Download CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE

CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE
25.4.
25
(a) IDENTIFY: By definition, J = I/A and radius is one-half the diameter.
SET UP: Solve for the current: I = JA = Jπ(D/2)2
EXECUTE: I = (1.50 106 A/m2)(π)[(0.00102 m)/2]2 = 1.23 A
EVALUATE: This is a realistic current.
(b) IDENTIFY: The current density is J = nqvd
SET UP: Solve for the drift velocity: vd = J/nq
EXECUTE: Since most laboratory wire is copper, we use the value of n for copper, giving
28
3
vd (1.50 106 A/m2 ) /[(8.5 10 el/m )(1.60 10 19 C) = 1.1 10 4 m/s = 0.11 mm/s
EVALUATE: This is a typical drift velocity for ordinary currents and wires.
L
and solve for L.
A
SET UP: A
D2 / 4 , where D 0.462 mm .
RA (1.00 )( 4)(0.462 10 3 m) 2
EXECUTE: L
9.75 m.
1.72 10 8
m
EVALUATE: The resistance is proportional to the length of the wire.
25.16.
IDENTIFY:
Apply R
25.24.
IDENTIFY:
Apply R
L
and V
A
IR .
r2
RA VA (4.50 V) (6.54 10 4 m) 2
1.37 10 7
m.
EXECUTE:
L
IL
(17.6 A)(2.50 m)
EVALUATE: Our result for shows that the wire is made of a metal with resistivity greater than that of good
metallic conductors such as copper and aluminum.
SET UP:
A
25.32.
IDENTIFY: When current passes through a battery in the direction from the terminal toward the + terminal,
the terminal voltage Vab of the battery is Vab E Ir . Also, Vab IR, the potential across the circuit resistor.
SET UP: E 24.0 V . I 4.00 A.
E Vab 24.0 V 21.2 V
EXECUTE: (a) Vab E Ir gives r
0.700 .
I
4.00 A
Vab 21.2 V
(b) Vab IR 0 so R
5.30 .
I
4.00 A
EVALUATE: The voltage drop across the internal resistance of the battery causes the terminal voltage of the
24.0 V
battery to be less than its emf. The total resistance in the circuit is R r 6.00 . I
4.00 A, which
6.00
agrees with the value specified in the problem.
25.38.
IDENTIFY: The sum of the potential changes around the loop is zero.
SET UP: The voltmeter reads the IR voltage across the 9.0
resistor. The current in the circuit is
counterclockwise because the 16 V battery determines the direction of the current flow.
EXECUTE:
(a) Vbc 1.9 V gives I
(b) 16.0 V 8.0 V (1.6
Vbc Rbc 1.9 V 9.0
9.0
0.21 A.
R)(0.21A) and R
1.4
5.48 V
0.21 A
(c) The graph is sketched in Figure 25.38.
EVALUATE: In Exercise 25.36 the current is 0.47 A. When the 5.0
resistor the current decreases to 0.21 A.
26.1 .
resistor is replaced by the 26.1
Figure 25.38
25.47.
IDENTIFY and SET UP:
By definition p
P
. Use P VI , E VL and I
LA
JA to rewrite this expression in
terms of the specified variables.
EXECUTE:
V
L
E and
(a) E is related to V and J is related to I, so use P = VI. This gives p
I
A
J so p
(b) J is related to I and
I
L
so p
A
JA and R
(c) E is related to V and
V
EVALUATE:
25.78.
L
so p
A
EL and R
VI
LA
EJ
is related to R, so use P
IR2 . This gives p
I 2R
.
LA
J 2 A2 L 2
J
LA2
is related to R, so use P V 2 / R. This gives p
E 2 L2
LA
A
L
For a given material (
E2
V2
.
RLA
.
constant), p is proportional to J 2 or to E 2 .
V
. P VI .
I
SET UP: When the temperature increases the resistance increases and the current decreases.
V V
(1 [T T0 ]) . I 0 IT (1 [T T0 ]) .
EXECUTE: (a)
IT I 0
RT
IDENTIFY:
T T0
I0
IT
IT
(b) (i) P VI
EVALUATE:
R0 (1
[T T0 ]) . R
1.35 A 1.23 A
217 C° . T 20°C 217 C 237°C
(1.23 A)(4.5 10 4 (C°) 1 )
(120 V)(1.35 A) 162 W (ii) P (120 V)(1.23 A) 148 W
P V 2 / R shows that the power dissipated decreases when the resistance increases.