Download - Et101 - Electrical Technology
-
8/17/2019 Et101 - Electrical Technology
1/78
I 1 I AIAED IH BAIC EECICA AIIE
AIAA/JKE/IAE101/I 1 1
I ,
I , .
1.1 B I
E A A
K K
A
I
. A
. A . A 1.2.
1.2 I
D
100
1
(D) 101 10
102 100
K K 103 1 000
106 1 000 000
G G 109 1 000 000
1012 1 000 000 000 000
101 0.1
102 0.01
103 0.001
106 0.000 001
109 0.000 000 001
1012 0.000 000 000 001
:
.
56
.
4.5
.
150
.
3.3Ω Ω
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
2/78
I 1 I AIAED IH BAIC EECICA AIIE
AIAA/JKE/IAE101/I 1 2
.
1 = 1000
.
1 = 1000
.
1 = 1000
.
1Ω = 1000Ω
Ω
Ω
. ,
D ,
1 C, C
. = I
C,
C
2. ,
.
J, J
3. ,
.
E
.
,
4. E
F, ..
.
,
5. ,
.
, Ω
6. C, I E
.
A, A
7. C, G C .
,
8. E, E E .
E =
J, J
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
3/78
I 1 I AIAED IH BAIC EECICA AIIE
AIAA/JKE/IAE101/I 1 3
I 10A , .
, =I . I =10A = 4 60 = 240.
H
= 10 240 = 2400C
F :
.
25Ω .
20 Ω .
10Ω
a.
b.
c.
A ... 15 3A 6 . H
?
E = = I = (3)(15) = 45 = 6 60 = 360
H, E = = (45)(360)=16.2J
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
4/78
I 1 I AIAED IH BAIC EECICA AIIE
AIAA/JKE/IAE101/I 1 4
1.
:
.
1500 F = F
.
0.06 μF = F
.
10000 H = H
.
68 Ω = Ω
.
0.56 A =μA
2.
F :
.
100Ω
.
50 Ω
.
250Ω
3.
A 25 μ. ?
4.
A ... 220 5A. ?
5.
F 120 20 .
6.
I 15A 25 C?
7.
A 10A 15 . ?
8.
H 100A 80 C?
9.
A 1.5KΩ , 25 . D ( ) .
10.
A ... 10 5A 2 . H
?
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
5/78
2 CE AD BAE
A/F/E/AE101/ 2 1
G,
.
. .
. A . F
, 9 6 .
A () .
F 1, . A
. , . (.. .. ) . B
... ...
.
F 1
1.
C 1.
C
2.
2.
3.
C 3.
C .
1
.
C
.
A
.
.
E
.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
6/78
2 CE AD BAE
A/F/E/AE101/ 2 2
F 2
F 3
A ;
=
E = .. =
=
:
)
, E = E ()
)
, = (Ω)
)
, = +
)
A ;
=
E = ..
=
=
:
)
, E = E ()
)
,
)
,
)
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
7/78
2 CE AD BAE
A/F/E/AE101/ 2 3
F 4
. E 5Ω. E.. 1.5
0.2Ω. C .
.. , E = E = 3 1.5 = 4.5
= = 3 0.2 =
0.6Ω
, = +
= 0.6 + 5
= 5.6Ω
A ;
=
=
E = ..
=
=
:
)
,E = E ()
)
,
)
,
)
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
8/78
2 CE AD BAE
A/F/E/AE101/ 2 4
.. 1.5 0.2Ω
4Ω. .
.. , E = 1.5
= / = 0.2/2 =
0.1Ω
, = / +
= 0.1 + 4
= 4.1Ω
E . F
. E 1.5 0.6Ω. 5Ω
. C:
)
C
)
)
)
)
)
, = E
= 4(1.5) 1.16
= 4.84
@
= = 0.968 5 = 4.84
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
9/78
2 CE AD BAE
A/F/E/AE101/ 2 5
1.
A 10Ω 10
. E 1.5 0.2Ω. F .
2.
15 1.5 0.3Ω .
C , 5Ω .
3.
A 12
. . 1.5
0.2Ω. 4Ω. C .
4.
A . E 2.5 0.05Ω. A
15Ω. D:
.
.. .
.
5.
1.5 , 0.3Ω,
25Ω. D :
.
.
.
6.
, . E 1.45
0.04Ω. 5Ω ,
:
.
.
.
.
C
7.
. E 1.5
0.4Ω. F 5Ω .
8.
A 5 1.5 0.25Ω
. 1.5A,
.
9.
15 1.5 0.5Ω
. D :
.
C 15Ω
.
10.
20 1.45 0.5Ω 4 5 . 15Ω . C
.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
10/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 1
3
3.1
.
C
F
I C C B F
A FA
3.2
I
. , ,
.
.
( 1 = 6.24 1018
).
G, I , I .
Q ,
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
11/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 2
0.45 5 ?
:
Q , Q = I, :
I 20A , .
:
Q ,
3.3
V
A
Current
flow
F 1
F
(..) , ,
;
. .. ,
F 1
. C , ,
,
.
. , ,
.
.
A
. F 1
. .
A ..
.. . I F 1,
.. .
.
A . A ,
, , .
1
2
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
12/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 3
3.4
4 , :
)
R, R, , , , .. αααα .
)
R, R, , , . ..
αααα 1/
)
R α R α 1/ αααα /. B
.
ρ.,
ρρρρ
)
C 3
10 2. 0.03 10
6 Ω.
:
L,= 3 = 3000 ; , = 10 2 = 10 10
6
2 , ,ρ= 0.03 10
6
ρ
C , 2, , 25
0.30Ω . 0.03 106
Ω.
