oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
Question 1, chap 29, sect 5.
part 1 of 1
10 points
The wire is carrying a current I.
y
180◦
and r̂ are parallel meaning d~s × r̂ = 0 for this
part of the wire. It is now easy to see that the
right part, having a d~s antiparallel to r̂, also
~ at O.
gives no contribution to B
Let us go through the semicircle C. The
element d~s, which is along the wire, will now
be perpendicular to r̂, which is pointing along
the radius towards O. Therefore
| d~s × r̂| = ds
I
I
1
r
O
x
I
~
Find the magnitude of the magnetic field B
at O due to a current-carrying wire shown in
the figure, where the semicircle has radius r,
and the straight parts to the left and to the
right extend to infinity.
µ0 I
correct
4r
µ0 I
2. B =
3r
µ0 I
3. B =
r
µ0 I
4. B =
3πr
µ0 I
5. B =
2r
µ0 I
6. B =
πr
µ0 I
7. B =
4πr
µ0 I
8. B =
2πr
Explanation:
By the Biot-Savart Law,
Z
d~s × r̂
µ
I
0
~ =
.
B
4π
r2
1. B =
Consider the left straight part of the wire.
The line element d~s at this part, if we come
in from ∞, points towards O, i.e., in the xdirection. We need to find d~s × r̂ to use the
Biot-Savart Law. However, in this part of the
wire, r̂ is pointing towards O as well, so d~s
using the fact that r̂ is a unit vector. So the
Biot-Savart Law gives for the magnitude B of
the magnetic field at O:
Z
µ0 I
ds
B=
.
4 π C r2
Since the distance r to the element d~s is constant everywhere on the semicircle C, we will
be able to pull it out of the integral. The
integral is
Z
Z
1
1
ds
ds = 2 LC
= 2
2
r C
r
C r
where LC = π r is the length of the semicircle.
Thus the magnitude of the magnetic field is
B=
µ0 I 1
µ0 I
.
π
r
=
4 π r2
4r
Question 2, chap 29, sect 3.
part 1 of 1
10 points
A conductor consists of an infinite number
of adjacent wires, each infinitely long and
carrying a current I (whose direction is out-ofthe-page), thus forming a conducting plane.
A
C
If there are n wires per unit length, what is
~
the magnitude of B?
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
through the loop is n I l. Note that since there
~ in the direction of w, we
is no component of B
are only interested in the contributions along
sides l
I
~ · d~s = 2 B l = µ0 n l I
B
µ0 I
1. B =
4
µ0 I
2. B =
2
3. B = 2 µ0 I
B=
4. B = 4 µ0 I
5. B =
2
µ0 n I
4
µ0 n I
.
2
Question 3, chap 29, sect 3.
part 1 of 1
10 points
6. B = µ0 n I
A total current of 35 mA flows through an
infinitely long cylinderical conductor of radius
3 cm which has an infinitely long cylindrical
r
hole through it of diameter r centered at
2
along the x-axis as shown.
The permeability of free space is 4 π ×
10−7 T · m/A .
y
7. B = 2 µ0 n I
8. B = 4 µ0 n I
9. B = µ0 I
µ0 n I
correct
2
Explanation:
B
A
10. B =
l
x
W
C
B
By symmetry the magnetic fields are equal
and opposite through point A and C and horizontally oriented. Following the dashed curve
in
I a counter-clockwise direction, we calculate
~ · d~s, which by Ampere’s law is proporB
tional to the current through the dashed loop
coming out of the plane of the paper. In
this problem this is a positive current. Hence
~ along the horizontal legs points in the diB
rection in which we follow the dashed curve.
Ampere’s Law is
I
~ · d~s = µ0 I .
B
To evaluate this line integral, we use the rectangular path shown in the figure. The rectangle has dimensions l and w. The net current
What is the magnitude of the magnetic field
at a distance of 13 cm from the origin along
the positive x-axis?
Assume: The magnitude of the current
density is the same in the cylinder and in
the hole and that the currents in the cylinder
and the hole flow in opposite directions with
respect to each other.
Correct answer: 5.1505 × 10−8 T (tolerance
± 1 %).
Explanation:
Basic Concepts: Magnetic Field due to a
Long Cylinder
B =
µ0 I
.
2πr
Principle of Superposition.
