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Answer, Key – Homework 11 – David McIntyre
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The due time is Central time.
Chapters 27 and 28 problems.
001 (part 1 of 1) 0 points
It is known that about one electron per atom
of copper contributes to the current. The
atomic mass of copper is 63.54 g and its density is 8.92 g/cm3 .
Calculate the average drift speed of electrons traveling through a copper wire with a
cross-sectional area of 1 mm2 when carrying a
current of 1 A (values similar to those for the
electric wire to your study lamp).
Correct answer: 7.38624 × 10−5 m/s.
Explanation:
Given : N = 1 ,
M = 63.54 g ,
ρ = 8.92 g/cm3 ,
A = 1 mm2 = 1 × 10−6 m2 ,
I = 1 A , and
qe = 1.602 × 10−19 C/electron .
We first calculate n, the number of currentcarrying electrons per unit volume in copper.
Assuming one free conduction electron per
NA ρ
atom, n =
, where NA is Avogadro’s
M
number and ρ and M are the density and the
atomic weight of copper, respectively
¶
µ
electron NA ρ
.
n≡ 1
atom
M
¶
electron 6.02 × 1023 atoms
n= 1
atom
63.54 g
µ 6
¶
¡
¢ 10 cm3
3
· 8.92 g/cm
1 m3
= 8.45112 × 1028 electrons/m3 .
µ
=
1
1A
8.45112 × 1028 electrons/m3
1
·
−19
(1.602 × 10
C/electron)
1
·
(1 × 10−6 m2 )
= 7.38624 × 10−5 m/s .
002 (part 1 of 3) 0 points
The damage caused by electric shock depends
on the current flowing through the body;
1 mA can be felt and 5 mA is painful. Above
15 mA, a person loses muscle control, and 70
mA can be fatal. A person with dry skin has a
resistance from one arm to the other of about
100000 Ω. When skin is wet, the resistance
drops to about 5300 Ω.
What is the minimum voltage placed across
the arms that would produce a current that
could be felt by a person with dry skin?
Correct answer: 100 V.
Explanation:
Given : Imin = 1 mA , and
Rdry = 100000 Ω .
The minimum voltage depends on the minimum current for a given resistance, so
Vmin = Imin Rdry
µ
= (1 mA)
1A
1000 mA
¶
(100000 Ω)
= 100 V .
003 (part 2 of 3) 0 points
Using the same electric potential as in Part 1,
what would be the current if the person had
wet skin?
Correct answer: 18.8679 mA.
Explanation:
The drift speed is vd ,
vd =
I
n qe A
Given :
Vmin = 100 V , and
Rwet = 5300 Ω .
Answer, Key – Homework 11 – David McIntyre
For the voltage V from Part 1 and the resistance Rwet ,
Vmin
Rwet
µ
¶
100 V
1000 mA
=
5300 Ω
1A
= 18.8679 mA .
I=
2
Solution: The charge carrier density n is
unaffected. This is simply a property of the
material, and is related to the mass density
and the number of valence electrons available
per atom.
006 (part 2 of 4) 0 points
(b) What happens to the current density?
1. It doubles. correct
004 (part 3 of 3) 0 points
What would be the minimum voltage that
would produce a current that could be felt
when the skin is wet?
Correct answer: 5.3 V.
Explanation:
Given :
Imin = 1 mA , and
Rwet = 5300 Ω .
V1 = Imin Rwet
µ
= (1 mA)
1A
1000 mA
¶
(5300 Ω)
= 5.3 V .
005 (part 1 of 4) 0 points
If the current carried by a metallic conductor
is doubled,
(a) What happens to the charge carrier
density? Assume that the temperature of the
metallic conductor remains constant.
2. It is unchanged.
3. It is quartered.
4. It quadruples.
5. It is halved.
Explanation:
I
The current density J =
only depends
A
on the current I and the cross-sectional area
of the conductor A, so if I doubles and A
remains the same, J must also double:
I
2I
= 2 = 2J .
J0 =
A
A
007 (part 3 of 4) 0 points
(c) What happens to the electron drift velocity?
