zhang (rz4453) – mt2TakeHome – turner – (55160)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
15 V
6Ω
24 V
001 10.0 points
An electric field of 1430 V/m is applied to a
section of silver of uniform cross section.
Find the resulting current density if the
specimen is at a temperature of 4 ◦ C . The
resistivity ρ of silver is 1.59 × 10−8 Ω · m
at 20 ◦ C . and its temperature coefficient is
0.0038(◦ C)−1 .
Correct answer: 9.57593 × 1010 A/m2 .
Explanation:
a
b
38 V
8Ω
1
1Ω
4Ω
1Ω
9Ω
1Ω
5Ω
Find the magnitude of the potential difference between points a and b.
Correct answer: 1.12 V.
Explanation:
Let : E1
E2
E3
R1
R2
R3
R4
R5
r1
r2
r3
Let : ρ20 = 1.59 × 10−8 Ω · m ,
T = 20◦ C ,
α = 0.0038 (◦ C)−1 , and
E = 1430 V/m .
ρ = ρ20 [1 + α (T − T20 )]
= (1.59 × 10−8 Ω · m)
× {1 + 0.0038 (◦ C)−1 (4◦ C − 20◦ C)}
E1
R1
= 1.49333 × 10−8 Ω · m ,
= 15 V ,
= 24 V ,
= 38 V ,
= 6 Ω,
= 8 Ω,
= 4 Ω,
= 5 Ω,
= 9 Ω,
= 1 Ω,
= 1 Ω , and
= 1 Ω.
r1
i
i
so by Ohm’s Law,
a
J = σE =
E
1430 V/m
=
ρ
1.49333 × 10−8 Ω · m
= 9.57593 × 1010 A/m2 .
002
10.0 points
A network below consists of with three batteries, each having an internal resistance, and
five resistors.
b
R2
R3
E2
r2
R5
E3
r3
R4
i
i
There is no current flow between a and b,
so applying the loop rule to the outside path,
E1 − E3 − i (r1 + R1 + R2 + r3 + R4 + R3 ) = 0 .
Since
R1234 = 1 Ω + 6 Ω + 8 Ω
zhang (rz4453) – mt2TakeHome – turner – (55160)
+1 Ω+5Ω+4Ω
= 25 Ω , then
004
2
10.0 points
Consider an RC circuit
E1 − E3
r1 + R 1 + R 2 + r3 + R 4 + R 3
E1 − E3
15 V − 38 V
=
=
R1234
25 Ω
= −0.92 A .
We can then find the magnitude |Vab | of the
potential difference between a and b by using
the top loop:
Vab = E1 − E2 − i (R1 + r1 + R3 )
= (15 V) − (24 V)
− (−0.92 A) (6 Ω + 1 Ω + 4 Ω)
= −1.12 V
|Vab | = 1.12 V ,
where the current through r2 and R5 is zero.
003 10.0 points
The current in a wire decreases with time
according to the relationship
C
R
i=
E
S
where a battery, a resistor and a capacitor
are in series. As the switch is closed, at any
instant the current at any part of the circuit
is the same. Set the time when we close the
switch to t = 0.
After the switch is closed the current decreases exponentially with time according to
I(t) = I0 e−t/τ where I0 is the initial current
at t = 0, and τ is a constant having dimensions of time. Consider a fixed observation
point within the conductor.
How much charge Q(τ ) passes this point
between t = 0 and t = τ ? Notice: The
same amount of charge is deposited onto the
capacitor plate.
1. Q(τ ) = 0.632 I0
−a t
I = (1.62 mA) e
where a = 0.13328 s−1 .
Determine the total charge that passes
through the wire from t = 0 to the time the
current has diminished to zero.
Explanation:
q=
=
dq
dt
3. Q(τ ) = 0.632 I02 τ
0.632 I0
τ
6. Q(τ ) = 2.72 I0
t
7. Q(τ ) = I02
I dt
Zt=0
∞
2.72 I0
τ
5. Q(τ ) = 0.632 I0 τ correct
I=
Z
2. Q(τ ) =
4. Q(τ ) =
Correct answer: 0.0121549 C.
