Date post: | 16-Apr-2017 |
Category: |
Education |
Upload: | rohit-mohd |
View: | 214 times |
Download: | 5 times |
Electromagnetic Induction
2
Define and use magnetic flux,
20.1 Magnetic Flux
cosBAAB
• is defined as the scalar product between the magnetic flux density, B and the vector of the surface area, A
where,: magnetic flux
: angle between B and AB: magnetic flux density
A: area of the coil• Scalar quantity• Unit: T m2 or Wb
Magnetic Flux, : a measure of the number of magnetic field lines that cross a given area
A
B = 90
BA = 0
3
Direction of vector A always perpendicular to the surface area, A.
cosBAAB
4
A flat surface with area 3.0 cm2 is placed in a uniform magnetic field. If the magnetic field strength is 6.0 T, making an angle 30° with the surface area, find the magnetic flux through this area.
Example 20.1Solution:
3030
coil
B
Normal A
Using: cosAB23 m T 109.0
θ = 90 – 30 = 60°
5
REVISION: Magnetic Flux cosBAAB
6
• A solenoid 4.00 cm in diameter and 20.0 cm long has 250 turns and carries a current of 15.0 A. Calculate the magnetic flux through the circular cross sectional area of the solenoid
• A long, straight wire carrying a current of 2.0 A is placed along the axis of a cylinder of radius 0.50 m and a length of 3.0 m. Determine the total magnetic flux through the cylinder.
REVISION: Magnetic Flux
A
area90
II I
SN
= 2.96×10–5 Tm2
7
The three loops of wire are all in a region of space with a uniform magnetic field. Loop 1 swings back and forth as the bob on a simple pendulum. Loop 2 rotates about a vertical axis and loop 3 oscillates vertically on the end of a spring. Which loop or loops have a magnetic flux that changes with time? Explain your answer.
Example 20.2
8
Only loop 2 has a changing magnetic flux.Reason:Loop 1 moves back and forth, and loop 3 moves up and down, but since the magnetic field is uniform, the flux always constant with time.Loop 2 on the other hand changes its orientation relative to the field as it rotates, hence its flux does change with time.
Solution:
9
1A long, straight wire carrying a current of 2.0 A is placed along the axis of a cylinder of radius 0.50 m and a length of 3.0 m. Determine the total magnetic flux through the cylinder.[Zero]
2A solenoid 4.00 cm in diameter and 20.0 cm long has 250 turns and carries a current of 15.0 A. Calculate the magnetic flux through the circular cross sectional area of the solenoid.[2.96×10–5 T m2]
Exercise 20.1
10
(d)Derive and use induced emf:- in straight conductor,
- in coil,
- in rotating coil,
(a)Use Faraday’s experiment to explain induced emf
(b)State Faraday’s law and use Lenz’s law to determine the direction of induced current
(c) Use induced emf,
20.2 Induced emf
dtd
sinBlv
dtdBNA
dtdANB
tNAB sin
11
History – Faraday’s experiment to induced emf
• In this experiment, Faraday hoped by using a strong enough battery, a steady current in X would produce a current in a second coil Y but failed
• Current carrying conductor magnetic field
• Magnetic field electric current????