1. It is unchanged.
1. It quadruples.
2. It is halved.
2. It is unchanged. correct
3. It is quartered.
3. It is quartered.
4. It quadruples.
4. It is halved.
5. It doubles. correct
5. It doubles.
Explanation:
Basic Concept:
J = σ E = n q vd
vd =
qE
τ.
m
Explanation:
The electron drift velocity vd is given by
I
vd =
,
nqA
where q is the charge of an electron. So if I
doubles, vd must also double:
2I
I
vd0 =
=2
= 2vd .
nqA
nqA
Answer, Key – Homework 11 – David McIntyre
008 (part 4 of 4) 0 points
(d) What happens to the average time between collisions?
3
3. D > B = C > A
4. A = C > D > B
1. It doubles.
5. A = D > B = C correct
2. It is halved.
6. A = B = C = D
3. It is quartered.
7. A = B > D = C
4. It is unchanged. correct
8. A > B = C > D
5. It quadruples.
9. A > B > C > D
Explanation:
The average time τ between collisions is
given by
vd m
,
τ=
qE
where m is the mass of an electron and E
is the applied electric field (determined by
Ohm’s Law: J = σ E).
Substituting for E,
τ=
vd m σ
.
qJ
Explanation:
From the figure, the total resistance of the
5R
R
. Then
circuit is 2 R + =
2
2
IA = I D =
V
2V
V
=
= 0.4
5R
5R
R
2
1V
V
1
IA =
= 0.2
2
5R
R
The rankings of the currents are also the corresponding rankings of the brightness.
IB = I C =
Since vd and J are each multiplied by a factor
of two, the factors of two cancel (as long as
σ does not change due to heating). All the
other variables remain the same (as long as σ
does not change due to heating). Therefore τ
remains unchanged.
009 (part 1 of 2) 0 points
The circuit below shows four identical bulbs
connected to an ideal battery, which has negligible internal resistance. Rank the bulbs in
order from brightest to dimmest.
A
V
10. A = B = C > D
B
010 (part 2 of 2) 0 points
Suppose a switch has been added to the circuit
as shown. The switch is initially closed.
A
V
B
C
D
When the switch is opened, what happens
to the currents through bulbs A, B, and D?
C
D
1. IA decreases, IB remains the same, ID
decreases.
1. D > C > B > A
2. IA remains the same, IB decreases, ID
remains the same
2. B = C > A = D
3. IA increases, IB increases, ID increases.
Answer, Key – Homework 11 – David McIntyre
4. IA remains the same, IB increases, ID
remains the same.
5. IA decreases, IB increases, ID decreases.
correct
6. IA remains the same, IB remains the
same, ID remains the same.
7. IA increases, IB remains the same, ID
decreases.
8. IA increases, IB remains the same, ID
increases.
9. IA increases, IB decreases, ID increases.
10. IA decreases, IB decreases, ID decreases.
Explanation:
With the switch open, the total resistance
of the circuit is 3 R, so
IA = I B = I D =
V
V
= 0.33
3R
R
This is to be compared with the case that the
switch is closed, where from previous part,
V
V
and IB = 0.2 . So for
IA = ID = 0.4
R
R
the present case, IA and ID decrease and IB
increases.
011 (part 1 of 4) 0 points
A battery has an emf of 12 V and an internal resistance of 0.13 Ω. Its terminals are
connected to a load resistance of 2 Ω.
Find the current in the circuit.
Correct answer: 5.6338 A.
Explanation:
Given : E = 12 V ,
R = 2 Ω , and
r = 0.13 Ω .
The total resistance is R + r, so
I=
E
R+r
4
12 V
2 Ω + 0.13 Ω
= 5.6338 A .
=
012 (part 2 of 4) 0 points
Calculate the terminal voltage of the battery.
Correct answer: 11.2676 V.
Explanation:
The terminal voltage V of the battery is
equal to
Vr = E − I r
= 12 V − (5.6338 A) (0.13 Ω)
= 11.2676 V .
013 (part 3 of 4) 0 points
Find the power dissipated in the load resistor.
Correct answer: 63.4795 W.
Explanation:
The power dissipated in the load resistor is
PR = I 2 R
= (5.6338 A)2 (2 Ω)
= 63.4795 W .
014 (part 4 of 4) 0 points
Find the power dissipated in the battery.
Correct answer: 4.12617 W.