(0.00162 A) e(−0.13328 s
−1
)t
t=0
e−(0.13328 s ) t
= (0.00162 A)
−0.13328 s−1
−1
= 0.0121549 C .
dt
8. Q(τ ) =
1.72 I0
τ
9. Q(τ ) =
I0
τ
∞
0
i
10. Q(τ ) = 1.72 I0
zhang (rz4453) – mt2TakeHome – turner – (55160)
Explanation:
△Q
Current is
, where Q is the charge and
△t
t is the time. In differential form this gives
dQ
, which can be integrated as follows
I(t) =
dt
Z t
Z t
Q(t) =
I dt = I0
e−t/τ dt
0
0
−t/τ
= I0 τ (1 − e
) and
005 10.0 points
Birds resting on high-voltage power lines are
a common sight. A certain copper power line
carries a current of 116 A, and its resistance
per unit length is 7.86 × 10−5 Ω/m.
If a bird is standing on this line with its feet
2.5 cm apart, what is the potential difference
across the bird’s feet?
Correct answer: 0.00022794 V.
S
iB
iA
iC
i0
If bulb B is removed from the circuit, i.e.,
the switch S is opened, what happens to the
currents through
1) the battery,
2) bulb A, and
3) bulb D;
Notice that in the diagram the current
through the battery, ibattery , is labeled as i0 .
Hint: You may find it helpful to work out the
currents through bulb A , bulb D , and the
battery for both cases by using V = 1 volt
and R = 1 Ω.
1. iA decreases, iD decreases, ibattery remains
the same
Explanation:
Let : I = 116 A ,
R
= 7.86 × 10−5 Ω/m ,
d
d = 2.5 cm .
iD
V
Q(τ ) = I0 τ (1 − e−1 ) = 0.632 I0 τ .
3
2. iA increases, iD increases, ibattery increases
and
From Ohm’s Law, ∆V = I R, so
R
∆V = I R = I
d
d
= (116 A)(7.86 × 10−5 Ω/m)(0.025 m)
3. iA increases, iD increases, ibattery remains
the same
4. iA increases, iD decreases, ibattery decreases
5. iA increases, iD increases, ibattery decreases
= 0.00022794 V .
6. iA decreases, iD remains the same, ibattery
decreases correct
006
7. iA remains the same, iD increases, ibattery
increases
10.0 points
All the bulbs in the figure below have the
same resistance R. The switch S is initially
closed.
8. iA increases, iD remains the same, ibattery
increases
zhang (rz4453) – mt2TakeHome – turner – (55160)
9. iA decreases, iD decreases, ibattery decreases
10. iA remains the same, iD remains the same,
ibattery remains the same
in parallel with the battery; so iD remains
the same. Since the equivalent resistance
of the center branch increases, the equivalent
resistance of the entire circuit also increases.
Hence ibattery decreases.
007
Explanation:
1
RBC
RBC
1
1
2
= + =
R R
R
R
=
2
3R
R
=
2
2
1
1
1
2
1
5
=
+ =
+ =
Req
RL R
3R R
3R
3R
Req =
5
RL = RA + RBC = R +
10.0 points
When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge (on each plate) of magnitude
153 µC. While the battery connection is
maintained, a dielectric slab is inserted into
the space between the capacitor plates and
completely fills this region. This results in
the accumulation of an additional charge of
150 µC on each plate.
Therefore
2V
V
=
iA =
3R
3R
2
V
iD =
R
5V
ibattery =
3R
4
κ
κ
What is the dielectric constant of the slab?
Correct answer: 1.98039.
Explanation:
Without bulb B:
RL = 2 R
1
1
3
1
=
+ =
Req
2R R
2R
2R
Req =
3
Therefore
V
2R
V
iD =
R
3V
ibattery =
2R
iA =
Qualitatively, if bulb B is removed, this increases the equivalent resistance of the center
branch, so iA decreases. The voltage drop
across bulb D remains the same since it is
Let :
Qa = 153 µC ,
Qd = 153 µC + 150 µC ,
= 303 µC .
and
Since the capacitor is connected to the battery the whole time, we know that the potential drop across the capacitor is held constant.