• The diagram below shows the apparatus used by Faraday in his attempt to produce an electric current from a magnetic field
12
• This is Faraday’s apparatus for demonstrating that a magnetic field can produce a current
• A change in the field produced by the top coil induces an EMF and, hence, a current in the bottom coil
• When the switch is opened and closed, the galvanometer registers currents in opposite directions
• No current flows through the galvanometer when the switch remains closed or open
REVISION: Faraday’s experiment to induced emf
OBSERVATION
GALVANOMETER DEFLECTION
Switch on YESSwitch of YESSteady current
NO
Faraday’s experiment used a magnetic field that was changing because the current producing it was changing
13
• Faraday concluded that although a steady magnetic field produces no current, a changing magnetic field can produce an electric current
• Such a current is called an induced current
• We therefore say that an induced current is produced by a changing magnetic field
• The corresponding emf required to cause this current is called an induced emf
Faraday’s experiment to induced emf
OBSERVATION
GALVANOMETER DEFLECTION
Switch on YESSwitch of YESSteady current
NO
14
• a magnetic field that is changing because the magnet is moving
A changing magnetic field induces an emf
15
Electromagnetic Induction is the production of induced e.m.f.s or induced currents whenever the
magnetic flux through a loop, coil or circuit is changed
meaning of changing in magnetic flux
a relative motion of loop &
magnet field lines are ‘cut’
number of magnetic field lines passing through a coil are
increased or decreased
To increase induced e.m.fUse a
stronger
magnet can
increase magneti
c flux
Push the
magnet faster into the coil to increase speed
The area of the coil is
greater
The number of turns increa
sed
Electromagnetic Induction
16
• states that an induced electric current always flows in such a direction that it opposes the change producing it
Lenz’s Law
• states that the magnitude of the induced emf is proportional to the rate of change of the magnetic flux
Faraday’s Law
• These two laws are summed up in the relationship
• For a coil of N turns
• Since d = final - initial
• The negative sign indicates that the direction of induced emf always oppose the change of magnetic flux producing it (Lenz’s law)
dtdΦ
dtdN Φ
dt
N if ΦΦ
17
Solution:A coil of wire 8 cm in diameter has 50 turns and is placed in a B field of 1.8 T. If the B field is reduced to 0.6 T in 0.002 s , calculate the induced emf.
Example 20.3
tN
dtdΦN initialfinalB
tBB
NA initialfinal
tBBdN initialfinal
2
2
V 151
dtdN Φ
Note:
To calculate the magnitude of induced emf, the negative sign can be ignored
18
Solution:
By applying the Faraday’s law equation for a coil of N turns , thus
The magnetic flux passing through a single turn of a coil is increased quickly but steadily at a rate of 5.0102 Wb s1. If the coil have 500 turns, calculate the magnitude of the induced emf in the coil.
Example 20.412 s Wb1005
turns;500
.dtdN
dtdN Φ
2100.5500
V 25
19
• When an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it
• The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant
• In the examples below, if the B field is increasing, the induced field acts in opposition to it
• If it is decreasing, the induced field acts in the direction of the applied field to try to keep it constant
Lenz’s Law
20
Lenz’s Law
21
Lenz’s Law
N
I
Induced current flows in counterclockwise
N
S Induced current flows in clockwise
22
Lenz’s Law
(1834)
• an induced electric current always flows in such a direction that it opposes the change producing it
Faraday’s Law
(1831)
• the magnitude of the induced emf is proportional to the rate of change of the magnetic flux
Faraday’s Experime
nt
• a steady magnetic flux/ field produces no current
• a changing magnetic flux/ field can produce an electric current induce current
Magnetic flux,
dtdΦ
REVISION
23
• Faraday's Law : change in magnetic flux INDUCED CURRENT/ EMF, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it
• The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant
Lenz’s Law • In the examples below, if the B field is increasing, the induced
field acts in opposition to it• If it is decreasing, the induced field acts in the direction (SAME
DIRECTION) of the applied field to try to keep it constant
24
Lenz’s Law
N
I
Induced current flows in counterclockwise
25
Lenz’s Law
Induced current flows in clockwise
N
S
26
Another great example of Lenz's law is to take a copper tube (it's conductive but non-magnetic) and drop a piece of steel
down through the tube. The piece of steel will fall through, as you might expect. It accelerates very close to the acceleration
due to gravity.
Now take the same copper tube and drop a magnet through it. You will notice that the magnet falls very slowly. This is because the copper tube "sees" a changing magnetic field from the falling magnet. This changing magnetic field induces a current in the copper tube. The induced current in the copper tube creates its
own magnetic field that opposes the magnetic field that created it.
27
Faraday’s Law dtdΦ
How can we induce emf?By changing magnetic flux/
field
In a plane coil
By changing area in
magnetic field
By changing magnetic
field strength,
B
Straight conductor
moving through a magnetic
field
Rotating Coil
28
• From
Induced emf in a plane coil by changing area in B
dtdN Φ
cosBAΦand
dtcosBAdN
dtdAcosNB
Stretching the coil reduces the area of the coil magnetic flux through coil is decreased and, current is induced in the coil
Flux through coil is decreased
29
The flexible loop has a radius of 12 cm and is in a magnetic field of strength 0.15 T. The loop is grasped at point A and B and stretched until its area is nearly zero. If it takes 0.20 s to close the loop, find the magnitude of the average induced emf in it during this time.