Explanation:
The power dissipated in the battery is
Pr = I 2 r
= (5.6338 A)2 (0.13 Ω)
= 4.12617 W .
015 (part 1 of 2) 0 points
In the figure below the switch S is initially in
position “1”. Given, R1 = R2 = R3 . Neglect
the internal resistance of the battery.
Answer, Key – Homework 11 – David McIntyre
5
as when R1 was in the circuit.
R1
V
2
1
3
S
0
R1
R2
What happens to the current through R3
when the switch is moved to the open position
“2”.
Explanation:
Since R2 and R3 have the same terminal
voltage and resistance, the current through
R2 and R3 must now be the same.
017 (part 1 of 1) 0 points
Consider the following circuit containing identical bulbs.
V
A
B
C
D
E
1. The current through R3 decreases to twothirds its original value.
2. The current through R3 increases to twice
its original value.
3. The current through R3 increases to
three-halves its original value.
4. The current through R3 is reduced to
one-half its original value.
5. The current through R3 remains the
same. correct
1. A = B = C = D = E
2. A = D = E > B = C
3. B = C > A = D = E
4. A = B = C > D = E correct
5. A = B = C > D > E
6. A = B > A = D = E
Explanation:
The voltage across R3 is the E of the battery, and is unchanged. The current through
E
R3 remains the same,
.
R3
8. A = D = E > B > C
016 (part 2 of 2) 0 points
What happens when switch S is moved to
position “3”, leaving R2 and R3 parallel?
10. E = D > A > B = C
1. The current through R2 is half what it
was with R1 in the circuit.
2. The current through R2 and R3 are now
the same. correct
3. The current through R3 increases.
4. The current through R3 decreases.
5. The current through R2 remains the same
7. A = C > B > D = E
9. C > B > A > D > E
Explanation:
The potential across bulbs A, B and C is
V
V . The potential across bulbs D and E is .
2
Thus
V
iA = i B = i C =
R
V
iD = i E =
.
2R
018 (part 1 of 2) 0 points
Four identical light bulbs are connected either in series (circuit 1) or parallel (circuit 2)
Answer, Key – Homework 11 – David McIntyre
to a constant voltage battery with negligible
internal resistance, as shown.
6
We can see that the bulbs in circuit 2 are more
than 4 times brighter than the bulbs in circuit
1.
019 (part 2 of 2) 0 points
If one of the bulbs in circuit 2 is unscrewed
and removed from its socket, the remaining 3
bulbs
1. go out.
2. are unaffected. correct
circuit 1
3. become brighter.
4. become dimmer.
Compared to the individual bulbs in circuit
1, the individual bulbs in circuit 2 are
1. the same brightness.
1
as bright.
4
1
3. less than as bright.
4
2.
4. 4 times brighter.
5. more than 4 times brighter. correct
Explanation:
Solution: In circuit 1, the voltage across
each light bulb is
020 (part 1 of 3) 0 points
Hint: Apply the Kirchhoff’s law to the loop
ACDA.
Given: 2 R1 = 2 R4 = R2 = R3 , R1 = r. I
is the current entering and leaving the circuit.
Each current shown in the figure is denoted
by the same subscript as the resistor through
which it flows (e.g., i1 is the current flowing
through R1 ).
C
R
R1
3
V2
E2
=
.
R
16 R
In circuit 2, the voltage across each bulb is
identical; namely E. Hence the power of each
bulb in circuit 2 is
i1
R
A
E
E
V =IR=
R= ,
4R
4
so the power of each bulb in circuit 1 is
I
2
i2
Find the ratio
i1
.
i2
i1
= 2 correct
i2
1
i1
=
2.
i2
3
1.
B
R4
I
i4
D
P1 =
1
E2
=
P1 .
P2 =
R
16
i3
i5
circuit 2
Explanation:
Since the bulbs are parallel, after one of
the bulbs is unscrewed, the voltage across
each remaining bulb is unchanged, and the
brightness is unaffected.
Answer, Key – Homework 11 – David McIntyre
i1
=3
i2
i1
4.
=1
i2
1
i1
=
5.
i2
2
i1
1
6.
=
i2
4
i1
7.
