Capacitors and Inductors Mustafa Kemal Uyguroğlu Eastern Mediterranean University 1 Chap. 6, Capacitors and Inductors Introduction Capacitors Series and Parallel Capacitors Inductors Series and Parallel Inductors Eastern Mediterranean University 2 6.1 Introduction Resistor: a passive element which dissipates energy only Two important passive linear circuit elements: 1) Capacitor 2) Inductor Capacitor and inductor can store energy only and they can neither generate nor dissipate energy. Eastern Mediterranean University 3 Michael Faraday (1971-1867) Eastern Mediterranean University 4 6.2 Capacitors A capacitor consists of two conducting plates separated by an insulator (or dielectric). C εA d r 0 0 8.854 10 12 (F/m) Eastern Mediterranean University 5 εA C d Three factors affecting the value of capacitance: 1. Area: the larger the area, the greater the capacitance. 2. Spacing between the plates: the smaller the spacing, the greater the capacitance. 3. Material permittivity: the higher the permittivity, the greater the capacitance. Eastern Mediterranean University 6 Fig 6.4 (a) Polyester capacitor, (b) Ceramic capacitor, (c) Electrolytic capacitor Eastern Mediterranean University 7 Fig 6.5 Variable capacitors Eastern Mediterranean University 8 Fig 6.3 Eastern Mediterranean University 9 Fig 6.2 Eastern Mediterranean University 10 Charge in Capacitors The relation between the charge in plates and the voltage across a capacitor is given below. q Cv 1F 1 C/V q Linear Nonlinear v Eastern Mediterranean University 11 Voltage Limit on a Capacitor Since q=Cv, the plate charge increases as the voltage increases. The electric field intensity between two plates increases. If the voltage across the capacitor is so large that the field intensity is large enough to break down the insulation of the dielectric, the capacitor is out of work. Hence, every practical capacitor has a maximum limit on its operating voltage. Eastern Mediterranean University 12 I-V Relation of Capacitor + v i C dq dv q Cv, i C dt dt - Eastern Mediterranean University 13 Physical Meaning dv iC dt + v i C - • when v is a constant voltage, then i=0; a constant voltage across a capacitor creates no current through the capacitor, the capacitor in this case is the same as an open circuit. • If v is abruptly changed, then the current will have an infinite value that is practically impossible. Hence, a capacitor is impossible to have an abrupt change in its voltage except an infinite current is applied. Eastern Mediterranean University 14 Fig 6.7 A capacitor is an open circuit to dc. The voltage on a capacitor cannot change abruptly. Abrupt change Eastern Mediterranean University 15 dv iC dt 1 v(t ) C 1 t v(t ) idt v(to) C to t idt v() 0 v(to) q(to) / C + v i C - The charge on a capacitor is an integration of current through the capacitor. Hence, the memory effect counts. Eastern Mediterranean University 16 Energy Storing in Capacitor dv p vi Cv dt v (t ) dv 1 2 w pdt C v dt C v ( ) vdv Cv dt 2 t t 1 w(t ) Cv 2 (t ) 2 ( v( ) 0) q 2 (t ) w(t ) 2C + v v (t ) v ( ) i C - Eastern Mediterranean University 17 Model of Practical Capacitor Eastern Mediterranean University 18 Example 6.1 (a) Calculate the charge stored on a 3-pF capacitor with 20V across it. (b) Find the energy stored in the capacitor. Eastern Mediterranean University 19 Example 6.1 Solution: (a) Since q Cv, q 3 10 12 20 60pC (b) The energy stored is 1 2 1 12 w Cv 3 10 400 600 pJ 2 2 Eastern Mediterranean University 20 Example 6.2 The voltage across a 5- F capacitor is v(t ) 10 cos 6000t V Calculate the current through it. Solution: By definition, the current is dv 6 d iC 5 10 (10 cos 6000t ) dt dt 5 10 6 6000 10 sin 6000t 0.3 sin 6000t A Eastern Mediterranean University 21 Example 6.3 Determine the voltage across a 2-F capacitor if the current through it is i(t ) 6e 3000t mA Assume that the initial capacitor voltage is zero. Solution: 1 t v 0 idt v(0) and v(0) 0, Since C 3 t 1 3000 t 3 10 3000t t 3 v 6 e dt 10 e 6 0 0 2 10 3000 (1 e 3000t )V Eastern Mediterranean University 22 Example 6.4 Determine the current through a 200- F capacitor whose voltage is shown in Fig 6.9. Eastern Mediterranean University 23 Example 