ragsdale (zdr82) – HW9 – ditmire – (58335)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A coil has an inductance of 4.5 mH, and the
current through it changes from 0.22 A to
0.7 A in 0.32 s.
Find the magnitude of the average induced
emf in the coil during this period.
Correct answer: 6.75 mV.
n2 =
L
µ0 A ℓ
1
1.25664 × 10−6 N/A2
0.00085 H
×
(3.21699 × 10−5 m2 )(0.02 m)
= 1.05131 × 109 (turns/m)2
=
Therefore,
n=
Explanation:
1
q
1.05131 × 109 (turns/m)2
= 32423.9 turns/m
E=L
= 324.239 turns/cm .
∆I
∆t
= (0.0045 H)
0.7 A − 0.22 A
0.32 s
= 0.00675 V = 6.75 mV .
r
I
002 10.0 points
A small air-core solenoid has a length of 2 cm
and a radius of 0.32 cm.
The permeability of free space is
1.25664 × 10−6 N/A2 .
If the inductance is to be 0.85 mH, how
many turns per centimeter (whole turns plus
fractional turns) are required?
Correct answer: 324.239 turns/cm.
003 10.0 points
A long solenoid has inside a coil of fine wire
coaxial with it.
Inside coil has N total turns
Outside solenoid has n turns per meter
What is the mutual inductance between the
solenoid and the inner coil?
Explanation:
1. M = 2 π R µ0 n N
Let : ℓ = 2 cm = 0.02 m ,
µ0 = 1.25664 × 10−6 N/A2 ,
r = 0.32 cm = 0.0032 m , and
L = 0.85 mH = 0.02 m .
Let n be the number of turns per centimeter.
The area is
A = π r2
= π (0.0032 m)2
= 3.21699 × 10−5 m2 .
The inductance of the air-core solenoid is
L = µ0 n2 A ℓ
R
2. M = 2 π R2 µ20 n N
3. M = π R2 µ0 n N correct
4. M = π r 2 µ0 n
5. M = π R2 µ0 n
6. M = 2 π r 2 µ20 n N
7. M = 2 π r µ0 n N
8. M = π r 2 µ0 n N
9. M = 2 π R µ0 n
ragsdale (zdr82) – HW9 – ditmire – (58335)
10. M = 2 π r µ0 n
B = µ0 n I .
The magnetic flux is
ΦB = B · A = (µ0 n I)(πR2) ,
so the emf is
d ΦB
dI
= −πR2µ0 n N
.
dt
dt
(Because we are interested in the emf in the
inner coil, the area to use is the area of the
inner coils rather than the solenoid area.)
The emf created through a mutual inducdI
tance M is E = −M
, so we have
dt
M = π R2 µ0 n N .
keywords:
004 (part 1 of 5) 10.0 points
A circuit is set up as shown in the figure.
E
I
L
I2
R2
S
R1
I1
The switch is closed at t = 0. The current
I through the inductor takes the form
E 1 − e−t/τx
I=
Rx
where Rx and τx are to be determined.
Find I immediately after the circuit is
closed.
1. I =
E
R1
E
R1 + R2
E
3. I =
R2
2. I =
Explanation:
The magnetic field of a solenoid is
E = −N
2
4. I = 0 correct
Explanation:
Before the circuit is closed, no current is
flowing. When we have just closed the circuit
we are at “t = 0+ ”, a mathematical notation meaning a very short time ǫ after t = 0.
(Nothing happens in the circuit at t = 0, only
immediately after when the switch is, indeed,
closed. However, this is just a mathematical
detail.) There are two loops in the problem, one with E, R1 , R2 and one with E, R1 ,
L. So at t = 0+ , the battery “wants to”
drive a current through both loops. The first
loop presents no problem; since there is no inductance “working against us,” a current will
immediately be set up. The second loop, however, has an inductor which tries to prevent
any change in the current going through it,
and so goes up smoothly from I = 0, as can
be seen in the given solution (just put t = 0
to find I = 0). Therefore, at this instant, the
inductor L carries no current, and we can neglect it when we find the current through R2 .
The equivalent resistance is Req = R1 + R2
so, from E = Req I we find
I2 =
E
R1 + R2
005 (part 2 of 5) 10.0 points
Find I2 immediately after the circuit is
closed.