=4
i2
Explanation:
I
correct
3
I
5. i5 =
2
I
6. i5 =
5
I
7. i5 =
7
I
8. i5 =
4
Explanation:
4. i5 =
3.
Given :
R1
R2
R3
R4
= r,
= 2r,
= 2r,
= r.
i1 = 2 i 2
and
I = i 1 + i2 = 2 i 2 + i2 = 3 i 2
I
⇒ i2 =
3
2I
.
⇒ i1 =
3
C
2r
r
i3
2r
I
i2
i5
i1
A
B
r
I
i4
D
Basic Concept: DC Circuit.
Solution: Based on Kirchhoff’s law, the
equation for the loop ACDA is given by
−i1 R1 + i2 R2 = 0
i1
R2
⇒
=
i2
R1
2r
=
r
= 2.
021 (part 2 of 3) 0 points
Find the magnitude of the current i5 which
flows from C to D.
1. i5 = 0
I
8
I
3. i5 =
6
2. i5 =
7
Following a similar analysis, one finds that
i4
I
2I
= 2, so that i3 = and i4 =
.
i3
3
3
Note: The junction equation at D is
i2 + i 5 = i 4
⇒ i5 = i4 − i2 , or
= i3 − i1
2I
I
I
=−
= −
3
3
3
I
|i5 | = .
3
022 (part 3 of 3) 0 points
Find the resistance RAB .
1. RAB = 3 r
2. RAB = 0
3. RAB = r
4. RAB = 2 r
5. RAB =
4
r correct
3
Answer, Key – Homework 11 – David McIntyre
1
6. RAB = r
3
2
7. RAB = r
3
3
8. RAB = r
2
5
9. RAB = r
3
Explanation:
By inspection, the following circuit is equivalent to the original circuit.
8
Determine IAC , i.e. the current between
points A and C.
I
correct
3
I
2.
4
I
3.
9
I
4.
5
1.
5. I
I
6
I
7.
2
6.
R1
R3
A
B
R2
R4
Here
R1 R2
R1 + R 2
2 r2
=2
r + 2r
4
= r
3
RAB = 2
023 (part 1 of 3) 0 points
Consider a cubic resistor network shown,
where all the resistors are the same. Each
resistor has a resistance r. A current I comes
into the network at A. The same current
leaves the network at B.
I
B
C
D
Explanation:
Basic Concept: Symmetry in circuits.
Solution:
Go through the network systematically from
A to B. The current enters at A. Since the
network is completely symmetric, there can
be no difference in resistance for the three
possible paths the current can split into at A.
Thus, each of the three must carry identical
current, and they must add up to I (since no
current is lost at the junction):
IAC =
024 (part 2 of 3) 0 points
Determine ICD , i.e. the current between
points C and D.
I
9
I
2.
4
I
3.
2
1.
4. I
A
I
3
5.
I
5
Answer, Key – Homework 11 – David McIntyre
I
6. correct
6
I
7.
3
Explanation:
Now we are at C. The current coming in from
A has two possible choices of path, and again
there is complete symmetry between the two.
Thus the current going through C splits in
half:
I
IAC
=
ICD =
2
6
9
Now |VAB | = RAB I, so
5
RAB = r
6
026 (part 1 of 2) 0 points
Two identical light bulbs A and B are connected in series to a constant voltage source.
Suppose a wire is connected across bulb B as
shown.
A
B
025 (part 3 of 3) 0 points
Determine RAB , i.e. the effective resistance
between points A and B.
r
3
r
2.
6
5r
3.
correct
6
r
4.
2
V
1.
5. r
4r
3
2r
7.
3
3r
8.
2
7r
9.
6
Explanation:
Finally, we calculate the equivalent resistance.
We can step through the path ACDB to find
the potential VAB between A and B. Kirchoff’s Laws tell us that each time we cross a
resistor (moving with the current) we drop a
potential V = RI. Therefore
6.
VA − VB = rIAC + rICD + rIDB
Now note that IAC = IDB by symmetry, so
µ
¶
I
I
I
5
VA − VB = |VAB | = r
+ +
= rI
3 6 3
6
Bulb A
1. will burn half as brightly as before.
2. will burn as brightly as before.
3. will burn nearly four times as brightly as
before. correct
4. will burn twice as brightly as before.