Example 20.5
30
Solution:
dtd
dANBdt
final initial( )A ANBt
2(0 )rNB
t
2(0 (0.12) )(1)(0.15)0.20
V 104.3 2
31
• From
Induced emf in a plane coil by changing magnetic field, B strength
dtdN Φ
cosBAΦand
dtcosBAdN
As the magnetic field strength, B is increasing or decreasing with time, the magnetic flux through the area changes, therefore induces an emf in the coil
dtdBcosNA
32
A circular coil has 200 turns and diameter 36 cm. the resistance of the coil is 2.0 Ω. A uniform magnetic field is applied perpendicularly to the plane of the coil. If the field changes uniformly from 0.5 T to 0 T in 0.8s.(a) Find the induced e.m.f. &
current in the coil while the field is changed.
(b) Determine the direction of the current induced.
Example 20.6Solution:
1. Calculate area, A = r2
2. Determine emf,
3. Determine I from dtdBNA
IRA36.6
33
Solution:A narrow coil of 10 turns and diameter of 4.0 cm is placed perpendicular to a uniform magnetic field of 1.20 T. After 0.25 s, the diameter of the coil is increased to 5.3 cm.(a) Calculate the change in the
area of the coil.(b) If the coil has a resistance of
2.4 , determine the induced current in the coil.
Example 20.7
0
A
B
B
A
Initial Final
dtdAcosNB
34
Solution:A coil having an area of 8.0 cm2 and 50 turns lies perpendicular to a magnetic field of 0.20 T. If the magnetic flux density is steadily reduced to zero, taking 0.50 s, determine(a)the initial magnetic flux
linkage.(b)the induced emf
Example 20.8
B
0A
Wb108.0 3initial
V 106.1 2
dtdBcosNA
35
Calculate the current through a 37 Ω resistor connected to a single turn circular loop 10 cm in diameter, assuming that the magnetic field through the loop is increasing at a rate of 0.050 T/s. State the direction of the current.
Example 20.9
36
• Calculate, Solution:
dtdΦ
2
2
dA,
dtdBA
RI
I induced
I induced
V. 410933
A 10 x 1.06 5-R
I
37
• Induced emf in a plane coil:– by changing area in
magnetic field
– by changing magnetic field strength
dtdAcosNB
dtdBcosNA
How can we induce emf?By changing magnetic flux/ field
In a plane coil
By changing area in
magnetic field
By changing magnetic
field strength,
B
Straight conductor
moving through a magnetic
field
Rotating Coil
REVISION
dtdΦ
38
Induced emf in a straight conductor moving through a magnetic field
• As a conductor is moved through a magnetic field, current is induced and the bulb is lightened up
39
• Consider a straight conductor of length l is moved at a speed v to the right on a U-shaped conductor in a uniform magnetic field B that points out the paper
• This conductor travels a distance dx =vdt in a time dt
• The area of the loop increases by an amount:
Induced emf in a straight conductor moving through a magnetic field
ldxdA
40
• This induced emf is called motional induced emf
• The direction of the induced current or induced emf in the straight conductor can be determined by using the Lenz’s Law
• If B field is increasing, the induced field acts in opposition
to it• If B is decreasing, the induced field acts in the direction (SAME DIRECTION) of the applied field
to try to keep it constant
• According to Faraday’s law, the e.m.f. is induced in the conductor and its magnitude is given by
angle between v and B
Induced emf in a straight conductor moving through a magnetic field
dtd
dtdAB
dtdxBl v
dtdx
and
sinBlvBlv
41
42
Consider the arrangement shown below. Assume that R = 6 Ω, L = 1.2 m & a uniform 2.50 T magnetic field is directed into the page.(a)At what speed should the
bar be moved to produced a current of 0.5A in the resistor
(b)what is the direction of the induced current?
Example 20.10
43
(a) = Blv sin , = IR
(b)From Lenz’s Law
Solution:
1s m1 v
44
(a)Calculate the motional induced emf in the rod.
(b)If the rod is connected in series to the resistor of resistance 15 , determine
(i) the induced current and its direction
(ii) the total charge passing through the resistor in two minute
A 20 cm long metal rod CD is moved at speed of 25 m s1 across a uniform magnetic field of flux density 250 mT. The motion of the rod is perpendicular to the magnetic field as shown.