The charge changes only because the capacitance does. When the capacitor is filled with
air, we have Qa = Ca V . When the dielectric slab is inserted, the charge is given by
Qd = Cd V . The dielectric constant is then
κ=
Cd
Q
303 µC
= d =
= 1.98039 .
Ca
Qa
153 µC
008
10.0 points
zhang (rz4453) – mt2TakeHome – turner – (55160)
The resistance between points a and e drops
to one-half its original value when the switch
S2 is closed.
c
10 Ω
70 Ω
a
S2
10 Ω
E
S1
e
b
70 Ω
Req,o = R + R1234 .
With the switch closed,
c
R2
R1
a
S2
R3
R
d
5
E
S1
e
b
R4
R
d
Find the value of R.
R1 and R3 are in parallel, so
Correct answer: 5 Ω.
Explanation:
Let :
R1
R2
R3
R4
Req,c
= 70 Ω ,
= 10 Ω ,
= 10 Ω ,
= 70 Ω , and
1
= Req,o .
2
R2
a
S2
R3
E
e
S1
d
R13 =
1
1
+
R1 R3
=
1
1
+
70 Ω 10 Ω
−1
= 8.75 Ω .
R2 and R4 are in parallel, so
With the switch open,
c
R1
−1
b
R4
R
R1 and R2 are in series, so
−1
R24 =
1
1
+
R2 R4
=
1
1
+
10 Ω 70 Ω
−1
= 8.75 Ω .
R, R13 and R24 are in series, so
Req,c = R + R13 + R24 .
The new resistance is one-half the original, so
1
(R + R1234 )
2
= R + R1234
R + R13 + R24 =
2R + 2R13 + 2R24
R12 = R1 + R2 = 70 Ω + 10 Ω = 80 Ω .
R3 and R4 are in series, so
R34 = R3 + R4 = 10 Ω + 70 Ω = 80 Ω .
R12 is in parallel with R34 , so
1
R1234
R1234
1
1
R34 + R12
=
+
=
R12 R34
R12 R34
R12 R34
(80 Ω) (80 Ω)
=
=
R34 + R12
80 Ω + 80 Ω
= 40 Ω .
R = R1234 − 2R13 − 2R24
= 40 Ω − 2(8.75 Ω) − 2(8.75 Ω)
= 5Ω .
009 10.0 points
A 14.2 V battery is connected to a 6.8 pF
parallel-plate capacitor.
zhang (rz4453) – mt2TakeHome – turner – (55160)
What is the magnitude of the charge on
each plate?
Correct answer: 9.656 × 10−11 C.
Explanation:
Let :
∆V = 14.2 V and
C = 6.8 pF .
The capacitance is
Q
C=
∆V
Q = C ∆V = (6.8 × 10−12 F) (14.2 V)
= 9.656 × 10−11 C .
12 Ω
33 Ω
60 Ω
63 Ω
010 10.0 points
The following diagram shows part of an electrical circuit.
37 Ω
34 Ω
A
17 Ω
B
Find the equivalent resistance Req between
points A and B of the resistor network.
Correct answer: 36 Ω.
Explanation:
R5
R2
R6
R1
R4
R7
A
R3
6
Start from the right-hand side in determining
the equivalent resistances.
Step 1: R1 , R2 , and R3 are in series, so
R123 = R1 + R2 + R3
= 33 Ω + 34 Ω + 17 Ω
= 84 Ω .
Step 2: R123 is now parallel with R4 , so
1
R1234
=
1
1
R123 + R4
+
=
R4 R123
R123 R4
R123 R4
R123 + R4
(84 Ω) (60 Ω)
=
84 Ω + 60 Ω
= 35 Ω .
R1234 =
Step 3: R1234 is in series with R5 and R6 , so
Rs = R1234 + R5 + R6
= 35 Ω + 37 Ω + 12 Ω
= 84 Ω .
Step 4: Finally, Rs is parallel with R7 , so
1
1
R7 + R5
1
=
+
=
Req
Rs R7
R5 R7
Rs R7
Req =
Rs + R7
(84 Ω) (63 Ω)
=
84 Ω + 63 Ω
= 36 Ω .
