A device that can hold or store a reasonable amount of electric charge

It is made of two parallel plates separated by insulator( dielectric) or air

It has two leads that can be connected directly to other components

The ability to store amount of charges for a particular values of voltage is called capacity or capacitance and measured in Farad (F) . 1 Farad is defined as 1 Coulomb (C ) of electricity capacity at 1 V potential difference

picofarads (pF) = 10-12 F

nanofarads (nF) = 10-9 F

microfarads (F) = 10-6 F

millifarads (mF) = 10-3 F

Farad (F) = 100 F

dA

Insulator(dielectric)

d

Insulator(dielectric)

Cross section

C39nF

fixed capacitor

+

C100pF

Polarized capacitor

C470pF

Variable capacitor

Positive terminal of a battery repels the positive charges (positive ions) towards plate A and attracts negative charges (electrons) towards it – Plate A then becomes positive.

Negative terminal of a battery repels negative charges (electron) towards plate B and attracts positive charges (positive ion) towards it – Plate B then becomes negative charge.

Positive charges accumulate in plate A reduces more positive charges from the battery terminal to enter it and at the same time negative charges in also reduces more negative charges from the negative terminal of the battery.– the current flow from the battery the plate will be reduced.

R

V

Capacitor

Plate A Plate B

•Charges develop the potential different between the plates and increase with the increase of charges.•When the potential different is same as the voltage of the battery, the entering of charges stop. •Charges are stored in the capacitor plates after the connection to the battery is disconnected.•Ratio of Q:V is constant and is called as capacitance, thus

C = Q/V or Q = CV

R

V

Capacitor

Plate A Plate B

Capacitance of a capacitor depends on its physical construction and given as:

C = A/d [F]

where = ro

= absolute permittivity [F/m] r = relative permittivity [no unit]

o = free space permittivity [8.854 x 10-12 F/m]

A = area of a plate [m2] d = distance between two plates [m]

A capacitor made of two parallel plates of area 10 cm2 each. The plates are separated by an insulator of 0.5mm thickness and a relative permittivity equal to 5. Determine its capacitance.

C = roA/d

= (5 x 8.854 x 10-12 x 10 x 10-4)/0.5 x 10-3

= (5 x 8.854 x 10-12 x 10)/5

= 88.54 pF

When the voltage applied to the capacitor changing with time , the charges also will change with time . Thus Q = CV becomes

dq = C dv

Then differentiate with time , we havedt

dvC

dt

dq

dt

dvCi

idtC

1v

therefore

integrates

note idt

dq

dt

dvCvvip Power in capacitor is given by

Cvdvpdtdw Energy for time dt is given by

VvdvCW

0

Vv

C0

2

2

Total Energy

2

2

1CVW Thus Energy stored in capacitor is

n21T C

1......

C

1

C

1

C

1

idtC

1idt

C

1idt

C

1

21T

21T C

1

C

1

C

1

C1

C2

i(t)

vT(t)v1(t)

v2(t)

Using Kirchhoff‘s voltage law we have:

vT(t) = v1(t) + v2(t)

Generalized for n capacitors idtC

1v

Note from previous slides

Both side haveSame integration

Thus

2

1

1

2

C

C

V

V

VVV 21

21

21 CC

C

VV

21

12 CC

C

VV

2

1

1

1

C

C

V

V-V

12 VVV

C1

C2

i(t)

vT(t)v1(t)

v2(t)

22 C

QV

11 C

QV

Voltage across each capacitor

Taking the ratio (*)

But summation of voltage

or

Substitute in (*)

Then we have

Similarly

v(t)

iT(t)

i1(t) i2(t)

C1 C2

Using Kirchoff’s current law

iT(t) = i1(t) + i2(t)

CT = C1 + C2

To generalized for n capacitor , thus

CT = C1 + C2 + ……. + Cn

dt

dv(t)C

dt

dv(t)C

dt

dv(t)C 21T

dt

dvCi

Note from previous slides

C3470pF

C13.5nF

C2220pF

A

BpFCC

CCCs 150

470220

470220

32

32

nFpFpFpFCCC sT 65.3365015035001

Total of series capacitors C2 and C3

The overall total

Cs

C3392nF

C2980nF

C14.67uF

A

B

Charges in the capacitors C1 and C2 are Q1 = 20 C and Q2 = 5 C respectively, determine the energy stored in C1, C2 and C3 and the total energy in all capacitors.