5. will go out.
Explanation:
The electric power is given by
V2
.
R
Before the wire is connected,
P = I2 R =
IA = I B =
V
2R
so that
V2
.
4R
After the wire is connected,
PA =
V
and IA0 = 0 , so
R
IB0 = 0
V2
0
= 4 PA .
PA =
R
IA0 =
Answer, Key – Homework 11 – David McIntyre
027 (part 2 of 2) 0 points
and bulb B
1. will burn nearly four times as brightly as
before.
2. will burn half as brightly as before.
3. will go out. correct
4. will burn as brightly as before.
5. will burn twice as brightly as before.
Explanation:
Since there is no potential difference between the two ends of bulb B, it goes off.
028 (part 1 of 3) 0 points
Consider the circuit shown below.
E1
r1
B
A
E2
C
i1
r2
R
029 (part 2 of 3) 0 points
What equation does the loop DCFED yield?
1. −E2 − i2 r2 + i R = 0
2. −E2 + i2 r2 − i R = 0
3. −E2 − i2 r2 − i R = 0
4. E2 − i2 r2 − i R = 0 correct
6. E2 + i2 r2 + i R = 0
E
1. −E1 − E2 − i2 r2 + i1 r1 = 0
2. −E1 − E2 + i2 r2 − i1 r1 = 0
3. E1 − E2 − i2 r2 − i1 r1 = 0
4. −E1 + E2 + i2 r2 + i1 r1 = 0
5. E1 + E2 − i2 r2 + i1 r1 = 0
6. E1 − E2 + i2 r2 − i1 r1 = 0 correct
Explanation:
ABCDA : E1 − E2 + i2 r2 − i1 r1 = 0 .
D
i
Apply Kirchhoff’s rules. What equation
does the loop ABCDA yield?
7. E1 + E2 + i2 r2 − i1 r1 = 0
Recall that Kirchhoff’s loop rule states that
the sum of the potential differences across all
the elements around a closed circuit loop is
zero. If a resistor is traversed in the direction
of the current, the change in potential is −I R.
If a emf source is traversed from the − to
+ terminals, the change in potential is +E.
Apply the opposite sign for traversing the
elements in the opposite direction.
Hence, by inspection
5. E2 + i2 r2 − i R = 0
i2
F
10
7. E2 − i2 r2 + i R = 0
Explanation:
DCF ED : E2 − i R − i2 r2 = 0 .
030 (part 3 of 3) 0 points
Let: E1 = E2 = E, and r1 = r2 = r, where
E = 9 V and r = 2.4 Ω. Also R = 4.1 Ω.
Hint: From symmetry, one expects i1 = i2 .
Find the current i.
Correct answer: 1.69811 A.
Explanation:
Given : E1 = E2 = E = 9 V ,
r1 = r2 = 2.4 Ω , and
R = 4.1 Ω .
We are given that E1 = E2 and r1 = r2 . This
implies that i1 = i2 . (Why? Look at the loops
Answer, Key – Homework 11 – David McIntyre
DCFED and ABFEA and see if there are any
similarities.) Hence the junction rule yields
i1 + i2 = 2i2 = i
i
⇒ i2 = .
2
7. It=0 = (R1 + R2 ) Vo
Vo
correct
R1
Explanation:
When the switch is in position “a” at t = 0,
there is no potential drop across the capacitor.
Note: There is no current flowing through
R2 , so the entire potential drop, Vo , is across
the resistor R1 .
From Ohm’s law,
8. It=0 =
Substituting this into the loop equation
DCFED,
i
E2 − i R − r2 = 0
2
Solving for i yields
i=
E2
I=
r2
R+
2
9V
=
2.4 Ω
4.1 Ω +
2
= 1.69811 A .
1. τ =
3. τ =
E
S b
a
4. τ =
5. τ =
6. τ =
After S is switched to position “a”, the
initial current through R1 is
1. It=0 =
R1 R2
Vo
R1 + R 2
2. It=0 = R1 Vo
3. It=0 =
Vo
R1 + R 2
4. It=0 = R2 Vo
R1 + R 2
Vo
R1 R2
Vo
=
R2
5. It=0 =
6. It=0
R1
C
2. τ = R1 C
R2
C
Vo
.
