Download Document

MASSACHUSETTS INSTINUTE OF TECHNOLOGY
ESG Physics
8.02 with Kai
Spring 2003
Problem Set 5 Solution
Problem 1: 27.8
An electric current is given by the expression
I ( t ) = 100sin (120π t )
(1.1)
where I is in amperes and t is in seconds. What is the total charge carried by the current
1
s?
from t = 0 to t =
240
Solution:
Since I =
dq
,
dt
1
240
0
q = ∫ I dt = ∫
(100 ) sin (120π t ) dt =
−100 
π

cos − cos 0 

120π 
2

(1.2)
which gives
q = 0.265 C
2003 Spring 8.02 with Kai Problem set 5 Solution
(1.3)
1
Problem 2: 27.24
The rod in Figure P27.24 (not drawn to scale) is made of two materials. Both have a
square cross section of 3.00 mm on a side. The first material has a resistivity of
40.0 × 10−3 Ω ⋅ m and is 25.0 cm long, while the second material has a resistivity of
6.00 ×10−3 Ω ⋅ m and is 40.0 cm long. What is the resistance between the ends of the rod?
Solution:
Applying the formula,
R=
ρ1l1
A1
+
ρ 2l2
A2
=
ρ1l1 + ρ 2l2
d2
(2.1)
Substituting, we have
R=
( 4.00 ×10
−3
Ω ⋅ m ) ( 0.250 m ) + ( 6.00 × 10−3 Ω ⋅ m ) ( 0.400 m )
( 3.00 ×10−3 m )
2
(2.2)
which gives
R = 378 Ω
2003 Spring 8.02 with Kai Problem set 5 Solution
(2.3)
2
Problem 3: 27.53
A high-voltage transmission line with a diameter of 2.00 cm and a length of 200 km
carries a steady current of 1000 A. If the conductor is copper wire with a free charge
density of 8.00 × 1028 electrons/m3 , how long does it take one electron to travel the full
length of the cable?
Solution:
We find the drift velocity from
I = nqvd A = nqvd π r 2
(3.1)
Therefore,
vd =
I
nqπ r
2
=
(8.00 ×10
1000 A
28
m -3 )(1.60 ×10−19 C ) π (10−2 m )
2
(3.2)
which gives
vd = 2.49 × 10−4 ms -1
(3.3)
For the time that one electron takes to travel the full length of the cable, we have
t=
l
200 × 103 m
=
= 8.04×108 s
v 2.49 × 10−4 ms -1
(3.4)
which is equavalent to 25.5 years.
2003 Spring 8.02 with Kai Problem set 5 Solution
3
Problem 4: 27.66
A resistor is constructed by shaping a material of resistivity ρ into a hollow cylinder of
length L and with inner and outer radii ra and rb , respectively (Fig. P27.66).
In use, the application of a potential difference between the ends of the cylinder produces
a current parallel to the axis.
(a) Find a general expression for the resistance of such a device in terms of
L, ρ , ra and rb .
(b) Obtain a numeraical value for R when L = 4.00 cm, ra = 0.500 cm , rb = 1.20 cm ,
and ρ = 3.50 × 105 Ω ⋅ m
(c) Now suppose that the potential difference is applied between the inner and outer
surfaces so that the resulting current flows radially outward. Find a general
expression for the resistance of the device in terms of L, ρ , ra and rb .
(d) Calculate the value of R, using the parameter values given in part (b).
Solution:
(a)
R=
ρl
A
=
ρL
(4.1)
π ( rb2 − ra2 )
(b) Substituting the values given, we have
( 3.50 ×10 Ω ⋅ m ) ( 0.0400 m )
5
R=
2
2
π ( 0.0120 m ) − ( 0.00500 m ) 

(4.2)

which gives
R = 37.4 MΩ
(c)
Since
(4.3)
dR ρ
= , therefore
dl A
2003 Spring 8.02 with Kai Problem set 5 Solution
4
dR =
ρ dl
A
=
Therefore
R=
ρ dr
ρ dr
=
( 2π r ) L 2π L r
(4.4)
ρ r dr
2π L ∫r r
(4.5)
b
a
which gives
r 
ρ
ln  b 
2π L  ra 
(4.6)
(d) Substituting that the value given, we have
( 3.50 ×10 Ω ⋅ m ) ln  1.20
R=
5
2π ( 0.0400 m )
which gives
R = 1.22 MΩ
2003 Spring 8.02 with Kai Problem set 5 Solution



