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)= F nanofarads(nF)= F microfarads(  F)= F millifarads (mF)= F Farad (F)= 10 0 F

d A Insulator (dielectric) d Insulator (dielectric) Cross section fixed capacitor Polarized capacitor 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 APlate 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 orQ = CV R V Capacitor Plate APlate B

Capacitance of a capacitor depends on its physical construction and given as: C =  A/d[F] where  =  r  o  = absolute permittivity [F/m]  r = relative permittivity [no unit]  o = free space permittivity [8.854 x F/m] A = area of a plate [m 2 ] 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=  r  o A/d = (5 x x x 10 x )/0.5 x = (5 x x x 10)/5 = 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 have therefore integrates note

Power in capacitor is given by Energy for time dt is given by Total Energy Thus Energy stored in capacitor is

C1C1 C2C2 i(t) v T (t) v 1 (t) v 2 (t) Using Kirchhoff‘s voltage law we have: v T( t) = v 1 (t) + v 2 (t) Generalized for n capacitors Note from previous slides Both side have Same integration Thus

C1C1 C2C2 i(t) v T (t) v 1 (t) v 2 (t) Voltage across each capacitor Taking the ratio (*) But summation of voltage or Substitute in (*) Then we have Similarly

v(t) i T (t) i 1 (t)i 2 (t) C1C1 C2C2 Using Kirchoff’s current law i T (t) = i 1 (t) + i 2 (t) C T = C 1 + C 2 To generalized for n capacitor, thus C T = C 1 + C 2 + ……. + C n Note from previous slides

A B Total of series capacitors C 2 and C 3 The overall total CsCs

A B Charges in the capacitors C 1 and C 2 are Q 1 = 20  C and Q 2 = 5  C respectively, determine the energy stored in C 1, C 2 and C 3 and the total energy in all capacitors. C 1 = 4.67  F; Q 1 = 20  Q Therefore V 1 = Q 1 /C 1 = (20 x )/(4.67 x ) = 4.3 V AndW 1 = ½C 1 (V 1 ) 2 = ½ x 4.67 x x = 43.2  J

C 2 = 980 nF; Q 2 = 5  Q ThenV 2 = Q 2 /C 2 = (5 x )/(980 x ) = 5.1 V W2 = ½C 2 (V 2 ) 2 = ½ x 980 x x = 12.7  J W3= ½C 3 (V 2 ) 2 = ½ x 392 x x = 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)

where flux  is equivalent to current I m.m.f. NI equivalent to e.m.f. V reluctance S equivalent to resistance R and  is an absolute pemeability of core material and given by :  =  r  o where  o is a permeability of air (4  x H/m) Substituting S in equation  = NI/S: we have

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. The induced voltage is: 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= H mH= H H= 10 0 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 The length of the core is 30 mm and area of cross section is 78.5 mm 2. Calculate the value of the inductance.  r = 1500;  o = 4  x H/m; A = 78.5 mm 2 ; N = 180 ; l = 30 mm

Inductor stores energy in the form of magnetic fields. For duration of dt sec For current changes between i = 0 to i = I Voltage Power

v2v2 v1v1 i vTvT In general

i1i1 i2i2 iTiT

Effective inductance between A and B is: L e = L 2 + L 3 = = 40 mH L t = (L 1 x L e )/(L 1 + L e ) = (60 x 40)/( ) = 24 mH 25 mH 15 mH 60 mH A B

L1L1 L2L2 L3L3 R1R1 R2R2 I = 600 mA A B 350 mH 150 mH 100 mH W 1 = ½L 1 I 1 2 I 1 =  ( 2W 1 /L 1 ) =  [(2 x 28 x )/(350 x )] = 400 mA I 2 = I – I 1 ( Kirchoff’s current law) = 600 – 400 = 200 mA L e = L 2 + L 3 = = 250 mH W e = ½L e I 2 2 = ½ x 250 x x (200 x ) 2 = 5 mJ Total energy W = W 1 + W e = = 33 mJ By supplying a total of constant current at 600 mA, L 1 is found to store an energy of 28 mJ in its magnetic field. Calculate the total energy stored in all three inductors L 1, L 2 dan L 3 ?