Our goal is to model the given situation,
which is complex and lacks symmetry, by
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
3


adding together the fields from combinations
µ0 I  3 x − 2 r 
of simpler current configurations which to=
6π x x− r
gether match the given current distribution.
2
The combination of the currents in Fig. 2 will
−7
(4 π × 10 T m/A) (35 mA)
do so if we choose Icyl and Ihole correctly.
=
6π


y
Hole
r
 3 (13 cm) − 2 (3 cm) 
r
y

×

2
3 cm 
(13 cm) 13 cm −
2
x
x
= 5.1505 × 10−8 T .
4
Icyl = I
3
1
Ihole = − I
3
Question 4, chap -1, sect -1.
part 1 of 1
10 points
Since the current is uniform, the current
I
density J =
is constant. Then
In a region of space, the electric field varies
A
according to E = (0.07 N/C) sin(2400 s−1 ) t,
J = Icyl Acyl = −Ihole Ahole
where t is in seconds.
The permittivity of free space is 8.85 ×
2
π
r
10−12 C2 /N · m2 .
, so
Clearly, Acyl = π r 2 , and Ahole =
4
Find the maximum displacement current
Icyl
~
through a 2 m2 area perpendicular to E.
Ihole = −
.
−9
4
Correct answer: 2.9736 × 10
A (tolerance
Note: The minus sign means Ihole is flowing
± 1 %).
in the direction opposite Icyl and I, as it must
Explanation:
if it is going to cancel with Icyl to model the
hole.
We also require I = Icyl + Ihole . We then
Let : E0 = 0.07 N/C ,
4
1
have Icyl = I, and Ihole = − I. With these
ω = 2400 s−1 ,
3
3
currents, the combination of the two cylinders
A = 2 m2 , and
in figure 2 gives the same net current and
ǫ0 = 8.85 × 10−12 C2 /N · m2 .
current distribution as the conductor in our
problem.
The displacement current is
The magnetic fields are
d φe
Id = ǫ0
4
dt
µ0
I
d
3
= ǫ0 (E A)
Bcyl =
dt
2
πx d
1
= ǫ0 E0 A (sin ω t)
µ0 − I
dt
3
Bhole =
,
= ǫ0 E0 A ω cos ω t ,
2 π (x − r/2)
so the maximum displacement current is
so the total magnetic field is
Id,max = ǫ0 E0 A ω
Btotal = Bcyl + Bhole
= (8.85 × 10−12 C2 /N · m2 ) (0.07 N/C)


× (2 m2 ) (2400 s−1 )
µ0 I  4
1 
=
−
6π x x− r
= 2.9736 × 10−9 A .
2
+
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
B
Question 5, chap 31, sect 1.
part 1 of 1
10 points
A square piece of copper is rotated about
its center in a magnetic field B (into the page
⊗, out of the page ⊙). Shown below are different charge configurations associated with
this procedure.
Select the figure with an acceptable charge
distribution.
B
1.
2.
B
−
++
−
++
−− +
− −−
+ −
+
−
+
−
++
ω
B
B
B
ω
B
ω
6.
B
B
B
+ +
+ −− +
− −
− −
+ −− +
+ +
B
ω
B
ω
− −
− ++ −
+ +
+ +
− ++ −
− −
ω
B
Explanation:
Using the right-hand-rule, the only acceptable charge distribution is
B
ω
B
B
ω
B
ω
correct
ω
− −
− ++ −
+ +
+ +
− ++ −
− −
B
B
4.
B
B
B
3.
ω
ω
B
+++
−
−
+
−
+
− −−
−
− +
++
−
++
−
5.
4
B
− −
− ++ −
+ +
+ +
− ++ −
− −
B
B
− −
− ++ −
+ +
+ +
− ++ −
− −
ω
B
ω
Question 6, chap 31, sect 2.
part 1 of 2
10 points
B
A long, straight wire carries a current and
lies in the plane of a rectangular loops of wire,
as shown in the figure.
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
5
12 cm
140 loops (turns)
7.2 cm
h
i
(31 A) sin (300 rad/s) t + δ →
The magnitude of the magnetic field is
Determine the maximum emf, |E|, induced
in the loop by the magnetic field created by
the current in the straight wire.