E
1. I2 =
R1
E
2. I2 =
R2
E
3. I2 =
correct
R1 + R2
4. I2 = 0
Explanation:
See previous explanation.
ragsdale (zdr82) – HW9 – ditmire – (58335)
006 (part 3 of 5) 10.0 points
Find I after the circuit is closed for a long
time.
E
1. I =
R2
2. I = 0
E
R1 + R2
E
4. I =
correct
R1
3. I =
008 (part 5 of 5) 10.0 points
Determine the time constant τx .
L
R1 + R2
L
2.
R1
1.
3. L(R1 + R2 )
4. L R2
Explanation:
When we wait a long time, the current I
dI
though the inductor levels out, i.e.,
→ 0.
dt
The voltage over the inductor is
VL = L
3
dI
,
dt
at all times. Thus VL → 0 as t → ∞, and
we can replace the inductor in the figure by a
straight wire. When the current now comes
from R1 to the junction where I1 splits into
I2 and I, there is no resistance in the “I”
path but a nonzero resistance R2 in the other.
Naturally, the current takes the path with no
resistance. Since we do pass through R1 in
any case, the equivalent resistance is now R1 .
At t = ∞
E
I=
R1
L
1
1
+
R1 R2
6. L R1
5.
L
R2
1
1
+
correct
8. L
R1 R2
Explanation:
To find τx , we set up the loop equation
through L
7.
E − R 1 I1 − L
and the loop through R2
E − R1 I1 − R2 (I1 − I) = 0
since I2 = I1 − I. We see from the second
equation that
I2 = 0
007 (part 4 of 5) 10.0 points
Find I2 after the circuit is closed for a long
time.
E
1. I2 =
R2
E
2. I2 =
R1 + R2
E
3. I2 =
R1
4. I2 = 0 correct
Explanation:
See previous explanation.
dI
=0
dt
I1 =
E + I R2
.
R1 + R2
If we substitute this into the first equation,
we find
E + I R2
dI
E − R1
−L
=0
R1 + R2
dt
or
R2
E
R1 + R2
−
R1 R2
R1 + R2
I −L
dI
=0
dt
meaning the circuit is equivalent to a one-loop
circuit with a battery
R2
′
E = E
R1 + R2
ragsdale (zdr82) – HW9 – ditmire – (58335)
nor current around the upper loop while at
position b. At t1 = 0 the switch is set to
position a.
and a resistance
R′ =
R1 R2
R1 + R2
4
.
C
With these notations we see
E ′ − R′ I − L
dI
= 0,
dt
L
c
so the time constant is
τx =
L
R1 R2
R1 + R2
1
1
+
τx = L
R1 R2
E
L
=
R′
Alternative solution to Part 3:
The mathematically minded might realize
that an equation of the form
has a solution which is composed of a homogeneous solution and a particular solution.
The particular solution corresponds to the
first term in the I in the problem statement,
the homogeneous corresponds to the second
term. The homogeneous solution is therefore
the only one of relevance to the time constant τx , and we can disregard any batteries
E. This provides a simple solution to Part 3:
Just ignore the battery and find the equivalent resistance connected to L, in this case
so the time constant is
1
1
+
R1 R2
R
What is the potential difference Vca ≡ Va −
Vc at time t1 = 0?
1. Vca = −E correct
2. Vca = −E (1 − e−1 )
4. Vca = ∞
5. Vca = −∞
6. Vca = E e−1
7. Vca = 0
8. Vca = E
9. Vca = E (1 − e−1 )
10. Vca = ∞ − ∞
1
1
1
=
+
,
Req
R1 R2
a
3. Vca = −E e−1
dI
L
−RI −E = 0
dt
L
=L
τ=
Req
S b
as before.
Nobody seems to have proven that ignoring
the batteries always works, so the previous
solution is more reliable.
009 (part 1 of 4) 10.0 points
Consider the figure shown below. The switch
is initially set at position b. There is no charge
Explanation:
Since at time t1 = 0 the current is zero,
we can consider the potential difference Vca ≡
Va − Vc across the battery, which immediately
yields
Va − Vc = −E ,
Vc − Va = +E ,
or
where Vc > Va , therefore Vc −Va < 0 . Another
way of approaching this problem is by using
the formula
d I .
Vca ≡ Va − Vc = −L
dt t1 =0
ragsdale (zdr82) – HW9 – ditmire – (58335)
010 (part 2 of 4) 10.0 points
The switch is still in position a.
What is the power consumption through
1
the resistor R at time t1 = τL , where τL is
2
the time constant of the LR loop?