6.4 Solution: The voltage waveform can be described mathematically as 50t V 0 t 1 100 50t V 1 t 3 v(t ) 200 50t V 3t 4 0 otherwise Eastern Mediterranean University 24 Example 6.4 Since i = C dv/dt and C = 200 F, we take the derivative of to obtain 0 t 1 10mA 0 t 1 50 50 1 t 3 10mA 1 t 3 6 i (t ) 200 10 50 3 t 4 10mA 3t 4 0 otherwise 0 otherwise Thus the current waveform is shown in Fig.6.10. Eastern Mediterranean University 25 Example 6.4 Eastern Mediterranean University 26 Example 6.5 Obtain the energy stored in each capacitor in Fig. 6.12(a) under dc condition. Eastern Mediterranean University 27 Example 6.5 Solution: Under dc condition, we replace each capacitor with an open circuit. By current division, 3 i (6mA) 2mA 3 2 4 v1 2000 i 4 V, v 2 4000i 8 V 1 1 2 3 2 w1 C1v1 (2 10 )(4) 16mJ 2 2 1 1 2 3 2 w2 C2 v2 (4 10 )(8) 128mJ 2 2 Eastern Mediterranean University 28 Fig 6.14 Ceq C1 C2 C3 .... C N Eastern Mediterranean University 29 6.3 Series and Parallel Capacitors i i1 i2 i3 ... iN dv dv dv dv i C1 C2 C3 ... C N dt dt dt dt N dv dv C K Ceq dt k 1 dt Ceq C1 C2 C3 .... CN The equivalent capacitance of N parallelconnected capacitors is the sum of the individual capacitance. Eastern Mediterranean University 30 Fig 6.15 1 1 1 1 1 ... Ceq C1 C2 C3 CN Eastern Mediterranean University 31 Series Capacitors v(t ) v1 (t ) v2 (t ) ... vN (t ) 1 Ceq 1 1 1 1 t id ( C1 C2 C3 ... CN )id t q(t ) q(t ) q(t ) q(t ) Ceq C1 C2 CN The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances. 1 1 1 Ceq C1 C2 C1C2 Ceq C1 C2 Eastern Mediterranean University 32 Summary These results enable us to look the capacitor in this way: 1/C has the equivalent effect as the resistance. The equivalent capacitor of capacitors connected in parallel or series can be obtained via this point of view, so is the Y-△ connection and its transformation Eastern Mediterranean University 33 Example 6.6 Find the equivalent capacitance seen between terminals a and b of the circuit in Fig 6.16. Eastern Mediterranean University 34 Example 6.6 Solution: 20 F and 5 F capacitors are in series : 20 5 4 F 20 5 4 F capacitor is in parallel with the 6 F and 20 F capacitors : 4 6 20 30 F 30 F capacitor is in series with the 60 F capacitor. 30 60 Ceq F 20 F 30 60 Eastern Mediterranean University 35 Example 6.7 For the circuit in Fig 6.18, find the voltage across each capacitor. Eastern Mediterranean University 36 Example 6.7 Eastern Mediterranean University 37 Example 6.7 Solution: Two parallel capacitors: Ceq Total charge 1 1 1 1 60 30 20 mF 10mF 3 q Ceq v 10 10 30 0.3 C This is the charge on the 20-mF and 30-mF capacitors, because they are in series with the 30-v source. ( A crude way to see this is to imagine that charge acts like current, since i = dq/dt) Eastern Mediterranean University 38 Example 6.7 Therefore, v1 q 0.3 15 V, 3 C1 20 10 v2 q 0.3 10 V 3 C2 30 10 Having determined v1 and v2, we now use KVL to determine v3 by v3 30 v1 v2 5V Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage v3 and their combined capacitance is q 0 .3 40+20=60mF. v3 60mF 60 10 Eastern Mediterranean University 3 5V 39 Joseph Henry (1979-1878) Eastern Mediterranean University 40 6.4 Inductors An inductor is made of a coil of conducting wire N A L l 2 Eastern Mediterranean University 41 Fig 6.22 N 2 A L l r 0 0 4 10 7 (H/m) N : number of turns. l :length. A:cross sectional area. : permeability of the core Eastern Mediterranean University 42 Fig 6.23 (a) air-core (b) iron-core (c) variable iron-core Eastern Mediterranean University 43 Flux in Inductors The relation between the flux in inductor and the current through the inductor is given below. Li ψ 1H 1 Weber/A Linear Nonlinear i Eastern Mediterranean University 44 Energy Storage Form An inductor is a passive element designed to store energy in the magnetic field while a capacitor stores energy in the electric field. Eastern Mediterranean University 45 I-V Relation of Inductors An inductor consists of a coil of conducting wire. d di v L dt dt i + v L - Eastern Mediterranean University 46 Physical Meaning d di v L dt dt When the current through an inductor is a constant, then the voltage across the inductor is zero, same as a short circuit. No abrupt change of the current through an inductor is possible except an infinite voltage across the inductor is applied. The inductor can be used to generate a high voltage, for example, used as an igniting element. 