:
ρ
ρ
ρ
1
2
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
13/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 4
1
1. 5 600Ω .D:
) 8 ,
) 420Ω 960Ω,3.5
2. C 2
100 2. 0.0310
6Ω.
0.6Ω
3. 1.5 0.17 2 150Ω. D
. 0.017 10−6
Ω 0.017 Ω
4. D 1200 12
1.7 108
Ω. 0.18Ω
5. 2 2.5Ω. D () 7
() 6.25Ω.
() 8.75Ω () 5
6. 12 20Ω. D:
) 4
2.
)
32Ω.
() 5Ω () 0.625 2
7. F 800 20 2.
0.02 μΩ. 0.8Ω
8. C , 2, 100
2Ω. 0.03 106
Ω. 1.5 2
9. F 1 10
0.017 106
Ω. 0.216Ω
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
14/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 5
3.5
,
.
3.6 ,
3.6.1
F 2
:
) ,
) I ,
) , 1, 2 3
.
) 1, 2 3 , .
().
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
15/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 6
F 3, .
15kΩ
35kΩ50V
R1
R2
F 3:
3.6.2
F 4
:
a) , ,
b) ,
c) I1, I2 I3 ,
F O ,
1
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
16/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 7
()
C
().
F 5.
F 5
:
3.6.3
F 6
2
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
17/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 8
:
) , ,
) ,
) C
)
F 7, .
F 7
:
,
3
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
18/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 9
2
1. C
A B .
)
F 1
)
F 2
)
F 3
2. F 4,
.
F 4
3. F 5, 30Ω
.
F 5
4. F 6,
R, I, I1 I2.
F 6
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
19/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 10
5. F 7,
.
F 7
6. F 8, :
)
)
) C 6Ω
4Ω.
F 8
7. B 9, :
) , R
) , I
) C 16Ω
) A B
) P .
F 9
8. B 10, :
)
) 3Ω
D R
60.
) C I I
C D R
I 1.5A.
F 10
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
20/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 11
3.7
D ∆
F 8
F D
F D
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
21/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 12
C A B .
A
B
4Ω 2Ω
6Ω
3Ω 3Ω
F 9
:
F 10 F 11
1:
∆→Υ 4Ω, 2Ω 6Ω .
Ω
Ω
Ω
2:R ∆ Υ (F11).
3:
F ,R.
Ω
1
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
22/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 13
B , D :
) C 12Ω 10Ω
) 12Ω 10Ω
Figure 12
:
F 13 F 14
) C 10Ω 12Ω
1:
Ω
Ω
Ω
2
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
23/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 14
) 12Ω 10Ω
3.8
I. , . H,
1
F O ,
(1) :
A, O ,
I (1) :
I
.
3.9 .
) , , , ,
, ,
)
) , , , ,
,
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
24/78
NI 3 AN INRODCION O ELECRIC CIRCI
MARLIANA/JKE/POLIA/E101NI3 15
3
1. C
A B 11.
F 11
2. ,
:
) , R
) C, I
) C 15Ω
F 12
3. B 13,
15Ω
F 13
4. B F 14, ,
6Ω .
F 14
5. B F 15,
,
RL.
F 15
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
25/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 1
4
1.0
()
, Σ , 1:
() , 2
( )
.
F 2
1.1
A
.
:
1. A .
2. A .
3. .
∑ = ∑
1 + 2+ 3 = 4 + 5
1 + 2 + ( 3 ) + ( 4 ) + (5 ) = 0
∑ = 0
E = 1 + 2
E = (1 + 2 )
E + ( 1 ) + ( 2) = 0
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
26/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 2
1
F .
F 3
:
1: A .
F 4
2: A .
1:
1
2:
2 3: .
1 2
:
F C :
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
27/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 3
2
F .
F 5
:
1: A .
F 6
2: A .
1:
1
2:
2
3: .
1 2
:
F C :
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
28/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 4
1.2
A C
.
:
1. D .
2. A .
3. A C
4. .
5. D
.
3
F .
F 7
:
1: D (F 8).
1 () , C
2: A (F 8).
F 8
3: A C
C:
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
29/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 5
4: .
∴∴∴∴
5: D
4
F .
F 9
:
1: D (F 10).
1 () , C
2: A (F 10).
F 10
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
30/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 6
3: A C
C: 4: .
∴∴∴∴
5: D
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
31/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 7
1
F A
A.
)
)
)
R1
R2
R3
V1
V2
4kΩ
2kΩ
3kΩ
30V
25V
)
)
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
32/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 8
2.0
:
"
".
.
F 11
:
1. ().
2. F ()
.
3. D 11 .
F .
5
F 30Ω F 12.
F 12
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
33/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 9
:
1: ().
F 13
Ω
D
2: F ()
F 14
Ω
3: D
F 15
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
34/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 10
6
F 4.
60Ω
30Ω 90Ω 25Ω300mA
IsR1
R2
R3 R4
F 16
:
1 : ().
F 17
Ω
CD, 2
2: F () , .
F 18
Ω
3: D
RTH
RLVTH
4.5V
45Ω
25Ω
IL
F 19
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
35/78
2
1. 1,
12Ω
.
F 1
2. F 15Ω
2
.
F 2
3. C
.
F 3
4.
5Ω 4.
F 4
5. C 30Ω
5
.
F 5
6. 6,
50Ω .
F 6
7. ,
=10Ω.
F 7
8. ,
=10Ω.
F 8
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
36/78
3.0
:
"
".