2
E2 1 + e−1/2
R
E2
2. Pca =
R
2
E2 +1/2
3. Pca =
1−e
R
2
E2 1 − e−1/2 correct
4. Pca =
R
2
E2 5. Pca =
1 + e+1/2
R
E 2 +1/2 2
6. Pca =
e
R
E 2 −1/2 2
7. Pca =
e
R
Explanation:
The equation for the power dissipated is
given by
2
E2 2
−t1 /τL
P (t1 ) = I(t) R =
1−e
R
τ
At t1 = L , we then have
2
2
E2 −1/2
P =
1−e
.
R
1. Pca =
011 (part 3 of 4) 10.0 points
After a long time at position a, the switch is
set back to position b at time t2 = 0 .
What is maximum charge on the capacitor
while the switchris at position b?
E
2R
1. Qmax =
R
L
r
R
E
2. Qmax =
R
L
r
L
E
3. Qmax =
R R
E
1
√
4. Qmax =
R 2LC
5
1
5. Qmax = E R √
LC
r
L
6. Qmax = E R
2R
r
R
7. Qmax = E R
L
√
8. Qmax = E R 2 L C
E
1
√
R LC
E √
L C correct
10. Qmax =
R
Explanation:
By conservation of energy we have
9. Qmax =
Qmax 2
1
= L Imax 2 .
2C
2
Solving for Qmax yields
Qmax = Imax
√
LC =
E √
LC .
R
012 (part 4 of 4) 10.0 points
3
T , where T is the period
At time t2 =
4
of the LC circuit, the relationship between
Vb and Vc , and the direction of the current
through the inductor are given by
1. Vc > Vb ; flows from c through L to b
2. Vc > Vb ; flows from b through L to c
3. Vb = Vc ; no current flow
4. Vb = Vc ; flows from b through L to c
5. Vb > Vc ; no current flow
6. Vb > Vc ; flows from b through L to c
7. Vb = Vc ; flows from c through L to b
8. Vb > Vc ; flows from c through L to b
9. Vc > Vb ; no current flow correct
ragsdale (zdr82) – HW9 – ditmire – (58335)
Explanation:
Let’s track the oscillation. At t2 = 0 the
current is at its maximum value flowing leftto-right through the inductor; the capacitor
is uncharged.
At one quarter of the pe
T
riod i.e., t2 =
, the right plate is fully
4
T
the current
charged positive. At t2 =
2
is flowing right-to-left through the inductor
3T
, the left plate of the
while at time t2 =
4
capacitor is fully charged positive.
3
Hence at time t2 = T , I = 0 and Vc > Vb .
4
013 (part 1 of 3) 10.0 points
In the LC circuit below the capacitor is
charged to its maximum 23.8 µC while the
switch S is open. Then the switch is closed at
t = 0.
1 µF
2H
dq
= −ω qmax sin(ω t + δ)
dt
1
2π
, where T is the period
=√
with ω =
T
LC
of oscillation.
Solution: When the switch closes, initially
there is zero current in the circuit. Thus the
solution has the correct phase at t = 0
I=
I=
Now at t =
dQ
= −ω qmax sin(ω t) .
dt
T
,
4
I = −q0 ω sin
2π T
×
T
4
= −q0 ω
= −0.0168291 A .
Thus
1
1
L I 2 = L (−q0 ω)2 = ULmax
2
2
1 2
= UCmax =
q
2C 0
1
(2.38 × 10−5 C)2
=
2 (1 × 10−6 F)
106 µJ
×
J
= 283.22 µJ .
UL =
Q
S
Find the energy stored in the inductor at
the quater of a period of oscillations in the
circuit.
Correct answer: 283.22 µJ.
Explanation:
T
Let : t = ,
4
C = 1 µF = 1 × 10−6 F ,
L = 2 H , and
Q0 = 23.8 µC = 2.38 × 10−5 C .
C
6
L
Q
S
Basic concept: Oscillation in LC circuit:
q = qmax cos(ω t + δ)
T
Observe that at t = , the magnitude of
4
the current flow through the inductor is at a
maximum, and the charge q = qmax cos ω t is
T
π
zero at t =
=
. Thus, at this moment
4
2ω
there is no energy stored in the capacitor and
all the energy is stored in the inductor. At
any other time, one can show that for an
ideal LC circuit where there is no dissipation
from resistance, the total energy stored is the
sum of the energy in the capacitor and the
energy in the inductor, and this energy is
constant in time (although the energy in each
one oscillates).