B
Let :
R1
R2
R3
R4
R5
R6
R7
= 33 Ω ,
= 34 Ω ,
= 17 Ω ,
= 60 Ω ,
= 37 Ω ,
= 12 Ω ,
= 63 Ω .
and
011 10.0 points
A resistor is constructed by forming a material of resistivity 3.2 × 105 Ω · m into a shape
of a hollow cylindrical shell of length 3 cm
and inner radius of 0.45 cm and outer radius 1.75 cm . In use, a potential difference is
applied between the ends of the cylinder, producing a current parallel to the length of the
cylinder.
zhang (rz4453) – mt2TakeHome – turner – (55160)
7
L
rb
ra
b
Circuit B
Find the resistance of the cylinder.
Correct answer: 10.6845 MΩ.
Explanation:
Let : ρ = 3.2 × 105 Ω · m ,
L = 3 cm = 0.03 m ,
ra = 0.45 cm = 0.0045 m ,
rb = 1.75 cm = 0.0175 m .
and
The cross-sectional area of the conducting
material is
A = π rb2 − ra2 ,
R=
=
so
ρL
ρL
=
A
π rb2 − ra2
1 MΩ
(3.2 × 105 Ω · m)(0.03 m)
i· 6
h
π (0.0175 m)2 − (0.0045 m)2 10 Ω
= 10.6845 MΩ .
012
10.0 points
Four identical light bulbs are connected either in series (circuit A), or in a parallel-series
combination (circuit B), to a constant voltage
battery with negligible internal resistance, as
shown.
Circuit A
E
E
Assuming the battery has no internal resistance and the resistance of the bulbs is
temperature independent, what is the ratio of
the total power consumed bycircuit Ato that
PA,T otal
consumed by circuit B; i.e.,
?
PB,T otal
P
1
1. A = correct
PB
4
P
2. A = 16
PB
P
3. A = 2
PB
1
P
4. A =
PB
16
P
5. A = 4
PB
P
1
6. A = √
PB
8
P
7. A = 1
PB
P
8. A = 8
PB
P
1
9. A =
PB
2
Explanation:
In circuit A, the equivalent resistance is
RA = 4 R, so the electric current through
each bulb is
V
iA =
4R
and the power of each bulb is
2
V2
V
2
.
R=
PA = I R =
4R
16 R
Thus the total power consumed by all four
bulbs in circuit A is
PA,T otal
V2
.
= 4 PA =
4R
zhang (rz4453) – mt2TakeHome – turner – (55160)
In circuit B, the equivalent resistance is
8
Solution:
1
1
1
1
=
+
=
RB
2R 2R
R
∆V2
12 V
=
=2A
R2
6Ω
I=
Req = 9 Ω + 6 Ω = 15 Ω
RB = R ,
so the electric current through each bulb is
∆V = IReq
= (2 A)(15 Ω)
V
iB =
2R
= 30 V .
and the power of each bulb is
R=
V2
.
4R
Thus the total power consumed by all four
bulbs in circuit B is
PB,T otal = 4 PB =
and
V2
R
PA,T otal
PA
1
=
= .
PB,T otal
PB
4
013 10.0 points
A 9.0 Ω resistor and a 6.0 Ω resistor are connected in series with an emf source. The
potential difference across the 6.0 Ω resistor
is measured with a voltmeter to be 12 V.
Find the potential difference across the emf
source.
Correct answer: 30 V.
R1 = 9.0 Ω ,
R2 = 6.0 Ω ,
∆V 2 = 12 V .
Basic Concepts:
Req = R1 + R2
∆V = IR
I1 = I2 = I
z
9 µF
9 µF
What is the charge on the 5 µF capacitor
centered on the left directly between points y
and z?
Correct answer: 0.00061 C.