Correct answer: 44.8022 mV (tolerance ± 1
%).
Explanation:
Let : N = 140 ,
ω = 300 rad/s ,
ℓ = 12 cm ,
a = 7.2 cm ,
b = 23 cm , and
I0 = 31 A .
I = I0 sin(ω t + δ)
dr
a
ℓ
b
Magnetic field near a long wire is
B=
µ0 I
.
2πr
Faraday’s Law is
E =−
µ0 I
.
2πr
Thus the flux linkage is
ΦB = N ΦB1
Z
µ0 N I ℓ a+b dr
=
2π
r
a
a+b
µ0 N I0 ℓ
sin(ω t + δ) .
ln
=
2π
a
23 cm
r
B=
d ΦB
.
dt
Finally, the induced emf E is
d ΦB
dt
µ0 N I0 ℓ ω
a+b
cos(ω t + δ) ,
=−
ln
2π
a
E =−
which is a maximum when the cosine function
yields 1.
Emax
a+b
µ0 N
I0 ℓ ω ln
=
2π
a
−6
(1.25664 × 10 N/A2)(140)
=
2(3.1415926)
× (31 A)(12 cm)(300 rad/s)
(7.2 cm) + (23 cm)
× ln
(7.2 cm)
3 mV
−2 m
10
× 10
cm
V
= 44.8022 mV .
Question 7, chap 31, sect 2.
part 2 of 2
10 points
If at t = 0, δ=0, and a positive I denotes
an upward moving current in the figure, then
which statement below is correct at t = 0.
1. The current in the loop is counterclockwise. correct
2. There is zero current in the loop.
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
6
Ohm’s Law
3. The sense of current is not determined
with the information given.
4. The current in the loop is clockwise.
Explanation:
dI
Note:
= ω I0 .
dt (t=0)
At t = 0, the magnetic field from the long
straight wire is increasing into the loop from
above which leads to a finite time rate of
change of magnetic flux, although B(t=0) = 0 .
The induced current in the loop then attempts to produce magnetic field through the
loop that penetrates the loop from below (i.e.,
from Lenz’s law it is attempting to resist the
change of flux). By the right hand rule, this
requires counter-clockwise induced current in
the loop.
Question 8, chap 31, sect 1.
part 1 of 1
10 points
Given: Assume the bar and rails have negligible resistance and friction.
In the arrangement shown in the figure,
the resistor is 2 Ω and a 5 T magnetic field
is directed into the paper. The separation
between the rails is 8 m . Neglect the mass of
the bar.
An applied force moves the bar to the left
at a constant speed of 1 m/s .
V
.
R
Solution: The motional E induced in the
circuit is
I=
E = Bℓv
= (5 T) (8 m) (1 m/s)
= 40 V .
From Ohm’s law, the current flowing through
the resistor is
E
I=
R
Bℓv
=
R
(5 T) (8 m) (1 m/s)
=
R
= 20 A .
The power dissipated in the resistor is
P = I2 R
B 2 ℓ2 v 2
=
R
R2
B 2 ℓ2 v 2
=
R
(5 T)2 (8 m)2 (1 m/s)2
=
(2 Ω)
= 800 W .
Note: First of four versions.
Question 9, chap 31, sect 3.
part 1 of 1
10 points
1 m/s
m≪1 g
2Ω
8m
I
5T
In the figure shown, the magnet is first
moved downward toward the loop of wire,
then withdrawn upward from the loop of wire.
5T
N
At what rate is energy dissipated in the
resistor?
Correct answer: 800 W (tolerance ± 1 %).
Explanation:
Basic Concept: Motional E
E = Bℓv.
S
Clockwise
induced I
current
down
then
up
Counterclockwise
I induced
current
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
As viewed from above, the induced current
in the loop is
B
2. 2
1. first clockwise, then counter-clockwise.
correct
B
7
cm
B
2. first counter-clockwise, then clockwise.
P
3. for both cases clockwise with decreasing
magnitude.
What is the magnitude of the electric field
at point P (a distance 1.52 cm from the center) and when t = 1.84 s?
Correct answer: 0.83904 mV/m (tolerance ±
1 %).
4. for both cases clockwise with increasing
magnitude.
5. for both cases counterclockwise with decreasing magnitude.