Example 20.11
C
D
B
1s m 52
45
Induced emf in a rotating coil
An ac generator / dynamo: transforms mechanical energy into electric energy
46
• By applying the equation of Faraday’s law for a coil of N turns, thus the induced emf is given by
• Consider a coil of N turns each of area A and is being rotated about a horizontal axis in its own plane at right angle to a uniform magnetic field of flux density B.
• As the coil rotates with the angular speed ω, the orientation of the loop changes with time.
Induced emf in a rotating coil
cosBA t andtBA cos
dtdN
tBAdtdN cos
tdtdNBA cos
tNBA sin
47
• The emf induced in the loop varies sinusoidally in time
• The induced emf is maximum when hence
where
Induced emf in a rotating coil
NBAmax
Tf 22
Note:
This phenomenon was the important part in the
development of the electric generator or
dynamo.
48
iii) The emf induced in a coil varies sinusoidally with time.
iv) Maximum voltage (ξ max = NBA ω) is produced when the coil is parallel to the magnetic field (sin ωt = 1).
v) No voltage exists when the coil is perpendicular to the magnetic field
• The graph show that :i) The magnitude of
induced emf is depends on the angle between the field and the coil.
ii) The induced emf is an alternating voltage because has positive value as well as negative value.
Induced emf in a rotating coil
(a)Define self-inductance(b)Apply self-inductance,
for coil and solenoid
20.3 Self Inductance
dtdI
L
50
• Running a changing current (by changing R), creates a changing magnetic field, which creates an induced emf that fights the change
• Unit: Henry (V s A-1)
Self Induction – the production of e.m.f. in a circuit due to the change of current in the circuit itself
51
When the switch S is closed, a current I begins to flow in the solenoid
The current produces a magnetic field whose field lines through the solenoid and generate the magnetic flux linkage
If the resistance of the variable resistor changes, thus the current flows in the solenoid also changed, then so too does magnetic flux linkage
Self Induction
S RII
NS
52
• According to the Faraday’s law, an emf has to be induced in the solenoid itself since the flux linkage changes
• In accordance with Lenz’s law, the induced emf opposes the changes that has induced it and it is known as a back emf
• For an increasing current, the direction of the induced field and emf are opposite to that of the current, to try to decrease the current
• For the current I increases:Self Induction
indε- +
NSI
ISN
Direction of the induced emf is in the opposite direction of the current I.
53
• If the current is decreasing, the direction of the induced field and emf are in the same direction as the current, to try to increase the current
• This coil is said to have self-inductance (inductance)
• A coil that has inductance is called an inductor use to store energy in the form of magnetic field
• For the current I decreases:Self Induction
+ -indε
NSI IindI
indI
NS
Direction of the induced emf is in the same direction of the current I.
54
(a) A current in the coil produces a magnetic field directed to the left
Self Induction
(b) If the current increases, the increasing magnetic flux creates an induced emf having the polarity shown by the dashed battery
(a)The polarity of the induced emf reverses if the current decreases
55
• From the Faraday’s law, thus
• From the self-induction phenomenon, we get
whereL: self inductance of the coil,
unit: Henry (H) or Wb A-1
I: current
Self Inductance, L: the ratio of the self induced e.m.f. to the rate of change of current in the coil
ILΦ
LILΦdt
d L
LIdtd
dtdIL
dtdIL
/
• The value of the self-inductance depends on(a) the size and shape of the coil(b) the number of turn (N)(c) the permeability of the medium in the
coil ()
56
• Therefore the self-inductance of the solenoid is given by
• The magnetic flux density at the centre of the air-core solenoid is given by
• The magnetic flux passing through each turn of the solenoid always maximum and is given by
Self Inductance of a solenoid
lNI
B 0
0cosBA
AlNI
0
lNIA0
INL
lNIA
INL 0
lAN
L2
0
57
Suppose you wish to make a solenoid whose self-inductance is 1.4 mH. The inductor is to have a cross-sectional area of 1.2 x 10 -3 m2 and a length of 0.052 m. How many turns of wire needed?