47 Eastern Mediterranean University Fig 6.25 An inductor are like a short circuit to dc. The current through an inductor cannot change instantaneously. Eastern Mediterranean University 48 1 di vdt L 1 t i v(t )dt L 1 t i v(t )dt i (to ) L to + v L - The inductor has memory. Eastern Mediterranean University 49 Energy Stored in an Inductor di P vi L i dt + v di L w pdt L idt dt i (t ) 1 2 1 2 L i ( ) i di Li (t ) Li ( ) i () 0, 2 2 t t The energy stored in an inductor Eastern Mediterranean University 1 2 w(t ) Li (t ) 2 50 Model of a Practical Inductor Eastern Mediterranean University 51 Example 6.8 The current through a 0.1-H inductor is i(t) = 10te5t A. Find the voltage across the inductor and the energy stored in it. Solution: di Since v L and L 0.1H, dt d v 0.1 (10te 5t ) e 5t t (5)e 5t e 5t (1 5t )V dt The energy stored is 1 2 1 w Li (0.1)100t 2 e 10t 5t 2 e 10t J 2 2 Eastern Mediterranean University 52 Example 6.9 Find the current through a 5-H inductor if the voltage across it is 2 30t , t 0 v(t ) t0 0, Also find the energy stored within 0 < t < 5s. Assume i(0)=0. Solution: 1 t Since i t v(t )dt i (t0 ) and L 5H. L 3 t 1 t i 0 30t 2 dt 0 6 2t 3 A 3 5 0 Eastern Mediterranean University 53 Example 6.9 The power p vi 60t , and the energy stored is then 5 6 t 5 w pdt 0 60t dt 60 156.25 kJ 60 Alternatively, we can obtain the energy stored using 5 5 Eq.(6.13), by writing 1 2 1 w(5) w(0) Li (5) Li (0) 2 2 1 3 2 (5)(2 5 ) 0 156.25 kJ 2 as obtained before. Eastern Mediterranean University 54 Example 6.10 Consider the circuit in Fig 6.27(a). Under dc conditions, find: (a) i, vC, and iL. (b) the energy stored in the capacitor and inductor. Eastern Mediterranean University 55 Example 6.10 Solution: (a ) Under dc condition : capacitor open circuit inductor short circuit 12 i iL 2 A, vc 5i 10 V 1 5 (b ) 1 1 wc Cvc (1)(10 2 ) 50J, 2 2 1 2 1 wL Li (2)(2 2 ) 4J 2 2 2 Eastern Mediterranean University 56 Inductors in Series Leq L1 L2 L3 ... LN Eastern Mediterranean University 57 Inductors in Parallel 1 1 1 1 Leq L1 L2 LN Eastern Mediterranean University 58 6.5 Series and Parallel Inductors Applying KVL to the loop, v v1 v2 v3 ... vN Substituting vk = Lk di/dt results in di di di di v L1 L2 L3 ... LN dt dt dt dt di ( L1 L2 L3 ... LN ) dt N di di LK Leq dt K 1 dt Leq L1 L2 L3 ... LN Eastern Mediterranean University 59 Parallel Inductors Using KCL, But 1 i Lk t t 0 1 ik Lk i i1 i2 i3 ... iN t t vdt ik (t0 ) o 1 1 t vdt i1 (t0 ) t vdt is (t0 ) ... LN L2 0 t t vdt iN (t0 ) 0 1 1 1 t ... t vdt i1 (t0 ) i2 (t0 ) ... iN (t0 ) LN L1 L2 0 N N 1t 1 t t vdt ik (t0 ) vdt i (t0 ) t Leq k 1 k 1 Lk 0 0 Eastern Mediterranean University 60 The inductor in various connection has the same effect as the resistor. Hence, the Y-Δ transformation of inductors can be similarly derived. Eastern Mediterranean University 61 Table 6.1 Eastern Mediterranean University 62 Example 6.11 Find the equivalent inductance of the circuit shown in Fig. 6.31. Eastern Mediterranean University 63 Example 6.11 Solution: Series : 20H, 12H, 10H 42H 7 42 6H Parallel : 7 42 Leq 4 6 8 18H Eastern Mediterranean University 64 Practice Problem 6.11 Eastern Mediterranean University 65 Example 6.12 Find the circuit in Fig. 6.33, i (t ) 4(2 e If find : (a) i (0) i2 (0) 1 mA, (b) v(t ), v1 (t ), and v2 (t ); (c) i1 (t ) and i2 (t ) 10t )mA. 1 Eastern Mediterranean University 66 Example 6.12 Solution: 10 t (a ) i(t ) 4(2 e )mA i(0) 4(2 1) 4mA. i1 (0) i (0) i2 (0) 4 (1) 5mA (b) The equivalent inductance is Leq 2 4 || 12 2 3 5H di 10 t 10 t v(t ) Leq 5(4)(1)(10)e mV 200e mV dt di 10 t 10 t v1 (t ) 2 2(4)(10)e mV 80e mV dt 10 t v2 (t ) v(t ) v1(t ) 120e mV Eastern Mediterranean University 67 Example 6.12 1 t (c) i 0 v(t ) dt i (0) L 1 t 120 t 10t i1 (t ) 0 v2 dt i1 (0) e dt 5 mA 4 4 0 10 t t 3e 5 mA 3e 10t 3 5 8 3e 10t mA 0 1 t 120 t 10t i2 (t ) 0 v2 dt i2 (0) e dt 1mA 0 12 12 10 t t e 1mA e 10t 1 1 e 10t mA 0 Note that i1 (t ) i2 (t ) i (t ) Eastern Mediterranean University 68