ENGR-2150 CH-25 Homework-5 -SOLUTIONS
1. Exercise 25.3
2. Exercise 25.10
3. Exercise 25.12
4. Exercise 25.23
5. Exercise 25.39
6. Exercise 25.47
7. Problem 25.73
8. Problem 25.81
9. Exercise 25.52
10. Exercise 25.33
25.3.
IDENTIFY: I  Q/t. J  I /A. J  n q vd
SET UP:
A  ( /4) D2 , with D  205 103 m. The charge of an electron has magnitude e  160 1019 C.
EXECUTE: (a) Q  It  (500 A)(100 s)  500 C. The number of electrons is
(b) J 
(c) vd 
I
( /4) D
2

500 A
( /4)(205 103 m)2
Q
 312  1019.
e
 151 106 A/m2.
J
151106 A/m2

 111104 m/s  0111 mm/s.
n q (85  1028 m3 )(160 1019 C)
EVALUATE: (d) If I is the same, J  I /A would decrease and vd would decrease. The number of electrons
passing through the light bulb in 1.00 s would not change.
25.10.(a) IDENTIFY: Start with the definition of resistivity and solve for E.
SET UP: E  J  I / r 2
EXECUTE: E  (1.72  108   m)(2.75 A)/[ (0.001025 m)2 ]  1.43 102 V/m
EVALUATE: The field is quite weak, since the potential would drop only a volt in 70 m of wire.
(b) IDENTIFY: Take the ratio of the field in silver to the field in copper.
SET UP: Take the ratio and solve for the field in silver: ES  EC ( S/C )
EXECUTE: ES  (0.0143 V/m)[(1.47)/(1.72)]  1.22  102 V/m
EVALUATE: Since silver is a better conductor than copper, the field in silver is smaller than the field in copper.
25.12.IDENTIFY: E   J , where J  I /A. The drift velocity is given by I  n q vd A.
SET UP: For copper,   172 108  m. n  85 1028 /m3.
EXECUTE: (a) J 
I
36 A

 681  105 A/m2 .
A (23  103 m)2
(b) E  J  (172  108   m)(681  105 A/m2 )  0012 V/m.
(c) The time to travel the wire’s length l is
l ln q A (40 m)(85 1028 /m3 )(16 1019 C)(23 103 m) 2
t 

 80  104 s.
vd
I
36 A
t  1333 min  22 hrs!
EVALUATE: The currents propagate very quickly along the wire but the individual electrons travel very slowly.
25.23.IDENTIFY and SET UP: Eq. (25.5) relates the electric field that is given to the current density. V  EL gives the
potential difference across a length L of wire and Eq. (25.11) allows us to calculate R.
EXECUTE: (a) Eq. (25.5):   E /J so J  E /
From Table 25.1 the resistivity for gold is 244 108  m.
J
E


0.49 V/m
2.44  108  m
 2.008  107 A/m2
I  JA  J r 2  (2.008 107 A/m2 ) (0.42 103 m)2  11 A
(b) V  EL  (049 V/m)(64 m)  31 V
(c) We can use Ohm’s law (Eq. (25.11)): V  IR.
R
V 31 V

 028
I 11 A
EVALUATE: We can also calculate R from the resistivity and the dimensions of the wire (Eq. 25.10):
R
L
A

 L (2.44  108  m)(6.4 m)

 0.28, which checks.
 r2
 (0.42 103 m)2
25.39.IDENTIFY: The bulbs are each connected across a 120-V potential difference.
SET UP: Use P  V 2 /R to solve for R and Ohm’s law ( I  V /R) to find the current.
EXECUTE: (a) R  V 2 /P  (120 V)2 /(100 W)  144 .
(b) R  V 2 /P  (120 V)2 /(60 W)  240 
(c) For the 100-W bulb: I  V /R  (120 V)/(144 )  0.833 A
For the 60-W bulb: I  (120 V)/(240 )  0.500 A
EVALUATE: The 60-W bulb has more resistance than the 100-W bulb, so it draws less current.
25.47.(a) IDENTIFY and SET UP: P  VI and energy  (power)  (time).
EXECUTE: P  VI  (12 V)(60 A)  720 W
The battery can provide this for 1.0 h, so the energy the battery has stored is
U  Pt  (720 W)(3600 s)  26 106 J.
(b) IDENTIFY and SET UP: For gasoline the heat of combustion is Lc  46 106 J/kg. Solve for the mass m
required to supply the energy calculated in part (a) and use density   m/V to calculate V.
EXECUTE: The mass of gasoline that supplies 26  106 J is m 
The volume of this mass of gasoline is
m 00565 kg
 1000 L 
V 
 63  105 m3 
  0063 L.
 900 kg/m3
 1 m3 
26 106 J
46  106 J/kg
 00565 kg.
(c) IDENTIFY and SET UP: Energy  (power)  (time); the energy is that calculated in part (a).
U 26  106 J