C1 = 4.67 F; Q1 = 20 Q

Therefore V1 = Q1/C1

= (20 x 10-6)/(4.67 x 10-6)= 4.3 V

And W1 = ½C1(V1)2

= ½ x 4.67 x 10-6 x 4.32

= 43.2 J

C2 = 980 nF; Q2 = 5 Q

Then V2 = Q2/C2

= (5 x 10-6)/(980 x 10-9)= 5.1 V

W2 = ½C2(V2)2

= ½ x 980 x 10-9 x 5.12

= 12.7 J

W3 = ½C3(V2)2

= ½ x 392 x 10-9 x 5.12 = 5.1 J

Electric current passing through a conductor will produce magnetic field or flux around it as shown in Figure.

If the wire conductor is wound around a core, magnetic filed /flux will resemble like permanent magnetic bar.

Magnitude of flux produced depends on magnitude of current ,I, properties of core , ,and physical construction (length , area of cross-section A and number of turn N)

whereflux is equivalent to current Im.m.f. NI equivalent to e.m.f. Vreluctance S equivalent to resistance Rand is an absolute pemeability of core material and given by : = ro where o is a permeability of air (4 x 10-7 H/m)

Substituting S in equation = NI/S: we have

S

NI

R

VI

μAS

l

σAR

l

l

ANIor

If the current in the inductor is varied with time (t), flux will also varied with time. Variation of flux in the windings will induce voltage.

l

ANior

The induced voltage is: )(dt

dNv

l

ANi

dt

dN or

dt

diAN2

or

l

dt

diLv vdt

L

1i

l

2

or ANL

where:

then

By introducing

Or for current

The inductance L is measured in Henry defined as a coil that induce 1 V when rate of current variation is 1 A/s.

The units are: H = 10-6 HmH = 10-3 HH = 100 H

Symbol for inductors:

(a) – air cored inductor

(b) –iron cored inductor

(c) –ferrite-cored inductor

(d) – variable inductor

An inductor is built from a coil of 180 turns and the core is of iron having a relative permeability of 1500. The length of the core is 30 mm and area of cross section is 78.5 mm2. Calculate the value of the inductance.

r = 1500; o = 4 x 10-7 H/m; A = 78.5 mm2;N = 180; l = 30 mm

l

2

or ANL

3

267

1030

180105.781041500

mH160

dt

diLv

dt

diiLivp

dtdt

diLipdtdw Lidi

I

0

idiLWI

0

2

2

iL

2LI

2

1

Inductor stores energy in the form of magnetic fields.

I

L

For duration of dt sec

For current changes between i = 0 to i = I

Voltage

Power

nT LLLL ........21

21 vvvT

L2

L1

v2

v1

i

vT

21T LLL

dt

diL

dt

diL

dt

diL 21T

dt

diLv

In general

21 iiiT

vdtL

1vdt

L

1vdt

L

1

21T

vdtL

1i

L2L1

i1 i2iT

21

111

LLLT

nT LLLL

1........

111

21

In general

Effective inductance between A and B is:

Le = L2 + L3 = 25 + 15 = 40 mH

Lt = (L1 x Le)/(L1 + Le)

= (60 x 40)/(60 + 40) = 24 mH

L3

L2

L1

25 mH

15 mH

60 mH

A

B

L1

L2

L3

R1R2

I = 600 mAA

B

350 mH150 mH

100 mH

W1 = ½L1I12

I1= ( 2W1/L1) = [(2 x 28 x 10-3)/(350 x 10-3)] = 400 mA

I2 = I – I1( Kirchoff’s current law) = 600 – 400 = 200 mA

Le= L2 + L3 = 150 + 100 = 250 mH

We= ½LeI22 = ½ x 250 x 10-3 x (200 x 10-3)2 = 5 mJ

Total energy W = W1 + We = 28 + 5 = 33 mJ

By supplying a total of constant current at 600 mA, L1 is found to store an energy of 28 mJ in its magnetic field. Calculate the total energy stored in all three inductors L1, L2 dan L3?