 0.500 
(4.7)
(4.8)
5
Problem 5: 27.70
A material of resistivity ρ is formed into the shape of a truncated cone of altitude h, as
shown in Figure P27.70.
The bottom end has a radius b, and the top end has a radius a. Assuming that the current
is distributed uniformly over any particular cross-section of the cone so that the current
density is not a function of radial position (although it does vary with position along the
axis of the cone), show that the resistance between the two ends is given by the
expression
ρ h 
R=  
(5.1)
π  ab 
Solution:
From the geometry of the longitudinal section of the resistor shown in
the figure, we see that
b−r b−a
=
y
h
(5.2)
From this, the radius at a distance y from the base is
r = (a − b)
y
+b
h
(5.3)
For a disk-shaped element of volume,
dR =
ρ dy
π r2
(5.4)
Therefore
R=
ρ h
π ∫0 
dy
y

( a − b ) h + b 
2003 Spring 8.02 with Kai Problem set 5 Solution
2
(5.5)
6
Using the integral formula
du
∫ ( au + b )
2
=
1
a ( au + b )
(5.6)
We have
R=
ρ h
π ab
2003 Spring 8.02 with Kai Problem set 5 Solution
(5.7)
7
Problem 6: 28.6
(a) Find the equivalent resistance between points a and b in Figure P28.6.
(b) Calculate the current in each resistor if a potential difference of 34.0 V is applied
between points a and b.
Solution:
(a)
Rp =
1
1
1
+
7.00 10.0
= 4.12Ω
Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω
(1.1)
(1.2)
(b) For the 4.00 Ω and 9.00 Ω resistors, since ∆V = IR , we have
I=
∆V 34.0
=
= 1.99 A
R
17.1
(1.3)
For the 7.00 Ω resistor, we have
8.18
= 1.17 A
7.00
(1.4)
8.18
= 0.818 A
10.0
(1.5)
I=
For the 10.0 Ω resistor, we have
I=
2003 Spring 8.02 with Kai Problem set 5 Solution
8
Problem 2: 28.16.
Two resistors connected in series have an equivalent resistance of 690 Ω . When they are
connected in parallel, their equivalent resistance is 150 Ω . Find the resistance of each
resistor.
Solution:
Denoting the two resistors as x and y,
x + y = 690
(1.6)
1
1 1
= +
150 x y
(1.7)
and
Substituting equation (2.1) to (2.2), we have
( 690 − x ) + x
1
1
1
= +
=
x ( 690 − x )
150 x 690 − x
(1.8)
x 2 − 690 x + 103,500 = 0
(1.9)
which gives
By solving this quadratic equation, we have
x = 470 Ω and y = 220 Ω
2003 Spring 8.02 with Kai Problem set 5 Solution
(1.10)
9
Problem 3: 28.17
In Figures 28.4 and 28.5, let R1 = 11.0 Ω , let R2 = 22.0 Ω , and let the battery have a
terminal voltage of 33.0 V.
(a) in the parallel circuit shown in Figure 28.5, which resistor uses more power?
(b) Verify that the sum of the power ( I 2 R ) used by each resistor equals the power
supplied by the battery ( I ∆V ) .
(c) In the series circuit, which resistor uses more power?
(d) Verify that the sum of the power ( I 2 R ) used by each resistor equals the power
supplied by the battery ( P = I ∆V ) .
(e) Which circuit configuration uses more power?
Solution:
(a) We have
∆V = IR
(1.11)
33.0 V=I1 (11.0 Ω ) and 33.0 V = I 2 ( 22.0 Ω )
(1.12)
I1 = 3.00 A and I 2 = 1.50 A
(1.13)
Therefore
Therefore,
Since P = I 2 R , we have
P1 = ( 3.00 A ) (11.0 Ω ) and P2 = (1.50 A ) ( 22.0 Ω )
2
2
(1.14)
which gives
2003 Spring 8.02 with Kai Problem set 5 Solution
10
P1 = 99.0 W and P2 = 49.5 W
(1.15)
Therefore, the 11.0- Ω resistor uses more power.
(b) Since P1 + P2 = 148 W , we have
P = I ( ∆V ) = ( 4.50 )( 33.0 ) = 148 W
(1.16)
Rs = R1 + R2 = 11.0 + 22.0 = 33.0 Ω
(1.17)
(c)
Since ∆V = IR ,
I=
V 33.0 Ω
=
= 1.00 A
P 33.0 V
(1.18)
Since P = I 2 R , we have
P1 = (1.00 A ) (11.0 Ω ) and P2 = (1.00 A ) ( 22.0 Ω )
2
2
(1.19)
and you will probably get
P1 = 110 W and P2 = 22.0 W
(1.20)
Therefore the 22.0- Ω resistor uses more power.
(d)
P1 + P2 = I 2 ( R1 + R2 ) = (1.00 A ) ( 33.0 Ω ) = 33.0 W
(1.21)
P = I ( ∆V ) = (1.00 A )( 33.0 V ) = 33.0 W
(1.22)
2
(e) The parallel configuration uses more power.
2003 Spring 8.02 with Kai Problem set 5 Solution
11
2003 Spring 8.02 with Kai Problem set 5 Solution
12
2003 Spring 8.02 with Kai Problem set 5 Solution
13