F 20
:1. . F .
2. F .
3. D 20 . F
.
7
F 30Ω F 21.
F 21
1: . F .
F 22
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
37/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 13
2: F .
F 23
3: D . F
.
F 24
CD,
6
F 4.
F 25
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
38/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 14
:
1: . F .
60Ω
30Ω 90Ω 25Ω
300mA
IsR1
R2
R3 R4IN
F 26
C 90Ω 0A, .
2: F .
60Ω
30Ω 90Ω
R1
R2
R3RN
F 27
3: D . F
.
F 28
CD,
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
39/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 15
3
1. 1,
12Ω
.
F 1
2. F 15Ω
2
.
F 2
3. C
.
F 3
4.
5Ω 4.
F 4
5. C 30Ω
5
.
F 5
6. 6,
50Ω .
F 6
7. ,
=10Ω.
F 7
8. ,
=10Ω.
F 8
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
40/78
4.0
:
F 29
F , :
F , :
:
1. (=).
2. .
3. ( = ).
4. ,η% = 50%.
:
η
7
30,
.
) 25Ω ) 50Ω ) 75Ω ) 100Ω ) 125Ω
F 30
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
41/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 17
:
)
)
)
)
)
= %ηηηη
75Ω 0 0.133A 10 0 0% 0
75Ω 25Ω 0.1A 10 2.5 25% 0.25
75Ω 50Ω 0.08A 10 4 40% 0.32
75Ω 75Ω 0.067A 10 5.0 50% 0.336
75Ω 100Ω 0.057A 10 5.7 57% 0.325
75Ω 125Ω 0.05A 10 6.5 65% 0.312
F 31
0, 0
25, 0.25
50, 0.32 75, 0.336 100, 0.325 125, 0.312
00.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 20 40 60 80 100 120 140
( )
(Ω)
() ()
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
42/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 18
8
F 32,
.
F 32
:
().
F 33
Ω D
F ()
F 34
Ω
D
F 35
=.
, 10Ω.
,
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
43/78
5.0
:
,
,
C
9
D 2=5Ω 36
.
F 36
:
1: , . .
F 37
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
44/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 20
2: , . .
F 38
CD
3: 2=5Ω.
1A 6A
10
F 39.
F 39
:
1: 1 , 2
F 40
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
45/78
2: 1 , 2
F 41
3:
1 ⇒ 1=0.429A 1=0.571A
2 ⇒ 2=0.286A 2=0.714A
3 ⇒ 3=0.143A 3=0.143A
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
46/78
4 DC EAE CC AD E EE
AAA/E/A/E1014 22
4
F .
)
)
)
R1
R2
R3
V1
V2
4kΩ
2kΩ
3kΩ
30V
25V
)
)
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
47/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 1
UNIT 5 CAPACITORS AND CAPACITANCE
5.1 Capacitor and Capacitance
5.1.1 Associated Quantities
- A capacitor is an electrical device that is used to store electrical energy.
- The unit of capacitance is Farad. The symbol of capacitance is C.
- Capacitance is defined to be the amount of charge Q stored in between the two
plates for a potential difference or voltage V existing across the plates.
In other words:
- A capacitor has capacitance of one Farad when current charging of one Ampere
flows in one second . This process causing a transferring of one volt in plates
potential.
- The Farad unit is too large for practical as charge ratio to its potential difference.
uses. Thus microfarad (μF) , nanofarad (nF) or Pico farad (pF) is used as a suitable
unit for capacitor:-
Microfarad (µF) 1µF = 1/1,000,000F = 10-6F
Nanofarad (nF) 1nF = 1/1,000,000,000 = 10-9F
Microfarad (pF) 1pF = 1/1,000,000,000,000 = 10-12F
5.1.2 Types of capacitors
Fixed
Unpolarised
Mica
Ceramic
Film
Air-gap
Paper
PolarizedAluminum
Tantalum
Variable Trimmer
http://modul2poli.blogspot.com/
http://images.search.yahoo.com/images/view;_ylt=A2KJkIbi5mdP_TkA5RSJzbkF;_ylu=X3oDMTBlMTQ4cGxyBHNlYwNzcgRzbGsDaW1n?back=http%3A%2F%2Fimages.search.yahoo.com%2Fsearch%2Fimages%3Fp%3Daluminium%2Bcapacitor%26ei%3DUTF-8%26fr%3Dyfp-t-521%26tab%3Dorganic%26ri%3D28&w=360&h=360&imgurl=www.germes-online.com%2Fdirect%2Fdbimage%2F50157825%2FAluminum_Electrolytic_Capacitor.jpg&rurl=http%3A%2F%2Fwww.germes-online.com%2Fcatalog%2F87%2F439%2Fpage4%2F&size=31.9+KB&name=Aluminum+Electrolytic+Capacitor&p=aluminium+capacitor&oid=b3a31563a944cba4779dbe6fd19e0424&fr2=&fr=yfp-t-521&tt=Aluminum%2BElectrolytic%2BCapacitor&b=0&ni=48&no=28&tab=organic&ts=&sigr=11i68dh5d&sigb=13asncshf&sigi=12h5adrra&.crumb=o6F.j.RvaGDhttp://images.search.yahoo.com/images/view;_ylt=A2KJkeua52dPNxYAo_GJzbkF;_ylu=X3oDMTBlMTQ4cGxyBHNlYwNzcgRzbGsDaW1n?back=http%3A%2F%2Fimages.search.yahoo.com%2Fsearch%2Fimages%3Fp%3Dmica%2Bcapacitor%26n%3D30%26ei%3Dutf-8%26fr%3Dyfp-t-521%26tab%3Dorganic%26ri%3D2&w=300&h=300&imgurl=theonlinetutorials.com%2Ftheonlinetutorials_files%2F2011%2F04%2FMica-capacitor.jpg&rurl=http%3A%2F%2Ftheonlinetutorials.com%2Fcapacitors.html&size=34.4+KB&name=Mica+capacitor&p=mica+capacitor&oid=0af65e5b6487c0babacfdf9cd0bb4c98&fr2=&fr=yfp-t-521&tt=Mica%2Bcapacitor&b=0&ni=60&no=2&tab=organic&ts=&sigr=11d6umt06&sigb=139d69uqd&sigi=12ap7qdjv&.crumb=o6F.j.RvaGD
-
8/17/2019 Et101 - Electrical Technology
48/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 2
5.1.3 Capacitor Construction
Figure 1: Capacitor construction
In its most elementary state a capacitor consists of two metal plates separated by a
certain distance d, in between the plates lies a dielectric material with dielectric constant =εoε,
where εo is the dielectric of air.
