Recitation Week 6
Chapters 25 and 26
Problem 25.57. An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter
and is 14.0 m long. The resistance between the ends is 0.104 Ω. (a) What is the resistivity of the material? (b) If the electricfield magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has 8.5 · 1028 free electrons per
cubic meter, find the average drift speed under the conditions of (b).
(a) Using the resistivity formula (Eqn. 25.10)
R=
ρ
A
L
ρ=
RA
Rπ(d/2)2
0.104 Ω · π · (1.25 · 10−3 m)2
=
=
= 3.65 · 10−8 Ω· .
L
L
14.0 m
(1)
(b) Because the electric field is constant inside the conductor, the voltage difference across the conductor is
V = EL = 17.9 V .
(2)
Using Ohm’s law for the current through the conductor
V = IR
I=
V
= 172 A .
R
(3)
(c) In a time ∆t, all the electrons within v∆t of a given cross section will drift through that cross section. Since we know
the cross sectional area of the wire is A, that corresponds to a volume of
V = Av∆t .
(4)
We know the density of electrons n = 8.5 · 1028 e− /m3 , so the number of electrons crossing the section is
N = nV = nAv∆t
(5)
Q = eN = enAv∆t ,
(6)
and the charge crossing is
where e = 1.60 · 10−18 C is the charge on one electron.
The current in the wire is thus
enAv∆t
Q
=
= enAv ,
∆t
∆t
(7)
I
172 A
=
= 2.5 mm/s .
enA
1.8 · 10−18 C · 8.5 · 1028 e− /m3 · π(1.25 · 10−3 m)2
(8)
I=
and the drift velocity is
v=
Problem 25.72. A typical cost for electric power is 12.0¢ per kilowatt-hour. (a) Some people leave their porch light on all
the time. What is the yearly cost to keep a 75 W bulb burning day and night? (b) Suppose your refrigerator uses 400 W of
power when it’s running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?
This is just a unit conversion problem, but it’s good to get a feel for how much a lightbulb costs. Note that PECO power in
Philly is currently around 15.4¢/kW · hour (http://www.jea.com/services/electric/rates_quarterly.asp).
(a)
75 W ·
24 hours 365 days
$0.120
·
· 3
= $78.84
1 day
1 year
10 W·hour
(9)
400 W ·
$0.120
8 hours 365 days
·
· 3
= $140.16
1 day
1 year
10 W·hour
(10)
(b)
Problem 26.6. For the circuit shown in Fig. 26.40 both meters are idealized, the battery has no appeciable internal resistance,
and the ammeter reads 1.25 A. (a) What does the voltmeter read? (b) What is the emf Eof the battery?
25.0
+---V---+ +-/\/\/-A-+
|
| | 15.0
|
+-/\/\/-+--+-/\/\/---+------+
| 45.0
| 15.0
Z 10.0 |
|
+-/\/\/---+
|
|
35.0
E
|
+-------/\/\/----i|---------+
It’s always a good idea to start off by labeling things.
R1=25.0 I1=1.25A
+---V---+ +-/\/\/--<-+
|
| |R2=15.0 I2|
+-/\/\/-+--+-/\/\/--<-a----------+
| 45.0=Rv |R3a=15.0 Z 10.0=R3b |
|
+-/\/\/--<-+
|
| It
35.0=Re
E I3
|
+--->---/\/\/----i|--------------+
(a) First, we’ll get a handle on the I2/I2/I3 situation, since that’s our only handle on current at the moment. Looping
counter-clockwise from a through wires 1 and 2,
0 = −I1 R1 + I2 R2
25.0 Ω
R1
1.25 A = 2.08 A .
I1 =
I2 =
R2
15.0 Ω
(11)
(12)
Looping counter-clockwise from a through wires 1 and 3,
0 = −I1 R1 + I3 R3a + I3 R3b = −I1 R1 + I3 (R3a + R3b )
R1
25.0 Ω
1.25 A = 1.25 A .
I2 =
I1 =
R3a + R3b
(15.0 + 10.0) Ω
(13)
(14)
Now that we know all about that little cluster, we can move out to the larger loop. Using the junction rule at junction a
0 = It − I1 − I2 − I3
(15)
It = I1 + I2 + I3 = 4.58 A .
(16)
Knowing the current in the larger loop, we just use Ohm’s law on RV
V = It RV = 206 V
(17)
(b) For this part, we do the full loop, heading counter clockwise from a and using wire 1 (although you could use any of wires
1, 2, or 3, since they must all have the same voltage across them)
0 = −I1 R1 − It RV − It RE + E
(18)
E = I1 R1 + It RV + It RE = 1.25 A · 25.0 Ω + 206 V + 4.58 A · 35.0 Ω = 398 V
(19)
Problem 26.21. In the circuit shown in Fig. 26.49 find (a) the current in resistor R; (b) the resistance R; (c) the unknown
emf E. (d) If the circuit is broken at point x, what is the current in resistor R?