R1
032 (part 2 of 3) 0 points
Leave the switch at position “a” for a long
time, then move the switch from “a” to “b”.
When the switch is in position “b”, what is
the time constant, τ , of the circuit?
031 (part 1 of 3) 0 points
Consider the circuit shown below, the capacitor is initially uncharged.
R1
11
7. τ =
8. τ =
R2
C
R1 R2
C
R1 + R 2
R1 R2
(R1 + R2 ) C
(R1 + R2 )
C
R1 + R 2
R1 R2 C
R1 + R 2
C
R1 R2
9. τ = R2 C
10. τ = (R1 + R2 ) C correct
Explanation:
When the switch is in position “b”, R1 and
R2 are now in series so the equivalent resistance is R = R1 + R2 . By definition, the time
constant is
τ = R C = (R1 + R2 ) C .
Answer, Key – Homework 11 – David McIntyre
033 (part 3 of 3) 0 points
Let: Vo = 9 V, C = 1 µF, R1 = 7 Ω, and
R2 = 20 Ω.
3
At a time τ after S has been switched to
2
position “b”, what is the power consumption
of the circuit?
Correct answer: 0.149361 W.
Explanation:
Given : Vo
C
R1
R2
The switch on the circuit is closed at t = 0.
Find the charge on the capacitor at 4.73 s.
Correct answer: 19.626 µC.
Explanation:
Given : t = 4.73 s ,
R = 1.1 MΩ = 1.1 × 106 Ω ,
C = 1.4 µF = 1.4 × 10−6 F ,
E = 14.7 V .
= 9 V,
= 1 µF ,
= 7 Ω , and
= 20 Ω .
R
E
While the capacitor is discharging, the
magnitude of the current decreases as a function of time as
I(t) =
Vo
e−t/τ .
R1 + R 2
Noting that R = R1 + R2 , the power con3
sumed by the circuit at t = τ is
2
P = I 2 (R1 + R2 )
¸2
·
Vo
−3/2
e
(R1 + R2 )
=
R1 + R 2
Vo2
=
e−3
R1 + R 2
(9 V)2
e−3
=
7 Ω + 20 Ω
= 0.149361 W .
034 (part 1 of 3) 0 points
Given: The RC circuit in the figure below.
1.1 MΩ
14.7 V
1.4 µF
S
12
and
C
S
At t = 4.73 s,
³
´
q = CE 1 − e−t/(R C)
= (1.4 × 10−6 F)(14.7 V)×
¸¾
½
·
4.73 s
1 − exp −
(1.1 × 106 Ω) (1.4 × 10−6 F)
= 1.9626 × 10−5 C
= 19.626 µC .
035 (part 2 of 3) 0 points
Find the current in the resistor at 4.73 s.
Correct answer: 0.61947 µA.
Explanation:
At t = 4.73 s,
E −t/(R C)
e
R
14.7 V
×
=
6 Ω
1.1 ×
10
·
I=
4.73 s
exp −
6
(1.1 × 10 Ω) (1.4 × 10−6 F)
= 6.1947 × 10−7 A
= 0.61947 µA .
036 (part 3 of 3) 0 points
¸
Answer, Key – Homework 11 – David McIntyre
At 4.73 s the current in the resistor is I (Part
2) and the charge on the capacitor is q (Part
1).
What is the power delivered by the battery?
Correct answer: 9.10621 µW.
Explanation:
In the time interval ∆t, work done by the
battery in pushing charge ∆q across the battery is
∆Wbattery = ∆q · E .
Correspondingly, the power is
d Wbattery
dq
=E
=IE.
dt
dt
The power dissipated in a resistor is
d Wresistor
= I2 R .
dt
The power to create the electric field in a
capacitor is
d Wcapacitor
q
=I .
dt
C
Thus the total power dissipated in the capacitor and resistor, that is the power delivered
by the battery is
³
d Wbattery
q´
=I IR+
dt
C
= (6.1947 × 10−7 A)
h
× (6.1947 × 10−7 A) (1.1 × 106 Ω)
(1.9626 × 10−5 C) i
+
(1.4 × 10−6 F)
= 9.10621 × 10−6 W
= 9.10621 µW .
13