014 (part 2 of 3) 10.0 points
ragsdale (zdr82) – HW9 – ditmire – (58335)
T
Find total energy at t = .
12
Correct answer: 283.22 µJ.
Explanation:
The total energy does not change with respect
to time, so
Utotal = ULmax = UCmax = 283.22 µJ .
015 (part 3 of 3) 10.0 points
Consider a new initial condition at t = 0:
q0 = 0, I0 = +1.0 A. Let q = qmax cos(ω t+δ),
dq
I=
, where qmax is assumed to be positive.
dt
Find the new phase angle δ for this new
situation.
1. δ = 180◦
2. δ = 90◦
3. δ = 45◦
4. δ = 315◦
5. δ = 270◦ correct
6. δ = 225◦
7. δ = 135◦
8. δ = 0◦
Explanation:
We find the correct δ by fitting the choices
into the formulas below:
q = qmax cos(ω t + δ)
dq
= −ω qmax sin(ω t + δ)
dt
At t = 0, q should be 0, while I is positive.
Remember qmax is assumed to be positive.
Then only δ = 270◦ fits all the requirement;
i.e., at t = 0
I=
7
keywords:
016 10.0 points
In an RL series circuit, an inductor of 4.17 H
and a resistor of 7.51 Ω are connected to a
23.5 V battery. The switch of the circuit
is initially open. Next close the switch and
wait for a long time. Eventually the current
reaches its equilibrium value.
At this time, what is the corresponding
energy stored in the inductor?
Correct answer: 20.4156 J.
Explanation:
Let : L = 4.17 H ,
R = 7.51 Ω , and
E = 23.5 V .
The current in an RL circuit is
E I=
1 − e−Rt/L .
R
The final equilibrium value of the current,
which occurs as t → ∞ , is
E
23.5 V
I0 =
=
= 3.12916 A .
R
7.51 Ω
The energy stored in the inductor carrying a
current 3.12916 A is
1
U = L I2
2
1
= (4.17 H) (3.12916 A)2
2
= 20.4156 J .
017 10.0 points
In the circuit below, initially the switch S1 is
in position “a” and switch S2 is closed.
L
C2
S2
S1 b
C1
a
q = qmax cos 270◦ = 0 ,
E
and
I = −ω qmax sin 270◦ = ω qmax > 0 .
R
ragsdale (zdr82) – HW9 – ditmire – (58335)
When the switch S1 is then thrown to “b”,
the voltage will oscillate at a frequency which
is
1. the same whether switch S2 is open or
closed.
2. zero, since the circuit was not oscillating
to begin with.
3. higher when switch S2 is open. correct
4. lower when switch S2 is open.
5. zero, since the battery is no longer in the
circuit.
Explanation:
The oscillation frequency is
ω=√
1
1
.
=p
LC
L (C1 + C2 )
The capacitors are parallel C = C1 + C2 ,
thus when switch S2 is opened the capacitance
decreases and its inverse increases. Therefore,
the frequency ω is higher when only ONE of
the capacitors is connected.
018 (part 1 of 3) 10.0 points
The switch in the circuit in the figure below
is initially in position a with no connection to
position b. After a very long time, the switch
is thrown from position a to position b (with
no connection to position a) at time t1 . But,
while the switch is moving from position a
to position b, both positions are connected to
the switch. This allows the current through
the inductor to be maintained by the battery
while the switch is in the process of being
thrown until the switch is thrown all the way
to position b.
8
I(t)
R
L
S b
a
E
If I0 = I(t1 ) is the current in the inductor
at time t = t1 , what is the current I(t) as a
function of time?
−R (t−t1 )/L
1. I(t) = I0 1 − e
2. I(t) = I0 e−L (t−t1 )/R
3. I(t) = I0 e−R t/L
i
h
−R t/L
4. I(t) = I0 1 − e
5. I(t) = I0 e−L t/R
6. I(t) = I0 1 − e−L (t−t1 )/R
7. I(t) = I0 e−R (t−t1)/L correct
h
i
8. I(t) = I0 1 − e−L t/R
Explanation:
Basic Concepts:
LR circuits.
Magnetic energy stored in an inductor:
1
UM = L I 2 .
2
Power dissipated in a resistor:
V2
P = I V = I2 R =
.