Explanation:
Let : a = 1.66 T ,
b = 0.03 T/s2 ,
R = 2.2 cm , and
r = 1.52 cm .
6. for both cases counterclockwise with increasing magnitude.
Explanation:
When the south pole of the magnet is
moved downward toward the loop, there is
an increase in the upward magnetic flux at
the loop. This means that the direction of the
change in the magnetic flux is upward. Lenz’s
law implies that the loop generates an induced
magnetic field Bind to oppose this change, so
Bind is points downward. The right hand rule
implies that the induced current is clockwise
as viewed from above. As the magnet is withdrawn upward from the loop, following the
same reasoning, the induced current should
now flow in the opposite direction.
Faraday’s Law of Induction:
I
~ ind · d~s = − d ΦB
Eind = E
dt
Note: The magnetic field region is circular, so there is a symmetry for the induced
electric field; i.e., the electric field is equal in
magnitude for all points on the circle of radius
r.
So,
I
~ · d~s = 2 π r E ,
E
while
d ΦB
d (B A)
=
dt
dt
d (B π r 2 )
=
dt
dB
= π r2
dt
2
= πr · 2bt.
Question 10, chap -1, sect -1.
part 1 of 2
10 points
A magnetic field directed into the page
changes with time according to
From Faraday’s Law in general form, we get
B = a + b t2 ,
i.e.,
2
where a = 1.66 T, b = 0.03 T/s , and t is in
seconds.
The magnetic field pole has a circular cross
sectional radius of 2.2 cm.
2 π r E = π r2 · 2 b t ,
E = rbt
= (0.0152 m) (0.03 T/s2 ) (1.84 s)
= 0.83904 mV/m .
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
Question 11, chap -1, sect -1.
part 2 of 2
10 points
What is the direction of the electric field?
1. The electric field is parallel to r and directed to the center of the magnetic field.
2. Information is not sufficient to make a
decision.
3. The electric field is perpendicular to r and
directed counter-clockwise. correct
4. The electric field is parallel to r and directed away from the center of the magnetic
field.
5. The electric field is perpendicular to r and
directed clockwise.
Explanation:
Lenz’s Law – Induced emf’s always oppose
magnetic flux changes.
Use a “bracket”, “[ ]”, to denote the directions. The direction content of Faraday’s Law
then reads,
dΦ
[Eind ] = −
= −⊗ = ⊙ .
dt
The right hand rule then implies that the
direction of Eind is counter-clockwise. This
is consistent with Lenz’s Law: the magnetic
field is increasing into the page, so the induced
emf Eind creates a magnetic field out of the
page to counter this change.
Question 12, chap 31, sect 6.
part 1 of 2
10 points
At t = 0, a 14.7 V battery is connected to a
series circuit containing a 14 Ω resistor and a
3.1 H inductor.
What will be the current in the circuit, a
long time after the circuit is established?
Correct answer: 1.05 A (tolerance ± 1 %).
Explanation:
Let : E = 14.7 V
R = 14 Ω .
and
8
After a long time, we have a simple circuit of
a battery with E and a resistor of R. Then,
E = I R and thus the current is
I=
E
14.7 V
=
= 1.05 A .
R
14 Ω
Question 13, chap 31, sect 6.
part 2 of 2
10 points
How long will it take the current to reach
50 percent of its final value?
Correct answer: 0.153483 s (tolerance ± 1
%).
Explanation:
Let : L = 3.1 H and
I = 0.5 .
L
The time constant is τ = .
R
The current in an RL circuit is given by
I = Is 1 − e−t/τ ,
E
where Is is the steady state current
and τ
R
is the time constant of the circuit. Solving the
above equation for t, we obtain
I
t = −τ ln 1 −
Is
0.5 Is
L
= − ln 1 −
R
Is
3.1 H
=−
ln(1 − 0.5)
14 Ω
Therefore, the time at which I = 0.5 Is is
t1 = −(0.221429 s) ln (1 − 0.5)
= 0.153483 s .
Question 14, chap 31, sect 5.
part 1 of 3
10 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
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
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.
I(t)
R
L
E
S b
a
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?