 5800 s  97 min  1 6 h.
P
450 W
EVALUATE: The battery discharges at a rate of 720 W (for 60 A) and is charged at a rate of 450 W, so it takes
longer to charge than to discharge.
EXECUTE: U  Pt ,t 
25.73.IDENTIFY: Set the sum of the potential rises and drops around the circuit equal to zero and solve for I.
SET UP: The circuit is sketched in Figure 25.73.
EXECUTE:
  IR  V  0
  IR   I   I 2  0
 I 2  (R   )I    0
Figure 25.73
The quadratic formula gives I  (1/2 ) ( R   )  ( R   ) 2  4 


I must be positive, so take the  sign
I  (1/2 ) ( R   )  ( R   ) 2  4 


I  2692 A  4116 A  142 A
EVALUATE: For this I the voltage across the thermistor is 8.0 V. The voltage across the resistor must then be
126 V  80 V  46 V, and this agrees with Ohm’s law for the resistor.
25.81.IDENTIFY and SET UP: The terminal voltage is Vab    Ir  IR, where R is the resistance connected to the
battery. During the charging the terminal voltage is Vab    Ir. P  VI and energy is E  Pt. I 2r is the rate
at which energy is dissipated in the internal resistance of the battery.
EXECUTE: (a) Vab    Ir  120 V  (100 A) (024 )  144 V.
(b) E  Pt  IVt  (10 A) (144 V) (5) (3600 s)  259 106 J.
(c) Ediss  Pdisst  I 2rt  (10 A)2 (024 ) (5) (3600 s)  432 105 J.
(d) Discharged at 10 A: I 

rR
R
  Ir  120 V  (10 A) (024 )
I
10 A
 096 .
(e) E  Pt  IVt  (10 A) (96 V) (5) (3600 s)  173 106 J.
(f) Since the current through the internal resistance is the same as before, there is the same energy dissipated as
in (c): Ediss  432 105 J.
EVALUATE: (g) Part of the energy originally supplied was stored in the battery and part was lost in the internal
resistance. So the stored energy was less than what was supplied during charging. Then when discharging, even
more energy is lost in the internal resistance, and only what is left is dissipated by the external resistor.
25.52.IDENTIFY: The power delivered to the bulb is I 2 R. Energy  Pt.
SET UP: The circuit is sketched in Figure 25.52. rtotal is the combined internal resistance of both batteries.
EXECUTE: (a) rtotal  0. The sum of the potential changes around the circuit is zero, so
15 V  15 V  I (17)  0. I  01765 A. P  I 2 R  (01765 A)2 (17)  0530 W. This is also
(30 V)(01765 A).
(b) Energy  (0530 W)(50 h)(3600 s/h)  9540 J
0530 W
P
0265 W
 0265 W. P  I 2 R so I 

 0125 A.
2
R
17
The sum of the potential changes around the circuit is zero, so 15 V  15 V  IR  Irtotal  0.
(c) P 
30 V  (0125 A)(17)
 70.
0125 A
EVALUATE: When the power to the bulb has decreased to half its initial value, the total internal resistance of
the two batteries is nearly half the resistance of the bulb. Compared to a single battery, using two identical
batteries in series doubles the emf but also doubles the total internal resistance.
rtotal 
Figure 25.52
25.33.IDENTIFY: The voltmeter reads the potential difference Vab between the terminals of the battery.
SET UP: open circuit I  0. The circuit is sketched in Figure 25.33a.
EXECUTE: Vab    308 V
Figure 25.33a
SET UP: switch closed The circuit is sketched in Figure 25.33b.
EXECUTE: Vab    Ir  297 V
r
  297 V
I
308 V  297 V
r
 0067
165 A
Figure 25.33b
Vab 297 V

 180.
I
165 A
EVALUATE: When current flows through the battery there is a voltage drop across its internal resistance and its
terminal voltage V is less than its emf.
And Vab  IR so R 