R
For an LR-circuit, the current after the battery is disconnected is given by
I(t) = I0 e−(t−t1 )/τ ,
L
where τ =
and I0 is the current after the
R
circuit has been connected to the battery for
E
.
a long time. I0 =
R
The current is given by
′
I(t) = I0 e−t /τ .
ragsdale (zdr82) – HW9 – ditmire – (58335)
dt′
= 1 and
dt dI
1 −t′ /τ
e
.
= I0 −
dt
τ
If t′ = t − t1 ,
Thus for the power transferred from the inductor to the rest of the circuit,
d UM
1 −2 t′ /τ
2
e
= L I0 −
dt
τ
′
1
2
e−2t /τ
= −L I0
L/R
2
= −I (t) R .
Thus, we get the same results as we found using the simple argument from conservation of
energy. The minus sign tells us that energy is
being lost from the magnetic field as magnetic
energy is converted first into electrical energy
and then into heat energy in the resistor.
019 (part 2 of 3) 10.0 points
What is the instantaneous rate at which the
magnetic energy stored in the inductor’s magnetic field is converted into electrical energy
as a function of time after the switch is thrown
from position a to position b?
1. Pdiss (t) = L
dI
dt
dI
dt
dI
3. Pdiss (t) = 2 L I(t)
dt
2. Pdiss (t) = R I(t)
2
to electrical energy is dissipated in the resistor. There is nothing in the circuit to store
the electrical energy (i.e., there are no capacitors). This means that the rate at which
magnetic energy is converted into electrical
energy is equal to the rate at which electrical energy is converted into heat energy. The
rate of change of energy is power. This means
that the rate at which magnetic energy is converted into electrical energy is equal to the
power dissipated in the resistor, so
d UM
= Pdiss (t) = I 2 (t) R .
dt
The Complicated Way: The energy
stored in the magnetic field of the inductor
is given by
UM (t) =
R
I(t)
L
1
7. Pdiss (t) = L I 2 (t)
2
L
8. Pdiss (t) = I(t)
R
Explanation:
The Simple Way:
From conservation of energy, all of the energy which is converted from magnetic energy
6. Pdiss (t) =
1
L I 2 (t) .
2
The rate at which this energy is converted
into electrical energy is then
d UM
d 1
dI
2
=
L I (t) = L I
.
dt
dt 2
dt
020 (part 3 of 3) 10.0 points
What is the total amount of electrical energy
dissipated in the resistor after the switch is
thrown to position b?
1. ∆Ediss =
2. ∆Ediss
4. Pdiss (t) = I (t) R correct
5. Pdiss (t) = L I(t)
9
3. ∆Ediss
L
I0
R
d I = R I0
dt t=t1
d I = 2 L I0
dt t=t1
4. ∆Ediss
1
= L I02 correct
2
5. ∆Ediss = I02 R
R
I0
L
L d I =
dt t=t1
6. ∆Ediss =
7. ∆Ediss
8. ∆Ediss = L I0
Explanation:
ragsdale (zdr82) – HW9 – ditmire – (58335)
The Simple Way: Once again, we use
the principle of conservation of energy. The
initial energy stored in the magnetic field of
the inductor is
1
UM i = L I02 .
2
The final energy stored in the inductor is
1
1
UM f = L If2 = L × 0 = 0 .
2
2
The change in the magnetic energy is thus
1
∆UM = UM f − UM i = − L I02 .
2
The minus sign again tells us that magnetic
energy is decreasing as it is converted into
electrical energy. From conservation of energy, all of this energy is dissipated in the
resistor, so the amount of energy dissipated in
the resistor is
1
Ediss = L I02 .
2
The Complicated Way: The power dissipated in the resistor is Pdiss (t) = I 2 (t) R .
d Ediss
From the definition of power, Pdiss =
.
dt
We can solve for the energy dissipated by integrating this with respect to time;
Z ∞
Pdiss (t) dt
∆Ediss =
t1
Z ∞
I 2 (t) dt
=R
t1
Z ∞h
i2
2
= I0 R
e−(t−t1)/τ dt
Zt1∞
e−2 (t−t1)/τ dt .
= I02 R
t1
′
Let t = t − t1 ; then so dt′ = dt, and
Z ∞
′
2
∆Ediss = I0 R
e−2 t /τ dt′
i∞
0 τ h
−2 t′ /τ
2
e
= I0 R −
2 0
L
1 2
[0 − 1]
= − I0 R
2
R
1
= L I02 .
2
This is the same result we got using conservation of energy directly.
10