Example 20.14Induced emf of 5.0 V is developed across a coil when the current flowing through it changes at 25 A s-1. Determine the self inductance of the coil.
Example 20.12
H.20
dtdIL
Example 20.13If the current in a 230 mH coil changes steadily from 20.0 mA to 28.0 mA in 140 ms, what is the induced emf?
220 turnsl
ANL
20
58
Solution
a. The change in the current is
Therefore the inductance of the solenoid is given by
A 500 turns of solenoid is 8.0 cm long. When the current in the solenoid is increased from 0 to 2.5 A in 0.35 s, the magnitude of the induced emf is 0.012 V. Calculate(a)the inductance of the
solenoid,(b)the cross-sectional area of
the solenoid,(c) the final magnetic flux
linkage through the solenoid.
Example 20.15;A 5.2;0 m;10 0.8 turns;500 fi
2 IIlNV 012.0 s; 35.0 dt
if IIdI 05.2 dIA 5.2dI
dtdIL
35.05.2012.0 L
H 1068.1 3L
59
c. The final magnetic flux linkage is given by
b. By using the equation of self-inductance for the solenoid, thus
lAN
L2
0
2
273
100.85001041068.1
A
24 m 1028.4 A
ffL LI
5.21068.1 3 Wb102.4 3
fL
• An inductor is a circuit component (coil or solenoid) which produced an self induced emf
• Function of an inductor:(1)to control current(2)store energy in form of
magnetic field• Back emf produce in an
inductor is given by:
Derive and use the energy stored in an inductor,
20.4 Energy Stored In Inductor
2
21 LIU
Inductor
dtdIL
61
• The total work done while the current is changed from zero to its final value is given by
and analogous toin capacitor
• For a long air-core solenoid, the self-inductance is
• Therefore the energy stored in the solenoid is given by
• The electrical power P in overcoming the back emf in the circuit is given by
Energy Stored In Inductor
IP
dtdILIP
dILIPdt dILIdU
IUIdILdU
00
2
21 LIU
2
21 CVU
lAN
L2
0
2
21 LIU
lAIN
U22
0
21
62
A 400 turns solenoid has a cross sectional area 1.81×10-3 m2 and length 20 cm carrying a current of 3.4 A.(i) Calculate the inductance of
the solenoid(ii) Calculate the energy stored
in the solenoid.(iii)Calculate the induced emf in
the solenoid if the current drops uniformly to zero in 55 ms.
How much energy is stored in a 0.085-H inductor that carries a current of 2.5 A?
Example 20.16
2
21 LIU
Example 20.17 A steady current of 2.5 A in a coil of 500 turns causes a flux of 1.4 x 10-4 Wb to link (pass through) the loops of the coil. Calculate(a)the average back emf induced
in the coil if the current is stopped in 0.08 s
(b)the inductance of the coil and the energy stored in the coil (inductor).
Example 20.18
HL 31082.1
J21005.1
V1125.0
63
Solution:(a)1. Calculate A2. Use
(b) Use
A solenoid of length 25 cm with an air-core consists of 100 turns and diameter of 2.7 cm. Calculate(a)the self-inductance of the
solenoid, and (b)the energy stored in the
solenoid, if the current flows in it is 1.6 A.
(Given 0 = 4 107 H m1)
Example 20.19
lANL
20
2
21 LIU
H 1088.2 5LJ 1069.3 5U
• Mutual Induction: emf induced in a circuit by a changing current in another nearby circuit
(a)Define mutual inductance
(b)Use mutual inductance
between two coaxial solenoids or a coaxial coil and a solenoid
20.5 Mutual Inductance
lANNM 210
65
The induced current in this loop that caused by the change of current in neighbouring loop Mutual Induction
NS N S
Increasing current in loop 1 produces a change in magnetic flux. This changes is experienced by loop 2 that placed nearby
According to Faraday’s law, emf is induced in loop 2 to oppose the changes.
