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Appendix C
29 E
240 V V 38 104 N/C 64 10 3 m d 12 10 14 N F F E , so q 32 10 19 C E 38 104 N/C q q # electrons qe 32 10 19 C 160 10 19 C
Lamp
Battery Switch
23 a I b E
V R 120 V 15
2 31 q q 33 q C V, so the larger capacitor has a greater charge (68 10 C V C( V2 (22 10 20 10
6 5 6
Potentiometer
80 A 29 104 J
F)(24 V)
16 10
I2Rt
(80 A)2(15 )(300 s)
so q
V1) F)(150 V 60 V) C
c 29 104 J, because all electrical energy is converted to thermal energy 25 a E Pt (022)(1000 J/s)(10 min) 1 min 13 103 J b E Pt (078)(1000 J/s)(10 min) 1 min 47 103 J 27 E IVt = I(2V)(
t ) 2 60 s 60 s
22
1 P 3 P I 5 P I 7 V 9 a R b P IV IV
P V 75 W 125 V
(050 A)(125 V)
63 J/s
63 W
For a given amount of energy, doubling the voltage will divide the time by 2 t 29 a I
22 h 2 V R
060 A
11 h
115 V 12,0 00
Solutions for Practice Problems
P V 090 W 30 V
96 10
A 11 W
030 A 12 102 V
b P
(115 V)(96 10 (11 10 $010
c Cost IR
kW)($012/kWh)
(38 A)(32 )
120 V 050 A
(30 days)(24 h/day)
24 102 60 101 W 31 Echarge t
E IV
(13)IVt
(13)(55 A)(12 V)(10 h) 82 h
858 Wh
(050 A)(120 V)
858 Wh (75 A) (14 V)
11 a The new value of the current is
060 A 2
030 A IR (030 A)(21 102 ) 63 101 V
So V
23
1 R R1 20 60 I
V R 120 V 60
b The total resistance of the circuit is now Rtotal Rres
V I 125 V 030 A
R3 20 20
42 102
Therefore, Rtotal Rlamp 21 102 42 102 21 102 c P 13 IV (030 A)(63 101 V) 19 W
3 a It will increase b I
V , so it will decrease R
c No It does not depend on the resistance 5 V1 V2 IR1 IR2 IR3 V2 (3 A)(10 ) (3 A)(15 ) (3 A)(5 ) V3 30 V 30 V 45 V 15 V 45 V 15 V 90 V
A 85 mA 45 V I
V I 45 V 0085 A
V3 V1
voltage of battery 7 a I
V R 170 V 2550
53
667 mA
Appendix C
b First, find the total resistance, then solve for voltage R V c P PA PB RA IR IV RB 255 292 547 365 V 243 W 113 W 130 W
19 Neither They both reach maximum dissipation at the same voltage P V
V2 R
(667 mA)(547 ) (667 mA)(365 V)
The voltage is equal across parallel resistors, so: V P1R1 P2R2 (4 W)(60 )
I2RA I2RB
(667 mA)2(255 ) (667 mA)2(292 )
(2 W)(12 ) 5 V maximum
d Yes The law of conservation of energy states that energy cannot be created or destroyed; therefore, the rate at which energy is converted, or power is dissipated, will equal the sum of all parts 9 The resistor with the lower resistance will dissipate less power; thus, it will be cooler
25 By conservation of energy (and power): PT P1 P2 P3 30 W 15 W
20 W 65 W PT IV
PT V
11 a R
R1 22 55
R2 33
65 W 12 V
054 A
27 Then, all of the working lights are in series The 12 working lights will burn with equal intensity 22 A
b I c V
120 V 55
24
1 a repulsive 48 V b attractive 3 the bottom (the point) 72 V 5 a from south to north b west 7 the pointed end 9 Yes Connect the potentiometer in series with the power supply and the coil Adjusting the potentiometer for more resistance will decrease the current flow and the strength of the field 17 F
1 150
Appendix C
Across a 22- resistor, V 120 V R1 (22 ) V1 IR1
R 55
Across a 33- resistor, V 120 V V2 IR2 R2 (33 )
R 55
d V 13 VB
48 V
VRB RA RB
72 V
120 102 V
(45 V)(235 k ) 475 k 235 k
15 V 15 a
1 R 1 R1 1 R2 1 R3 1 150
1 150 3 150
BIL (040 N/A m)(80 A)(050 m) 16 N BIL, F
F IL
19 F B
weight of wire
035 N (60 A)(0400 m)
015 T
R b I c I
500
V R V R1 300 V 500 300 V 150
21 Opposite to the direction of the electron motion 600 A 200 A 23 F Bqv (90 10 86 10 25 F Bqv (50 10 64 10
2 16 2 16
T)(2)(160 10 N
C)(30 104 m/s)
17 a Yes, it gets smaller b Yes, it gets larger c No, it remains the same Currents are independent
T)(2)(160 10 N
C)(40 104 m/s)
Appendix C
25
1 a EMF BLv (04 T)(05 m)(20 m/s) 4V b I
EMF R 4V 60
(180 104 V)(050 A) 600 V
15 102 V
26
1 Bqv r
mv 2 r mv Bq (167 10 27 kg)(75 103 m/s) (060 T)(160 10 19 C)
07 A 3 a EMF BLv (10 T)(300 m)(20 m/s) 60 101V b I 5 a Veff
EMF R 60 101 V 150
13 10 3 Bqv
mv 2 r mv Bq (911 10 (60 10
40 A
31 2
(0707)Vmax (0707)(170 V) 12 102 V 5 m
kg)(50 104 m/s) T)(160 10 19 C)
47 10
b Ieff
(0707)Imax (0707)(070 A) 049 A
Vmax
B2 r2q 2V (72 10
T)2(0085 m)2(160 10 2(110 V)
Solutions for Practice Problems
27 10
2 Imax 2
c R
Veff Ieff 170 V 070 A
Vmax Imax
7 Bqv v
E B 60 10 2 N/C 15 10 3 T
40 105 m/s 15 All electromagnetic waves travel through air or a vacuum at c, 300 108 m/s 17
c f 300 108 m/s 82 10 14 Hz
24 102 7 a Veff
Vmax 2 425 V 2
301 102 V b Ieff
Veff R 301 102 V 50 102
37 10 19 v
060 A 17
VP VS NP NS VP NS NP (600 V)(90,000) 300
299,792,458 m/s 100054
299712 108 m/s 21 v so K
c K c ( v )2 ( 300 2 43
180 104 V VP IP IP VSIS
VS IS VP
108 m/s 2 10 8 m/s
Appendix C
27
1 (23 eV) 3 m so v
10 ( 1601 eV
28
37 10
1 2 mv 2
1 En E2 E3 E4 3 E
911 10
2KE m
kg, KE
2(3 7 10 19 J) 911 10 31 kg
136 eV n2 136 eV (2)2 136 eV (3)2 136 eV (4)2
340 eV 151 eV 0850 eV
90 105 m/s 5 KE qV0 ( 160 10 51 10 7 KEmax
19 19
1 ) 22 1 ) 4
( 136 eV)( C)(32 J/C)
1 42 1 ( 136 eV)( 16
J hf0 196 eV 5
255 eV
x 0075 m 5 10 11 m 25 10 15 m
1240 eV nm 1240 eV nm 425 nm
x 7 a
2 103 m E 882 eV 215 eV 667 eV
0960 eV 9 hf0 Thus, 450 eV, so
450 eV b
1240 eV nm 450 eV
hc E (663 10 34 J s)(300 108 m/s) (215 eV)(160 10 19 J eV)
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