1. I(t) = I0 e−L (t−t1 )/R
2. I(t) = I0 e−R t/L
i
h
−R t/L
3. I(t) = I0 1 − e
4. I(t) = I0 e−R (t−t1 )/L correct
−L (t−t1 )/R
5. I(t) = I0 1 − e
i
h
−L t/R
6. I(t) = I0 1 − e
−R (t−t1 )/L
7. I(t) = I0 1 − e
8. I(t) = I0 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 bat-
9
tery 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 .
dt′
If t′ = t − t1 ,
= 1 and
dt dI
1 −t′ /τ
e
.
= I0 −
dt
τ
Thus for the power transferred from the inductor to the rest of the circuit,
1 −2 t′ /τ
d UM
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 the 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.
Question 15, chap 31, sect 5.
part 2 of 3
10 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
L I 2 (t)
2
L
2. Pdiss (t) = I(t)
R
dI
3. Pdiss (t) = L
dt
dI
4. Pdiss (t) = 2 L I(t)
dt
1. Pdiss (t) =
5. Pdiss (t) = L I(t)
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
dI
6. Pdiss (t) = R I(t)
dt
R
7. Pdiss (t) = I(t)
L
8. Pdiss (t) = I 2 (t) R correct
Explanation:
The Simple Way:
From conservation of energy, all of the energy which is converted from magnetic energy
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) =
1
L I 2(t) .
2
The rate at which this energy is converted
into electrical energy is then
d 1
dI
d UM
2
=
L I (t) = L I
.
dt
dt 2
dt
Question 16, chap 31, sect 5.
part 3 of 3
10 points
What is the total amount of electrical energy dissipated in the resistor after the switch
is thrown to position b?
1. ∆Ediss = I02 R
R
I0
L
L
=
I0
R
2. ∆Ediss =
3. ∆Ediss
4. ∆Ediss
10
L d I =
dt t=t1
5. ∆Ediss = L I0
6. ∆Ediss
7. ∆Ediss
8. ∆Ediss
d I = R I0
dt t=t1
1
= L I02 correct
2
d I = 2 L I0
dt t=t1
Explanation:
The Simple Way: Once again, we use the
principle of the 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 the 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
e−(t−t1 )/τ dt
= I0 R
Zt1∞
e−2 (t−t1 )/τ dt .
= I02 R
t1
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
Let t′ = t − t1 ; then so dt′ = dt, and
Z ∞
′
2
∆Ediss = I0 R
e−2 t /τ dt′
0 τ h
i∞
2
−2 t′ /τ
= I0 R −
e
0
2 1 2
L
= − I0 R
[0 − 1]
2
R
1
= L I02 .
2
This is the same result we got using the conservation of energy directly.
Question 17, chap 32, sect 4.
part 1 of 1
10 points
Consider the figure shown below. The
switch is initially set at position b. There
is no charge nor current around the right loop
while at position b. At t = t0 the switch is set
to position a.
C
L
b S
a
E
d
c
R
After a long time at position a, the switch
is set back to position b. Denote this time as
t = 0.
What is the maximum energy stored in the
capacitor, expressed in terms of L, E, and R.
(Hint: Use conservation of energy.)
1. Umax
2. Umax
3. Umax
4. Umax
5. Umax
E
=L
R
2
E
=L
R
E
=L 2
R
E
1
= L
2 R
2
E
1
correct
= L
2
R
11
E2
R
1 E2
= L
2
R
2
1 2 E
= L
2
R
1
E
= L 2
2 R
2
2 E
=L
R
6. Umax = L
7. Umax
8. Umax
9. Umax
10. Umax
Explanation:
By conservation of energy
Umax =
1 Q2max
1
2
L Imax
=
.
2
2 C
Since
Imax =
E
,
R
the maximum stored energy in the capacitor
is
2
1
E
Umax = L
.
2
R
Question 18, chap 32, sect 4.
part 1 of 2
10 points
At some time after the switch S is closed,
there is a current I flowing through the resistor and inductor which is increasing in time
dI
> 0 (see the figure below). At this time
dt
there is a charge of magnitude q on each plate
of the capacitor.
