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/


2/78



.

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



3/78



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



4/78



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

?



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

.



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 ()

)

,

)

,

)



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 ()

)

,

)

,

)



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



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

.



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 ,



11/78



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



12/78



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



13/78



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



14/78



3.5

,

.

3.6 ,

3.6.1

F 2

:

) ,

) I ,

) , 1, 2 3

.

) 1, 2 3 , .

().



15/78



F 3, .

15kΩ

35kΩ50V

R1

R2

F 3:

3.6.2

F 4

:

a) , ,

b) ,

c) I1, I2 I3 ,

F O ,

1



16/78



()

C

().

F 5.

F 5

:

3.6.3

F 6

2



17/78



:

) , ,

) ,

) C

)

F 7, .

F 7

:

,

3



18/78



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



19/78



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



20/78



3.7

D ∆

F 8

F D

F D



21/78



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



22/78



B , D :

) C 12Ω 10Ω

) 12Ω 10Ω

Figure 12

:

F 13 F 14

) C 10Ω 12Ω

1:

Ω

Ω

Ω

2



23/78



) 12Ω 10Ω

3.8

I. , . H,

1

F O ,

(1) :

A, O ,

I (1) :

I

.

3.9 .

) , , , ,

, ,

)

) , , , ,

,



24/78



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



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



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 :



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 :



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:



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



30/78

4 DC EAE CC AD E EE

AAA/E/A/E1014 6

3: A C

C: 4: .

∴∴∴∴

5: D



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

)

)



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



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



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



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



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



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



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,



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



40/78

4.0

:

F 29

F , :

F , :

:

1. (=).

2. .

3. ( = ).

4. ,η% = 50%.

:

η

7

30,

.

) 25Ω ) 50Ω ) 75Ω ) 100Ω ) 125Ω

F 30



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)

() ()



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.

,



43/78

5.0

:

,

,

C

9

D 2=5Ω 36

.

F 36

:

1: , . .

F 37



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



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



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

)

)



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://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


48/78



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.



49/78



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



50/78



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:



51/78



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]



52/78



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).



53/78



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.



54/78



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

+ -



55/78



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

+ -



56/78



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



57/78



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.



58/78



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]



59/78

UNIT 6 INDUCTOR AND INDUCTANCE

1MARLIANA/JKE/POLISAS/ET101-UNIT6


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’.



60/78



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



61/78



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



62/78



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



63/78



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.



64/78



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



65/78



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



66/78



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 .



67/78



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



68/78



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]



69/78

NI 7 MAGNEIC CICI, ELECOMAGNEIM AND ELECOMAGNEIC INDCION

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F 1() :



70/78



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71/78



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72/78



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73/78



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74/78



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75/78

NI 7 MAGNEI

MALIANA/JKE/POLIA/E101NI 7

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76/78

NI 7 MAGNEI

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77/78



:

( )

11

F 11

I ,

. B F ... ... . A

. ... E F

12 :

=

B, , , ,

, , , ,

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F 12

I θ0 ( 90

0

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θ θθ θ

1

A 300 4 /

1.25 . D

() ,() 20 Ω .

... ...

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I ... E = B =(1.25)(300/1000)(4) = 1.5



78/78


()

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


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Et101 - Electrical Technology