12
66
• In mutual induction, the e.m.f. induced in one coil is proportional to the rate at which the current in the other coil is changing
• If we assume that the current in coil 1 changes at a rate of dI1/dt, the magnetic flux will change by dΦ1/dt and this changes is experienced by coil 2
• The induced e.m.f. in coil 2 is
• If vice versa, the induced emf in coil 1, 1 is given by where
Mutual Inductance, M: the ratio of induced emf in a coil to the rate of change of current in another coil
dtdI1
2 dtdIM 1
122
dtdIM 2
211 MMM 2112
Keep In mind emf induced in coil 2 is due to the current change in coil 1
emf induced in coil 1 by changing current in coil 2
67
• From Faraday’s Law
and
• Since M12 = M21 = M
• Rearrange,
Mutual Inductance, M
dtdIM 1
122 dtdIM 1
2
dtdIM 2
211 dtdIM 2
2
dtdI
dtdIM
2
1
1
2
dtdN 2
22
dtdN
dtdIM 2
21
12
22112 dNdIM
1
2212 I
NM
22112 NIM
2
1121 I
NM
2
11
1
22
IN
INM
68
Solution:(a)Using
(b)From
Two coils, X & Y are magnetically coupled. The emf induced in coil Y is 2.5 V when the current flowing through coil X changes at the rate of 5 A s-1. Determine:(a)the mutual inductance of the
coils(b)the emf induced in coil X if
there is a current flowing through coil Y which changes at the rate of 1.5 A s-1.
Example 20.201552 As
dtdI,V. x
Y
dtdI
MX
Y
H5.0M
dtdIM Y
X
V75.0
69
• Consider a long solenoid with length l and cross sectional area A is closely wound with N1 turns of wire
• A coil with N2 turns surrounds it at its centre
• When a current I1 flows in the primary coil (N1), it produces a magnetic field B1,
• Magnetic flux,
Mutual inductance, M between two coaxial solenoids
lIN
B 1101
l
I1 I1
N1N2
A
N1: primary coil
N2: secondary coil
0cos11 ABl
AIN 1101
70
• If no magnetic flux leakage, thus
• If the current I1 changes, an emf is induced in the secondary coils, therefore the mutual inductance occurs and is given by
Mutual inductance, M between two coaxial solenoids
21
lAIN
INM 110
1
212
1
2212 I
NM
lANN
MM 21012
71
Solution:N1= 1000; l = 50×10-2 m;d1 = 3×10-2 m, N2 = 50;
(a)Using
(b)In secondary coil,
Primary coil of a cylindrical former with the length of 50 cm and diameter 3 cm has 1000 turns. If the secondary coil has 50 turns, calculate:(a)its mutual inductance(b)the induced emf in the
secondary coil if the current flowing in the primary coil is changing at the rate of 4.8 A s-1.
Example 20.2111 84 As.
dtdI
lANNM 210
HM 51088.8
dtdIM 1
2
V41025.4
72
Solution:(a)Using
(b)Using
The primary coil of a solenoid of radius 2.0 cm has 500 turns and length of 24 cm. If the secondary coil with 80 turns surrounds the primary coil at its centre, calculate(a) the mutual inductance of the
coils (b) the magnitude of induced
e.m.f. in secondary coil if the current in primary coil changes at the rate 4.8 A s-1.
Example 20.22
P
PS
lANNM 0
dtdIM P
S
73
A current of 3.0 A flows in coil C and is produced a magnetic flux of 0.75 Wb in it. When a coil D is moved near to coil C coaxially, a flux of 0.25 Wb is produced in coil D. If coil C has 1000 turns and coil D has 5000 turns.(a)Calculate self-inductance of coil C and the energy stored
in C before D is moved near to it(b)Calculate the mutual inductance of the coils(c) If the current in C decreasing uniformly from 3.0 A to
zero in 0.25 s, calculate the induced emf in coil D.
Example 20.23
74
(b)The mutual inductance of the coils is given by
(a)The self-inductance of coil C is given by
and the energy stored in C is
Solution Wb;25.0 Wb;75.0 A; 0.3 DCC I
turns5000 turns;1000 DC NN
C
CCC I
NL
0.375.01000
C L
H 250C L
2CCC 2
1 ILU
20.325021
J 1125C U
C
DD
INM
0.3
25.05000
H 417M
75
Given The induced emf in coil D is given by
Solution Wb;25.0 Wb;75.0 A; 0.3 DCC I
turns5000 turns;1000 DC NN A 0.30.30 s; 25.0 C dIdt
dtdI
M CD
25.0
0.3417
V 5004D
76
Next Chapter…CHAPTER 21 :Alternating current