The dielectric material allows for charge to accumulate between the capacitor plates.
Air (actually vacuum) has the lowest dielectric value of εo = 8.854 x 10-12 Farads/meter. All other
materials have higher dielectric values, since they are higher in density and can therefore
accumulate more charge.
The Physical meaning of capacitance can be seen by relating it to the physical
characteristics of the two plates, so that, the capacitance is related to the dielectric of the
material in between the plates, the square area of a plate and the distance between the plates
by the formula:
Clearly, the larger the area of the plate the more charge can be accumulated and hence
the larger the capacitance. Also, note that as the distance d increases the Capacitance
decreases since the charge cannot be contained as 'densely' as before.
By applying a voltage to a capacitor and measuring the charge on the plates, the ratio of
the charge Q to the voltage V will give the capacitance value of the capacitor and is therefore
given as: C = Q/V this equation can also be re-arranged to give the more familiar formula for the
quantity of charge on the plates as: Q = C x V.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
49/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 3
5.2 Capacitor equivalents circuit
5.2.1 Capacitor connected in series
C3
e2
C2C1
e1 e3
E
Figure 2: Capacitors in series
Total voltage
VT = e1 + e2 + e3
Since then
Where CT is the total equivalent circuit
capacitance
It follows that for n series-connected capacitors:
5.2.3 Capacitor connected in parallel
C2C1
E
C3
Figure 3: Capacitor in parallel
Total charge,
Q T = Q 1 + Q 2 + Q 3
CTE = C1V1 + C2 V2 + C3 V3
Total voltage
ET = e1 = e2 = e3
Total equivalent circuit capacitance
CT = C1 + C2 + C3
It follows that for n parallel connected
capacitors:
CT = C1 + C2 + C3 +……+ Cn
5.2.3 Capacitor connected in series-parallel
C2
C1E C3
Figure 4: Capacitor in series-parallel
Total equivalent circuit capacitance
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
50/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 4
Example 1
Calculate the equivalent capacitance of two capacitors of 3μF and 6μF connected:
(a) in parallel (b) in series.
Solution:
(a)
In parallel, equivalent capacitances:
(b) In series, equivalent capacitance :
Example 2
Find the capacitance to be connected in series with a 10μF capacitor for the equivalent capacitance tobe 6μF.
Solution:
, ,
For two capacitance in series:
Example 3
Find the total capacitance of the circuit:
C3C1E
C4
C2
3µF1µF 4µF
2µF
Solution:
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
51/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 5
TUTORIAL 1
1. Capacitors of 15μF and 10μF are connected (a) in parallel and (b) in series. Determine the equivalent
capacitance in each case.
2. Find the capacitance to be connected in series with a 25μF capacitor for the equivalent capacitance
to be 10µF.
3. Find the capacitance to be connected in parallel with a 25μF capacitor for the equivalent
capacitance to be 10µF.
4. Find the total capacitor the circuit.
a)
C2C4E
C1
100µF100µF 100µF
C5 50µF C3 200µF [40µF]
b)
C1
30µF
C2
60µF
C3
60µF
C4
30µF
C5
20µF
C6
40µF[11.14µF]
c)
C1
30µF
C2
30µF
C3
30µFC4
30µFC5
30µF
C6
30µF
[1.2µF]
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
52/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 6
5.3 Circuit with capacitive load
Figure 5
Mathematically, the capacitance of the device relates the voltage difference between
the plates and the charge accumulation associated with this voltage:
q(t)=CV(t) equation (1)
Capacitors which obey the relationship of equation (1) are linear capacitors, since the
potential difference between the conductive surfaces is linearly related to the charge on the
surfaces. Note that the charges on the right and left plate of the capacitor in Figure 5 are equal
and opposite. Thus, if we increase the charge on one plate, the charge on the other plate must
decrease by the same amount. This is consistent with our previous assumption electrical circuit
elements cannot accumulate charge, and current entering one terminal of a capacitor must
leave the other terminal of the capacitor.
So, current is defined as the time rate of change of charge,
5.3.2 Elements related to capacitance
a. Electric field:
Area that surrounds the electric charge or charges system where the
increasing and decreasing of electric force exists.
b. Line of electric force: A line of electric force is known as line or curve that pointed out from
positive charge (+) to negative charge (-) in a magnetic field.
c. Electric flux:
Known as amount of electric force line pointed out from positive charge (+)
to negative charge (-) in a magnetic field. Flux symbol is Ψ(phi).