R
I
E
S
L
C
Which of the following equations is correct?
dI
q
+ =0
dt
C
dI
q
2. E + I R − L
− =0
dt
C
1. E + I R − L
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
3. E + I R + L
4. E + I R + L
5. E − I R − L
6. E − I R − L
7. E − I R + L
8. E − I R + L
dI
dt
dI
dt
dI
dt
dI
dt
dI
dt
dI
dt
−
+
+
−
−
+
q
C
q
C
q
C
q
C
q
C
q
C
=0
the current will drop to zero. Setting I = 0 in
the loop equation from Part 1 gives
E−
=0
so
= 0 correct
=0
qmax = C E
= (5 × 10
−6
F)
1 × 10−6 F
µF
(11.1 V)
= 5.55 × 10−5 C .
=0
Question 20, chap 32, sect 2.
part 1 of 2
10 points
Explanation:
Apply Kirchoff’s loop rule to the above cirdI
cuit. Here we are told that
> 0; so the emf
dt
dI
. Summing
induced by the inductor is −L
dt
the potential differences around the loop gives
dI
q
E −IR−L
− = 0.
dt
C
Question 19, chap 32, sect 4.
part 2 of 2
10 points
A 1.28 µF capacitor is connected across
an alternating voltage with an rms value of
12.8 V. The rms current through the capacitance is 12.1 mA.
What is the source frequency?
Correct answer: 117.54 Hz (tolerance ± 1 %).
Explanation:
Let : C = 1.28 µF = 1.28 × 10−6 F ,
Vrms = 12.8 V , and
I = 0.0121 A .
The capacitive reactance is
5 mH
XC =
11.1 V
qmax
= 0,
C
=0
9. none of these
Consider the circuit
4 MΩ
12
5 µF
S
What is the charge on the capacitor after
the switch has been closed for a long time?
Correct answer: 5.55 × 10−5 C (tolerance ±
1 %).
Explanation:
Let : E = 11.1 V and
C = 5 µF = 5 × 10−6 F .
After a long time t the capacitor will reach
its maximum charge qmax . When this occurs
Vrms
1
=
I
ωC
1
.
ω=
XC C
Thus the source frequency is
ω
1
=
2π
2 π XC C
I
=
2 π Vrms C
0.0121 A
=
2 π (12.8 V) (1.28 × 10−6 F)
f=
= 117.54 Hz .
Question 21, chap 32, sect 2.
part 2 of 2
10 points
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 14 2007, 6:00 pm
If the capacitor is replaced by an ideal coil
with an inductance of 0.136 H, what is the
rms current through the coil?
Correct answer: 127.44 mA (tolerance ± 1
%).
Explanation:
Let : L = 0.136 H .
The capacitive reactance is
XL = ω L =
L
XC C
,
so the rms current is
Vrms
Vrms
=
XL
ωL
Vrms XC C
V2 C
=
= rms
L
IL
2
(12.8 V) (1.28 × 10−6 F) 103 mA
=
·
(0.0121 A) (0.136 H)
1A
Irms =
= 127.44 mA .
Question 22, chap -1, sect -1.
part 1 of 1
10 points
Calculate the average power delivered to
the series RLC circuit with resistance of
355 Ω, inductance of 0.57 H, capacitor of
2.68 µF, and frequency of 370 s−1 , if the maximum value of the applied voltage equal to
178 V.
Correct answer: 7.37906 W (tolerance ± 1
%).
Explanation:
ω = 370 s−1 ,
L = 0.57 H ,
C = 2.68 µF = 2.68 × 10−6 F ,
R = 355 Ω , and
Vmax = 178 V .
Let :
The reactances are
XL = ω L
= (370 s−1 ) (0.57 H)
= 210.9 Ω ,
XC =
1
ωC
=
13
1
s−1 ) (2.68 ×
(370
= 1008.47 Ω .
10−6 F)
Therefore, the impedance is
q
Z = R2 + (XL − XC )2
q
= (355 Ω)2 + (210.9 Ω − 1008.47 Ω)2
= 873.009 Ω .
The rms voltage and current are
178 V
Vmax
Vrms = √ = √ = 125.865 V ,
2
2
Vrms
125.865 V
Irms =
=
= 0.144174 A .
Z
873.009 Ω
The phase angle is
XL − XC
R
210.9 Ω − 1008.47 Ω
= arctan
355 Ω
◦
= −66.0061 .
φ = arctan
Thus the averaged power is
Pav = Irms Vrms cos φ
= (0.144174 A) (125.865 V)
× cos(−66.0061◦ )
= 7.37906 W .