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Physics 104
Assignment 25
Q
.
t
25.1.
IDENTIFY and SET UP: The lightning is a current that lasts for a brief time. I 
25.3.
EXECUTE: Q  I t  (25,000 A)(40  106 s)  10 C.
EVALUATE: Even though it lasts for only 40 µs, the lightning carries a huge amount of charge since it is an
enormous current.
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 
25.8.
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  10 A/m 2

 111 104 m/s  0111 mm/s.
n q (85  1028 m 3 )(160  1019 C)
6
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.
IDENTIFY: I  Q/t. Positive charge flowing in one direction is equivalent to negative charge flowing in the
opposite direction, so the two currents due to Cl and Na  are in the same direction and add.
SET UP: Na  and Cl each have magnitude of charge q   e.
EXECUTE: (a) Qtotal  (nCl  nNa )e  (392 1016  268 1016 )(160 1019 C)  00106 C. Then
I
Qtotal 00106 C

 00106A  106 mA.
t
100 s
(b) Current flows, by convention, in the direction of positive charge. Thus, current flows with Na  toward the
negative electrode.
25.10.
EVALUATE: The Cl ions have negative charge and move in the direction opposite to the conventional current
direction.
(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 )
25.17.
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.
L
IDENTIFY: R 
.
A
SET UP: For copper,   172 108  m. A   r 2 .
EXECUTE: R 
(172  108  m)(240 m)
 0125
 (1025  103 m) 2
EVALUATE: The resistance is proportional to the length of the piece of wire.
L
25.21.IDENTIFY: R 
.
A
SET UP: L  1.80 m, the length of one side of the cube. A  L2 .
1
2.75  108  m
 1.53  108
A
1.80 m
L2 L
EVALUATE: The resistance is very small because A is very much larger than the typical value for a wire.
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 /
EXECUTE: R 
25.23.
L
L




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
25.29.
L
A

 L (2.44  108  m)(6.4 m)

 0.28, which checks.
 r2
 (0.42  103 m) 2
IDENTIFY: Use R 
L
A
to calculate R and then apply V  IR. P  VI and energy  Pt.
SET UP: For copper,   172 108  m. A   r 2 , where r  0050 m.
EXECUTE: (a) R 
25.38.
25.39.
L
A

(1.72  108   m)(100  103 m)
 (0.050 m) 2
 0.219. V  IR  (125 A)(0219)  274 V.
(b) P  VI  (274 V)(125 A)  3422 W  3422 J/s and energy  Pt  (3422 J/s)(3600 s)  123 107 J.
EVALUATE: The rate of electrical energy loss in the cable is large, over 3 kW.
IDENTIFY and SET UP: There is a single current path so the current is the same at all points in the circuit.
Assume the current is counterclockwise and apply Kirchhoff’s loop rule.
EXECUTE: (a) Apply the loop rule, traveling around the circuit in the direction of the current.
160 V  80 V
 0471 A. Our calculated
160 V  I (16   50   14   90 )  80 V  0. I 
170 
I is positive so I is counterclockwise, as we assumed.
(b) Vb  160 V  I (16 )  Va . Vab  160 V  (0471 A)(16 )  152 V.
EVALUATE: If we traveled around the circuit in the direction opposite to the current, the final answers would
be the same.
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 .
25.42.
25.49.
(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.
IDENTIFY: P  VI . Energy  Pt.
SET UP: P  (90 V)(013 A)  117 W
EXECUTE: Energy  (117 W)(15 h)(3600 s/h)  6320 J
EVALUATE: The energy consumed is proportional to the voltage, to the current and to the time.
IDENTIFY: Some of the power generated by the internal emf of the battery is dissipated across the battery’s
internal resistance, so it is not available to the bulb.
2
SET UP: Use P  I 2 R and take the ratio of the power dissipated in the internal resistance r to the total power.
Pr
I 2r
r
35
 2


 0123  123%
EXECUTE:
PTotal I (r  R) r  R 285
25.51.
EVALUATE: About 88% of the power of the battery goes to the bulb. The rest appears as heat in the internal
resistance.
IDENTIFY: Solve for the current I in the circuit. Apply Eq. (25.17) to the specified circuit elements to find the
rates of energy conversion.
SET UP: The circuit is sketched in Figure 25.51
EXECUTE: Compute I:
  Ir  IR  0
  120 V  200 A
I
r  R 10   50
Figure 25.51
(a) The rate of conversion of chemical energy to electrical energy in the emf of the battery is
P   I  (120 V)(200 A)  240 W.
(b) The rate of dissipation of electrical energy in the internal resistance of the battery is
25.53.
P  I 2r  (200 A)2 (10)  40 W.
(c) The rate of dissipation of electrical energy in the external resistor
R is P  I 2R  (200 A)2 (50)  200 W.
EVALUATE: The rate of production of electrical energy in the circuit is 24.0 W. The total rate of consumption
of electrical energy in the circuit is 4.00 W  20.0 W  24.0 W. Equal rate of production and consumption of
electrical energy are required by energy conservation.
V2
 VI . V  IR.
IDENTIFY: P  I 2 R 
R
SET UP: The heater consumes 540 W when V  120 V. Energy  Pt.
V 2 (120 V) 2
V2

 267
so R 
P
540 W
R
P 540 W
(b) P  VI so I  
 450 A
V 120 V
EXECUTE: (a) P 
V 2 (110 V) 2

 453 W. P is smaller by a factor of (110/120)2 .
R
267
EVALUATE: (d) With the lower line voltage the current will decrease and the operating temperature will
decrease. R will be less than 267 and the power consumed will be greater than the value calculated in part
(c).
(c) Assuming that R remains 267, P 
3