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
53/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 7
d. Electric flux density (D):
Electric flux density is a measurement of electric flux that pass through a
unit of plate’s area with a coincide angle, that is an area of 1 meter².
The ratio between the charge of the capacitor and capacitor plates.
The symbol used is D. Based on Figure 1, if the area of capacitor is A, then
the flux density is given as:-
where Q= charge(Coulomb), A = surface area of capacitor
Figure6: Area and distance of capacitor plates
e. Electric field strength:
When two metal plates are charged and separated in a certain distance, a
potential difference will exists between the plates.
A force was also generated, known as electric force and the symbol is E. The
magnetic strength depends on the potential difference and distance
between plates.
Where; V = potential difference d = thickness of dielectric
f.
Dielectric:
Insulator that is used between the two plates of a capacitance is known as
dielectric.
Electric field exists in the dielectric and the flux density depends on the types
of insulator used.
g.
Absolute permittivity (ε):
Permittivity is a capacitance or ability to store energy of a capacitor.
A force was also generated, known as electric force and the symbol. It
depends on the dielectric substance, and the symbol is ε.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
54/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 8
5.3.3 Factor that effecting capacitance
Based on Figure 7, the factor that effecting capacitance is:
Figure 7
a. Capacitance between two plates proportional to the surface area
b. Capacitance between two plates inversely proportional to the thickness of
dielectric
c. Increasing the dielectric constant of the material between the plates
5.4 Process charging and discharging in a capacitor
5.4.1 Charging process in capacitor
Figure 8: Capacitor circuit for charging process
In initial state, a capacitor is uncharged (Vc = 0V). When a capacitor start
charged, maximum current will be flowing (i = Imax). The current would be decreased by
exponent, while voltage will be rising by exponent also. This state will continue until full
state (steady) achieved. In this full state, current had decreased to zero value, while
voltage increased until maximum value. The capacitor is said in fully charge.
A
VR VC
+ E -
R C
ib
a S
+ -
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
55/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 9
Figure 9 show both curves current and voltage for capacitor during charging
process, in x-axis (t, time). The current and voltage curve may be represented by
exponent equations respectively.
Figure 9: Current and voltage curve in capacitor during charging process
Time constant, = CR
- The times taken for voltage achieve value of 0.632Vmax and current achieve value of
0.371Imax
Initial current,
5.4.2 Discharging process in capacitor
Figure 10: Capacitor circuit for discharging process
When capacitor fully charge and then switch being transformed to ‘b’, discharge
process for capacitor will happen. The time taken to recharge and fully discharge is
5 =CxR. Figure 11 show the curve for discharging process in capacitor.
v c, i
t
IMax
VMax A
v c
= 0.632 VMax
i
= 0.371 IMax
= CR
Voltage through capacitor:
v c = Vmax (1 – e –t/
)
vc = Vmaxi = 0
Current flow:i = Imax (e
–t/)
A
VR VC
+ E -
R C
ib
a S
+ -
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
56/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 10
Figure 11: Current and voltage curve in capacitor during discharging process
5.4.3 Energy stored in a capacitor
During charging process through capacitor, it will get energy. Energy is kept in
static form. The voltage in capacitor will increase from 0 volt to E volt.
Example 1
One capacitor 0.326µF connected in series with 680kΩ and dc voltage 120V. Determine:
i. Time constant
ii. Initial current charge
iii. Current through capacitor, 100ms after charge to the source.
iv. Energy stored in capacitor.
Solution:
i.
ii.
iii.
iv.
t, time
Voltage through capacitor:
v c = Vmax ( e –t/
)
Vmax
v, i
Discharging current flow:
i =-Imax (e –t/
)Imax
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
57/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 11
Example 2
When switch are connected, calculate:
i. Time constant
ii.
Initial current charge
iii. Time taken for voltage through capacitor
increase to 160V.
iv. Current and potential difference through
capacitor, 4 second after charge to the
source.
v. Energy stored in capacitor.
Solution:
i. ii.
iii.
vc = 160V
47s
iv.
Example 3
When switch are connected, calculate:
i. Initial current charge
ii. Initial potential difference through
capacitor.
iii. Time constant
iv. Time taken for capacitor fully charges.
Solution:
i.
ii.
iii.
iv.
200kΩ 20µF
150V
R C
S
+ -
100kΩ 150µF
220V
R CS
+ -
iv.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
58/78
UNIT 5 CAPACITOR AND CAPACITANCE
MARLIANA/JKE/POLISAS/ET101-UNIT5 12
TUTORIAL 2
1. A 20µF capacitor is connected in series with a 50 kΩ resistor and the circuit is connected to a 20
V, d.c. supply. Determine:
a) The initial value of the current flowing,
b)
The time constant of the circuit,
c) The value of the current one second after connection,
d) The value of the capacitor voltage two seconds after connection,
e) The time after connection when the resistor voltage is 15 v
[0.4mA,1s,0.147mA,17.3V,0.288s]
2. A circuit consists of a resistor connected in series with a 0.5µF capacitor and has a time
constant of 12 ms. Determine:
(a)
The value of the resistor
(b)
The capacitor voltage 7 ms after connecting the circuit to a 10 V supply
*24kΩ,4.42V]
3. An 12µF capacitor is connected in series to a 0.5MΩ resistor across the dc voltage supply of
240V. Determine:
(a) Time constant
(b)
Initial charging current
(c)
Time for capacitor voltage increase to 150V
(d) The current flowing through the capacitor after 4 seconds
(e) Energy stored in the capacitor when it is fully charged
(f) Sketch the current and voltage (IV) curve to show the process of charging the capacitor.
