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) nanofarads (nF) microfarads (F) millifarads (mF) Farad = 10-12 F = 10-9 F = 10-6 F = 10-3 F (F) = 100 F d d A Insulator (dielectric) C 39nF fixed capacitor Insulator (dielectric) C 100pF + Polarized capacitor Cross section C 470pF Variable capacitor Plate A Plate B Capacitor R V 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. Plate A Plate B Capacitor R V •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 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 have note dq i dt integrates therefore 1 v idt C dq dv C dt dt dv iC dt Power in capacitor is given by dv p vi Cv dt Energy for time dt is given by dw pdt Cvdv V Total Energy W C vdv Thus Energy stored in capacitor is 0 V v C 2 0 2 W 1 CV 2 2 i(t) Using Kirchhoff‘s voltage law we have: 1 1 1 Both side have idt idt idt C1 C2 Same integration CT 1 1 1 C T C1 C 2 Generalized for n capacitors 1 1 1 1 ...... C T C1 C 2 Cn C1 v2(t) C2 vT(t) vT(t) = v1(t) + v2(t) Thus v1(t) Note from previous slides 1 v idt C i(t) Voltage across each capacitor Q Q V2 V1 C2 C1 Taking the ratio V2 C1 V1 C 2 (*) v1(t) C1 v2(t) C2 vT(t) But summation of voltage V1 V2 V or Substitute in (*) V2 V V1 V-V1 C1 V1 C2 Then we have V1 V C2 C1 C 2 Similarly V2 V C1 C1 C 2 iT(t) Using Kirchoff’s current law i1(t) iT(t) = i1(t) + i2(t) CT v(t) i2(t) C1 dv(t) dv(t) dv(t) C1 C2 dt dt dt CT = C1 + C2 To generalized for n capacitor , thus CT = C1 + C2 + ……. + Cn Note from previous slides dv iC dt C2 A C2 220pF C1 3.5nF Total of series capacitors C2 and C3 C2C3 220 470 Cs 150 pF C2 C3 220 470 Cs C3 470pF B The overall total CT C1 Cs 3500 pF 150 pF 3650 pF 3.65nF A C1 4.67uF 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 B energy in all capacitors. C1 = 4.67 F; Q1 = 20 Q Therefore V1 And W1 = Q1/C1 = (20 x 10-6)/(4.67 x 10-6) = 4.3 V = ½C1(V1)2 = ½ x 4.67 x 10-6 x 4.32 = 43.2 J C2 980nF C3 392nF C2 = 980 nF; Q2 = 5 Q Then V2 W2 W3 = Q2/C2 = (5 x 10-6)/(980 x 10-9) = 5.1 V = ½C2(V2)2 = ½ x 980 x 10-9 x 5.12 = 12.7 J = 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) NI S V I R l S μA R l σA 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 : = ro where o is a permeability of air (4 x 10-7 H/m) Substituting S in equation = NI/S: we have r o ANI l 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. d The induced voltage is: v N () dt r o AN 2 di l dt r o ANi l where: By introducing d ANi N r o dt l r o AN 2 L l then di vL dt 1 Or for current i vdt L The inductance L is measured in Henry defined as a coil that induce 1 V when rate of current variation is 1 A/s. H mH H Symbol for inductors: The units are: (a) – air cored inductor (c) –ferritecored inductor = 10-6 H = 10-3 H = 100 H (b) –iron 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 r o AN 2 1500 4 10 7 78.5 10 6 180 2 L 30 10 3 l 160 mH Voltage Power di vL dt I di p iv iL dt L For duration of dt sec di dw pdt Li dt dt Lidi For current changes between i = 0 to i = I I W L idi 0 I 1 2 i2 L LI 2 2 0 Inductor stores energy in the form of magnetic fields. di vL dt vT v1 v2 di di di L T L1 L 2 dt dt dt L T L1 L 2 In general LT L1 L2 ........Ln i v1 L1 vT v2 L2 1 i vdt L iT i1 i2 iT i1 i2 1 1 1 vdt vdt vdt LT L1 L2 1 1 1 LT L1 L2 In general 1 1 1 1 ........ LT L1 L2 Ln L1 L2 Effective inductance between A and B is: A 25 mH L2 15 mH L3 60 mH L1 Le = L2 + L3 = 25 + 15 = 40 mH Lt = (L1 x Le)/(L1 + Le) = (60 x 40)/(60 + 40) = 24 mH B A I = 600 mA By supplying a total of constant current L2 at 600 mA, L1 is found to store an 350 mH L energy of 28 mJ in its magnetic field. L3 Calculate the total energy stored in all R1 R2 three inductors L1, L2 dan L3? W1 = ½L1I12 B I1= ( 2W1/L1) = [(2 x 28 x 10-3)/(350 x 10-3)] = 400 mA 150 mH 1 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 100 mH