28.0V
R
I_1
+----|i-----/\/\/---<-------+
|
E
6.00 4.00A=I_2 |
+----|i--x--/\/\/---<-------a
|
3.00
6.00A=I_3 |
+---------/\/\/---->--------+
(a) Labeling the currents as above, we use Kirchoff’s junction rule summing the currents entering node a.
0 = −I1 − I2 + I3
I1 = I3 − I2 = 2.00 A
(20)
(b) We’ll use a loop rule for this part. We still don’t know Ein wire 2, so lets loop around 1 and 3 counter-clockwise starting
from a.
0 = −I1 R + 28.0 V − I3 R3
28.0 V − I3 R3
28.0 V − 6.00 A · 3.00 Ω
R=
=
= 5.00 Ω
I1
2.00 A
(21)
(22)
(c) Now we’ll use a loop involving wire 2, say 2 and 3 counter-clockwise starting from a.
0 = −I2 R2 + E − I3 R3
(23)
E = I2 R2 + I3 R3 = 4.00 A · 6.00 Ω + 6.00 A · 3.00 Ω = 42.0 V
(24)
(d) Breaking the circuit at x means that I2 → 0 so I1 = I2 = I. Applying our same loop rule from (b)
0 = −IR + 28.0 V − IR3
28.0 V
28.0 V
I=
=
= 3.50 A
R + R3
8.00 Ω
(25)
(26)
Problem 26.22. Find the emfs E1 and E2 in the circuit of Fig. 26.50, and find the potential difference of point b relative to
point a.
1.00 20.0V 6.00
+---/\/\/--|i----/\/\/---+
1.00A v
4.00
1.00
E1
|
a---/\/\/----/\/\/--|i---b
2.00A v
1.00
E2
2.00
|
+---/\/\/--|i----/\/\/---+
First we’ll simplify resistors in series, leading to the equivalent circuit
I_3=1.00A 20.0V 7.00=R_3
+---<-------|i----/\/\/---+
| I_1
R_1=5.00 E1
|
a---<--------/\/\/--|i----b
| I_2=2.00A E2
3.00=R_2|
+--->-------|i----/\/\/---+
We know everything about wire 3, so Vba is given by
Vba = −20.0 V + I3 R3 = −20.0 V + 1.00 A · 7.00 Ω = −13.0 V
(27)
We also know everything about wire 2, so we can find E2 by looping clockwise from a through 3 and 2.
0 = Vba + I2 R2 + E2
E2 = −(Vba + I2 R2 ) = 7.00 V
(28)
(29)
To find E1 we’ll need I1 . Applying Kirchoff’s junction rule to a
0 = I3 + I1 − I2
I1 = I2 − I3 = 2.00 A − 1.00 A = 1.00 A .
(30)
(31)
Looping clockwise from a through 3 and 1
0 = Vba + E1 − I1 R1
E1 = I1 R1 − Vba = 18.0 V
(32)
(33)
Problem 26.48. In the circuit shown in Fig. 26.61, C = 5.90 µF, E = 28.0 V, and the emf has negligible resistance. Initially
the capacitor is uncharged and the switch S is in position 1, The switch is then moved to position 2, so that the capacitor
begins to charge. (a) What will be the charge on the capacitor a long time after the switch is moved to position 2? (b) After
the switch has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 µC. What is the value of the
resistance R? (c) How long after the switch is moved to position 2 will the charge on the capacitor be equal to 99.0% of the
final value found in (a).
+---------.
|
/
|
S1 S2
|
|
|
=== C
|
--- E
|
|
|
|
|
+-/\/\-+----+
R
(a) After a long time in position 2, the capacitor will have become fully charged, no more current will flow through the
circuit, and the voltage across the capacitor will balance that across the battery.
Q = CV = CE = 5.90 µF · 28.0 V = 165 µC
(b) Here we use the charging capacitor formula (Eqn. 26.12)
q = CE 1 − e−t/RC
q
= 1 − e−t/RC
CE
q
e−t/RC = 1 −
CE
−t
q = ln 1 −
RC
CE
−3.00 ms
−t
=
= 463 Ω
R=
q
µC
C ln 1 − CE
5.90 µF · ln 1 − 110
165 µC
(34)
(35)
(36)
(37)
(38)
(39)
(c) Again, we use the charging-capacitor formula
−t
q = ln 1 −
RC
CE
q t = −RC ln 1 −
= −463 Ω · 5.90 µF · ln(1 − 0.99) = 12.6 ms
CE
(40)
(41)