[6s,0.48mA,5.88s,0.246mA,0.346J]
4.
A capacitor is charged to 100 V and then discharged through a 50 kΩ resistor. If the timeconstant of the circuit is 0.8 s, determine:
(a) The value of the capacitor,
(b) The time for the capacitor voltage to fall to 20 v,
(c) The current flowing when the capacitor has been discharging for 0.5 s
(d) The voltage drop across the resistor when the capacitor has been discharging for one
second.
[16µF,1.29s,1.07mA,28.7V]
5. A 0.1µF capacitor is charged to 200 V before being connected across a 4 kΩ resistor. Determine:
(a)
The initial discharge current
(b) The time constant of the circuit
[50mA,0.4s]
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
59/78
UNIT 6 INDUCTOR AND INDUCTANCE
1MARLIANA/JKE/POLISAS/ET101-UNIT6
UNIT 6 INDUCTOR AND INDUCTANCE
6.1 Inductor and inductance
6.1.1 Associated quantities
- Inductor, also called a choke, is another passive type electrical component designed
to take advantage of this relationship by producing a much stronger magnetic field
than one that would be produced by a simple coil.
- Symbol of inductance is L.
- Unit of inductance is Henry.
- Inductance – the property of an electric circuit by which an electromotive force is
induced in it as the result of changing magnetic flux.
- Electromagnet – temporary magnet production due to flow of electric current.
- Electromagnetic induction - production process electric form magnet.
6.1.2 Types of inductor
Fixed
Air core
Iron core
Ferrite core
Variable Core loss
6.1.3 Construction of inductor
An inductor is usually constructed as a coil of conducting material, typicallycopper wire, wrapped around a core either of air or ferrous material.
Core materials with higher permeability than air confine the magnetic field
closely to the inductor, thereby increasing the inductance. Inductors come in many
shapes. Most are constructed as enamel coated wire wrapped around a ferrite with wire
exposed on the outside, while some enclose the wire completely in ferrite and are called
‘shielded’.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
60/78
UNIT 6 INDUCTOR AND INDUCTANCE
2MARLIANA/JKE/POLISAS/ET101-UNIT6
Some inductors have an adjustable core, which enables changing of the
inductance. Small inductors can be etched directly onto a printed board by laying out
the trace in a spiral pattern.
Figure 1
6.2 Inductance equivalents circuit for series and parallel connection
6.2.1 Inductors connected in series
Figure 2: Inductor in series
Total voltage,
ET = e1 + e2
Total inductance,
LT = L1 + L2
It follows that for n series connected
inductors
LT = L1 + L2 +…….+ Ln
Current, I = IL1 = IL2
6.2.2 Inductors connected in parallel
Figure 3: Inductor in parallel
Total voltage,
ET = e1 = e2Total current,
IT = I1 + I2
Total inductance
It follows that for n parallel connected
inductors
e2
IT
L1 L2
ET
e1
I1 I2
L1
L2 ET
e1
e2
I
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
61/78
UNIT 6 INDUCTOR AND INDUCTANCE
3MARLIANA/JKE/POLISAS/ET101-UNIT6
6.2.3 Inductors connected in series-parallel
Figure 5: Inductor in series-
parallel connection
Total current,
IT = I1 +I2
Total inductance,LT = L1 + L2 // L3
Example 1
Calculate the equivalent inductance of two inductors of 3H and 5H connected:
(a) in series (b) in parallel.
Solution:
a) Total inductance in series, b) Total inductance in parallel,
Example 2
Find the inductance to be connected in parallel with a 10H capacitor for the equivalent capacitance to
be 6H.
Solution:
, ,For two capacitance in series:
Example 3
Find the total inductance for the circuit below Total inductance,LT = L1 + L2 // L3
L2 L3 ET
IT
I1 I2
L1
5H
2H 3HL2 L3 ET
IT
I1 I2
L1
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
62/78
UNIT 6 INDUCTOR AND INDUCTANCE
4MARLIANA/JKE/POLISAS/ET101-UNIT6
TUTORIAL 1
Find the total inductance of the circuit.
1.
[Ans: 0.788H]
2.
[Ans: 8.33mH]
3.
[Ans: 3.54H]
4.
[Ans: 20mH]
3mH
4mH
6mH
2mH 5mHA B
6H
2H
4H
2HA B
2H 5H
3H
5H 3H
12mH
40mH
5.2mH
12mH
4mH
12mH
2mH
4mH
ET
2H
3H 4H1H
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
63/78
UNIT 6 INDUCTOR AND INDUCTANCE
5MARLIANA/JKE/POLISAS/ET101-UNIT6
6.3 Circuit with inductive load
6.3.1 Electromagnetic induction
When a conductor is moved across a magnetic field so as to cut through the lines of
force (or flux, an electromotive force (e.m.f) is produced in the conductor. If the conductor
forms part of a closed circuit then the e.m.f produced causes an electric current to flow round
the circuit. Hence an e.m.f is induced in the conductor as a result of its movement across the
magnetic field. This effect is known a ‘electromagnetic induction’.
6.3.2 Faraday’s Law
Faraday’s laws of electromagnetic induction state:
i) An induced e.m.f is setup whenever the magnetic field linking that circuit
changes
ii) The magnitude of the induced e.m.f in any circuit is proportional to the rate of
change of the magnetic flux linking the circuit.
6.3.3 Mathematical relationship between the induced e.m.f and the network
Faraday noted that the e.m.f induced in a loop is proportional to the rate of change of
magnetic flux through it:
Where; e is the electromotive force induced (in volts)
N is the number of turns of the coil
d Φ is the change of flux in Weber, Wb
dt is the time taken for the flux to change in seconds.