Explanation:
Let :
C≡
Explanation:
Let :
014 10.0 points
Consider the capacitor network
9 µF
9 µF
y
9 µF
2
9 µF
V
2R
5 µF
PB = I R =
122 V
2
Cx = 5 × 10−6 F
V = 122 V .
and
q
, so
V
q = Cx V = (5 × 10−6 F) (122 V)
= 0.00061 C .
and
015 10.0 points
If the current in a wire of a CD player is 8.90
mA, how long would it take for 12.0 C of
charge to pass a point in this wire?
Correct answer: 1348.31 s.
Explanation:
zhang (rz4453) – mt2TakeHome – turner – (55160)
Let : I = 8.90 × 10−3 A
∆Q = 12.0 C .
and
∆Q
∆t
∆Q
12 C
∆t =
=
= 1348.31 s .
I
0.0089 A
I=
016 10.0 points
A wire of uniform cross-section is stretched
along a meter stick, and a potential difference
of 0.8 V is maintained between the 35 cm and
65 cm marks.
How far apart on the wire are two points
that differ in potential by 120 mV ?
An air-filled cylindrical capacitor has a capacitance of 10 pF and is 4.5 cm in length.
If the radius of the outside conductor
is 1.8 cm, what is the radius of the inner conductor? The Coulomb constant is
8.98755 × 109 N · m2 /C2 .
Correct answer: 1.40136 cm.
Explanation:
Let :
ke = 8.98755 × 109 N · m2 /C2 ,
ℓ = 4.5 cm = 0.045 m ,
b = 1.8 cm , and
C = 10 pF = 1 × 10−11 F .
For an inner radius of a and an outer radius
of b, The capacitance of a cylindrical capacitor
is given by
Correct answer: 4.5 cm.
C=
Explanation:
Let :
∆V1
∆V2
L1
L2
∆V1
0.8 V
=
= 0.0266667 V/cm .
∆L1
65 cm − 35 cm
To get a change of potential of 120 mV, a
length of
∆L2 =
=
∆L1
∆V2
∆V
1
1
0.0266667 V/cm
(0.12 V)
ℓ
b
2 ke ln
a
b
ℓ
ln
=
a
2 ke C
b
= eℓ/(2 ke C)
a
a = b e−ℓ/(2 ke C)
= 0.8 V ,
= 120 mV = 0.12 V .
= 35 cm , and
= 65 cm .
I ρL
V = IR =
, and the voltage is diA
rectly proportional to the length, so there is
a uniform change in potential per unit length
along the wire of
9
Since
2 ke C = 2 (8.98755 × 109 N · m2 /C2 )
× (1 × 10−11 F)
= 0.179751 m ,
a = (1.8 cm) e−(0.045 m)/(0.179751
m)
= 1.40136 cm .
018 10.0 points
If the plate separation of an isolated parallel
plate capacitor is doubled,
= 4.5 cm
1. the charge density on each plate is doubled.
is needed.
017
10.0 points
2. the charge on the each plate is halved.
zhang (rz4453) – mt2TakeHome – turner – (55160)
Q = C V , so the energy is
3. the potential difference is halved.
U=
4. None of these correct
If Q is halved, U falls to one-quarter:
5. the electric field is doubled.
Explanation:
Q and A are constant, so
E=
E′ =
V′ =
and
Q′
2Q
Q
=
=2
= 2V′.
′
C
C
C
019 10.0 points
A potential difference of 111.0 V exists across
the plates of a capacitor when the charge on
each plate is 448.0 µC.
What is the capacitance?
Correct answer: 4.03604 × 10−6 F.
Explanation:
Let :
C=
∆V = 111.0 V and
Q = 448.0 µC .
Q
448.0 × 10−6 C
=
∆V
111 V
= 4.03604 × 10−6 F .
020 10.0 points
A fully charged capacitor stores 7.7 J of energy.
How much energy remains when its charge
has decreased to half its original value?
Correct answer: 1.925 J.
Explanation:
Let : U = 7.7 J .
1
2
1
=
4
Uf =
Q
= constant .
A ǫ0
Q
(2 Q)
=
A ǫ0
(2 A) ǫ0
1
1 Q2
CV2 =
.
2
2 C
2
Q
1 Q2
1 1 Q2
2
=
=
C
2 4C
4 2 C
1
U = (7.7 J) = 1.925 J .
4
10