and charge:
i= dq dt
= 0:25te
2000t
:
Therefore,dq= 0:25te
2000t
dt:
To nd the charge, we can integrate both sides of the last equation. Note that we substitutexforqon the left side of the integral, andyforton
the right side of the integral:
Z q(t) q(0)
dx= 0:25
Z t
ye 2000y
dy:
We solve the integral and make the substitutions for the limits of the
integral:
q(t)q(0) = 0:25
e 2000y
(2000)
2 (2000y1)
t
= 62:510
9 e 2000t
(2000t1) + 62:510
9
= 62:510
9 (12000te 2000t e 2000t
):
Butq(0) = 0 by hypothesis, so
q(t) = 62:5(12000te
2000t e 2000t
) nC:
[b]q(0:001) = (62:5)[12000(0:001)e
2000(0:001)
e
2000(0:001)
] = 37:12 nC.
AP 1.4n= 7510 6 C/s
1:602210
19 C/elec
= 4:68110
14
elec/s:
AP 1.5 Start by drawing a picture of the circuit described in the problem statement: Also sketch the four gures from Fig. 1.6: 2 / 3
Problems1{3 [a]Now we have to match the voltage and current shown in the rst gure 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.We get (a)v= 50 V; i=0:25 A; (b)v= 50 V,i= 0:25 A; (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.AP 1.6p=vi;w= Z t
p dx:
Since the energy is the area under the power vs. time plot, let us plotpvs.t.Note that in constructing the plot above, we used the fact that 60 hr = 216;000 s = 216 ks.p(0) = (6)(1510 3
) = 9010
3 W; p(216 ks) = (4)(1510 3
) = 6010
3 W; w= (6010 3
)(21610
3
- +
1 2 (9010 3 6010 3
)(21610
3
) = 16;200 J:
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: [b]wtotal= Z 1
p(x)dx= Z 1
0:6e
500x dx=
0:6
500 e 500x
1
=0:0012(e
1 e
- = 0:0012 = 1:2 mJ:
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