*Notice the negative sign is the induced current will now produce an induced magnetic
filed. The direction of that magnetic field will be opposite to the direction the flux is
changing.
6.3.4 Self-inductance and the induced e.m.f
Inductance is the name given to the property of a circuit whereby there is an e.m.f
induced into the circuit by change of flux linkages produced by a current change.
When the e.m.f is induced in the same circuit as that in which the current is changing,
the property is called self-inductance, L.
Induced e.m.f is the product of self-inductance and the rate of change in current
Where; e is induced e.m.f (in volts)
L is self-inductance in Henry
di is the change of current in Amperes
dt is the time taken for the current to change in seconds.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
64/78
UNIT 6 INDUCTOR AND INDUCTANCE
6MARLIANA/JKE/POLISAS/ET101-UNIT6
6.3.5 The mathematical of self-inductance
Where; N is number of turns of coil
L is length of coil
A is surface area
µ is permeability
6.3.6 The factors that influence inductance
A component called an inductor is used when the property of inductance is required in a
circuit. The basic form of an inductor is simply a coil of wire.
Factors which affect the inductance of an inductor include:
i) The number of turns of wire (N) – more turns the higher the inductance
ii) The cross-sectional area of the coil of wire (A) – the greater the cross-sectional area
the higher the inductanceiii) The presence of magnetic core - when the coil is wound on an iron core, the same
current sets up a more concentrated magnetic field and the inductance is increased
iv) The way turns are arranged – a short tick coil of wire has a higher inductance than
the along thin one.
6.4 Rise and decay of current goes through an inductor in the dc circuit
6.4.1 Rise and decay of current
Figure 5: Inductor circuit
Refer to Figure 5, when switch in ‘a’ position, inductor connected to DC supply.
The current had not achieved maximum value immediately. The current are going to
reach maximum value in a period of time that certain caused by production e.m.f
induced by inductor which always against the supply voltage. In other words, the
currant of the circuit is rise delayed.
When switch is being transformed to position ‘b’, inductor circuit had short
circuit (no supply voltage). The current is not decrease continue to zero but take a time
that certain from maximum value until zero value. Refer to figure 6 which is shown the
exponential graph changing of current in inductor circuit.
V
b
a
LR
S
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
65/78
UNIT 6 INDUCTOR AND INDUCTANCE
7MARLIANA/JKE/POLISAS/ET101-UNIT6
Figure 6: Graph changing of current in inductor circuit
6.4.2 Time constant
Time constant, defines as time for current achieve maximum (IM) if this maintain the
early promotion rate current.
i) Time constant at rise of current
Practically, the current did not rise by regular. By graphically, it achieves 63.2% from
maximum value (point ‘B’ in figure 7) in time constant. In other words, time constant,
also defines as time for current of inductor achieve 63.2% from the maximum value.
Figure 7: Graph rise of current through an inductor
From RL circuit, time constant, , given by equation:
IM
)1( L Rt
M e I i
i
t
Rise of current
IM
L
Rt
M e I i
i
t
Decay of current
)1( L Rt
M e I i
IM
63.2%
i
t
5
A B
C DFrom the graph:
Current will be rising from
minimum value (0) by
exponent headed for
maximum value, IM (steady
state).
Time for value of i achieve
63.2% from maximum value
is time constant, .
Time for value of I achieve
maximum value is 5
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
66/78
UNIT 6 INDUCTOR AND INDUCTANCE
8MARLIANA/JKE/POLISAS/ET101-UNIT6
ii) Time constant at decay of current
In decay of current through an inductor, a method to find values of time constant same
as in rise of current through an inductor. The differences are value of current decay
from maximum value (IM) to minimum (0), and value 63.2% replaced with 36.8% which is
100% - 63.2%. Figure 8 shown clear pictures for decay of current in inductor.
Figure 8: Graph decay of current through an inductor
6.4.3 Energy stored in an inductor
An inductor possesses an ability to store energy. The energy stored, W in the magnetic
field of an inductor is given by:
Example 1
One inductor 0.5H connected in series with resistor 20Ω and dc voltage 120V. Determine:
i) Time constant
ii) Current at time 0.025s
iii) Energy stored in inductor
Solution
i) Time constant
ii)
iii) Energy store,
L
Rt
M e I i
IM
36.8%
0
i
t
5
From the graph:
Current will reduce from
maximum value (IM) by
exponentially until minimum
value (0).
Time for i to reach 36.8% from
maximum value (reducing of
63.2% from origin value, IM) is
time constant:
Time for i to reach final value
(zero) is 5 .
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
67/78
UNIT 6 INDUCTOR AND INDUCTANCE
9MARLIANA/JKE/POLISAS/ET101-UNIT6
Example 2
When switch connected to ‘a’,
calculate:
i) Time constant
ii) Time taken for current achieve
maximum value
iii) Maximum current if the current is
2.5A in 0.38s.
Solution
i) Time constant,
ii)
iii)
Example 3
One circuit has resistor 40Ω connected in series with inductor 15H and dc voltage 220V.
Calculate:
i) Time constant
ii) Current at time (i)
iii) Current at time 0.05s
iv) Energy stored in inductor
Solution
i) Time constant
ii)
iii) t = 0.05s
iv)
b
a
L=7.5HR=10Ω
S
100V
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
68/78
UNIT 6 INDUCTOR AND INDUCTANCE
10MARLIANA/JKE/POLISAS/ET101-UNIT6
TUTORIAL 2
1. A coil of inductance 0.04 H and resistance 10Ω is connected to a 120 V, d.c. supply. Determine
(a) the final value of current
(b) the time constant of the circuit
(c)
the value of current after a time equal to the time constant from the instant the supply
voltage is connected.
[12A,4ms,7.58A]
2. The winding of an electromagnet has an inductance of 3H and a resistance of 15Ω. When it is
connected to a 120 V d.c. supply, calculate:
(a) the steady state value of current flowing in the winding
(b) the time constant of the circuit
(c)
the value of the induced e.m.f. after 0.1s
(d)
the time for the current to rise to 85% of its final value
(e)
the value of the current after 0.3 s
[8A,0.2s,72.78V,0.379s,6.215A]
3. A coil has an inductance of 1.2H and a resistance of 40Ω and is connected to a 200 V, d.c. supply.
Determine the approximate value of the current flowing 60 ms after connecting the coil to the
supply. [4.3 A]
4. A 25 V d.c. supply is connected to a coil of inductance 1H and resistance 5Ω. Determine the
approximate value of the current flowing 100 ms after being connected to the supply. [2 A]
5. The field winding of a 200 V d.c. machine has a resistance of 20Ω and an inductance of 500mH.
Calculate:
(a)
the time constant of the field winding(b) the value of current flow one time constant after being connected to the supply
(c) the current flowing 50 ms after the supply has been switched on.
[(a) 25 ms (b) 6.32 A (c) 8.65 A]
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
69/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
MALIANA/JKE/POLIA/E101NI 7 1
7 A , A A
A
7.1
7.1.1
() . A
.
A ( , )
. A
.
, , , .
. ,
.
. I F 1(),
, . L
. ,
. I F 1(), (.. ),
, .. ,
.
F 1() :
F 1() :
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
70/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
MALIANA/JKE/POLIA/E101NI 7 2
7.1.2 M
. A
, , . M F
,
,
.
:
.
C
.
I
7.1.3 C / :
.
F
.
D
.
H .
.
H
F 2: M
7.1.4 . M , F
M ()
.
, F = NI A
E ()
D
E ()
C
E ()
E ()
E ()
C
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
71/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
MALIANA/JKE/POLIA/E101NI 7 3
. ,
.
1/H A/
. M , H
M ( )
A l .
. M
M (
) . Φ.
, .
M
:
B.
,.
. P BH
P
.
: μ (μ) / (/A)
μμ= B / H
μ =
μ = , 4π 107
μ = , μ = B () / B ()
H, μ = B / B
•••• μ ( μ μ )
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
72/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
MALIANA/JKE/POLIA/E101NI 7 4
F 3: BH
7.2
F , :
= 1 + 2 +..+
( )
1
A 6 12
2 0.5 2. A 200 6
0.4 A . D 2 , 750.
:
F 6 :
,
F 2
F , B 2
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
73/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
MALIANA/JKE/POLIA/E101NI 7 5
,
7.3
.. , E () , F (A)
C, I (A) F,Φ ()
, (Ω) , (H1
)
ρ
7.4
B
H.
H, B
F 4, . F 4
.
F F 4, O , O
, PP .
H .
.
. ,
, .
F 4: H
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
74/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
MALIANA/JKE/POLIA/E101NI 7 6
7.5
7.5.1 M
L
, , F 5().
I ,
, . B
F 5().
F 5() F 5()
I , .
. .
. I
, ,
.
M ,
(F 6()).
. F 6() ,
( + ). M, F 6()
, ( •,
).
C
M
C
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
75/78
NI 7 MAGNEI
MALIANA/JKE/POLIA/E101NI 7
F 6()
F 6:
D
:
.
C
.
C
.
C
A
7(),
.
F 7(): M
C CICI, ELECOMAGNEIM AND ELECOMAG
F 6() F 6()
P
.
.
. C
, , F 7(
. I , ,
F 7(): M
NEIC INDCION
7
,
)
F
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
76/78
NI 7 MAGNEI
MALIANA/JKE/POLIA/E101NI 7
)
)
)
7.6
7.6.1 L
.
.
.
:
C CICI, ELECOMAGNEIM AND ELECOMAG
:
F 8
.
F 9
F 10
:
.
NEIC INDCION
8
.
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
77/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
MALIANA/JKE/POLIA/E101NI 7 9
:
( )
11
F 11
I ,
. B F ... ... . A
. ... E F
12 :
=
B, , , ,
, , , ,
.
F 12
I θ0 ( 90
0
)
θ θθ θ
1
A 300 4 /
1.25 . D
() ,() 20 Ω .
... ...
.
I ... E = B =(1.25)(300/1000)(4) = 1.5
http://modul2poli.blogspot.com/
-
8/17/2019 Et101 - Electrical Technology
78/78
NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION
()
I 1.5
.
() F O , I = E/ =1.5/20 = 0.075 A 75 A
2A 75 0.6 ... 9
? A ,
.
I ... E = B , = E/B
H = 9/(0.6)(75103
)=(9 103)/(0.6 75)= 200 /
3
A 15 / () 90, () 60 () 30
2 . I
5 μ, ... .
= 15 /; , = 2 = 0.02 ;
A = 2 2 2 = 4 104 2, Φ = 5 106
() E90 = B 90 =(Φ/A) ( 5 106)(0.02)(15)(1)/(4104) = 3.75
() E60 = B 60 = E90 60 = 3.75 60 = 3.25
() E30 = B 30 = E90 30 = 3.75 30 = 1.875
http://modul2poli.blogspot.com/