CHAPTER FOUR MUTUAL INDUCTANCE

Transcription

1 CHAPTER FOUR MUTUAL INDUCTANCE

2 4.1 Inductance 4.2 Capacitance 4.3 Serial-Parallel Combination 4.4 Mutual Inductance

3 4.1 Inductance Inductance (L in Henry is the circuit parameter used to describe an inductor. The symbol a coil wire is remainder that inductance is a consequence of a conductor linking a magnetic field. L = (N 2 µa/l L-inductance (H, N-kebolehtelapan magnet, A-area,l-length Voltage drop across inductor, v = L di//dt where, v in volts, L in Henry, i in ampere and t in second. Voltage across the terminal is proportional to the time rate of change of the current in the inductor. If current is constant, voltage is zero, thus inductor behave as a short circuit. If current cannot change instantaneously in an inductor, that is, the current cannot change by a finite amount in zero time. Arcing is when switch is open but current initially continue to flow in the air across the switch and voltage surge must be controlled to prevent equipment damage.

4 EX : Current source, generate zero for t < 0 and pulse 10te- 5t A for t > 0. a. Waveform b. di/dt = 10(-5te -5t + e -5t = 10e -5t (1 5tA/s di/dt = 0 when t = 1/5s c. V L = Ldi/dt = 0.1(10e- 5t (1 5t for t > 0 VL = 0 for t < 0 i 1 L mh i(a 0.73 v(v t(s 0.2 t(s

5 The voltage across the terminals of an inductor as a function of the current in the inductor. To find I as the function of v ; multiply both side with dt : vdt = L x (di/dt x dt next integrate both side : L i(t i(t0dx = t t0d ז i(t = I/L t t0vd ז + i(t0 or normally i(t = I/L t 0vd ז + i(0

6 EX : v(t = 20te -10t V for t > 0 a. Waveform + 0 זdזe-10 ז 20 b. i = 1/0.1 t 0 = 200[(-e + ז 100(10 /( ז 10-1] t 0 =2(1 10te -10t e -10t A c. Current as function of time V L mh v(v ia t(s t(s

7 Power and Energy in the Inductor It can be derived directly from the current and voltage relationship. If the current reference is in direction of the voltage drop across the terminal of the inductor, the power p = vi, but v = Ldi/dt, so p = Lidi/dt or p = v[i/l t זt0vd + i(t0] Power is the time rate of expending energy, so, p = dw/dt = Lidi/dt multiply both with dt, dw = Lidi integrated with ref that zero energy is zero current in the inductor, so w 0dx = L w 0ydy energy, w = 1/2Li 2 assume w =0 at t = 0 i(t = 1/L t זt0vd + i(t0 or normally i(t = 1/L t ז 0vd + i(0 w( t w( t = p( τ dτ 0 = 1 2 t t 0 Li 2 ( t 1 2 Li 2 ( t 0

8 Ex : The current through 0.1H inductor is i(t = 10te -5t A. Find the voltage across the inductor and the energy stored in it. Since v = Ldi/dt and L = 0.1H V = 01.d/dt(10te -5t = e -5t + t(-5e -5t = e -5t (1 5t V Energy stored is = 1/2Li2 = ½(0.1100t 2 e -10t = 5t2e -10t J Ex : The current through 1mH inductor is i(t = 20cos100t ma. Find the terminal voltage and the energy stored.

9 4.2 Capacitance Capacitor consists of two conducting plate separated by an insulator (or dielectric. Capacitance occurs whenever electrical conductor are separate by a dielectric or insulating material. This condition applies that electric charge electric charge is to transport through the capacitor. Practical capacitor is in µf and pf. Although applying a voltage to the terminal of the capacitor cannot move a charge through the dielectric, it can displace a charge within the dielectric. As the voltage varies with time, the displacement of charge also varies with time, and causing the displacement current. charge, q(culomb = c(f x v(volt

10 Capacitance, C = ( A/d A surface area of the plate, the larger the area, the greater the capacitance. d spacing between plates, the smaller the spacing the greater the capacitance. - permittivity of the material, the higher, the greater the capacitance. There are two types, fix and variable. The fix type are polyester which light in weight, stable and their change with temperature is predictable. Other dielectric material such as mica and polystyrene may be used. Film capacitor are rolled and house in metal or plastic film. Electrolytic capacitor produce very high capacitance. Capacitance of a trimmer (or padder capacitor or a glass piston capacitor is varied by turning the screw. Usually parallel with another capacitor so the equ capacitance can be varied slightly.

11 capacitance of variable air capacitor (meshed plates is varied by turning the shaft. Variable capacitor are used in radio. Capacitor are used to block dc, pass ac, shift phase, store energy, start motor and suppress noise. q(t = Cv(t d/dtq(t = d/dtcv(t The current is proportional to the rate at which the voltage across the capacitor varies with times, i = dq/dt = Cdv/dt. (ז Cdv = (ז i i(t = Cdv(t/dt ; Voltage cannot change instantaneously across the terminals of a capacitor, coz such change would produce current infinite. If the voltage across the terminal is constant, the capacitor current is zero, coz a conduction current cannot be established in the dielectric material of the capacitor. Only a time-varying voltage can produce a displacement current. Thus a capacitor behaves as an open circuit in the presence of a constant voltage.

12 Expressing the voltage as current by multiply both with dt idt = Cdv or v(t v(t0dx = 1/C t t0idז Integration left hand of equation v(t = 1/C t t0i (זdז +v(t 0 So voltage capacitor, v(t = 1/C t - i (זdז Power absorb by capacitor, p = v(ti(t = v(txcdv(t/dt or p = i[1/c t t0idז +v(t0] So by combine the definition of energy, dw = Cvdv w 0dx = C t 0ydy or w = 1/2Cv 2 Energy store by capacitor from t to t 0 so capacitor energy, w(t = 1/2Cv(t ( ( 2 ( ( 0 ( 2 1 ( 2 1 ( 2 1 ( ( ( ( ( ( ( t v C t v C v C dv v C d d dv C v d p t w t w t v t v t v t v t t t t = = = = = τ τ τ τ τ τ τ τ τ

13 Capacitor is derive form it s capacity to store energy in an electric field ; 1. When the voltage across capacitor is not changing with time (i.e.dc voltage current through capacitor is zero, so capacitor is an open circuit to dc but if a battery connected across a capacitor, the capacitor charge. 2. Voltage on capacitor must continuous where it cannot change abruptly, but current through capacitor can change instantaneously. 3. Ideal capacitor does not dissipate energy. It take power from the circuit when storing energy in its filed and returns previously stored energy when delivering power to the circuit. 4. A real no ideal capacitor has a parallel model leakage resistance, as high as 100MΩ and can be neglected for most practical application.

14 Ex : Calculate the charge store on a 3-pF capacitor with 20V across it and find energy stored in the capacitor. Q = Cv = 3 x10-12 x 20 = 60pC W= 1/2Cv 2 = ½ x 3 x x 400 = 600 pj Ex: Voltage across 5 µf capacitor is v(t = 10cos600t, determine the current through capacitor. i(t = Cdv/dt = 5 x 10-6 d/dt(10cos6000t = -5 x 10-6 x 6000 x 10sin600t = -0.3sin6000t A Ex: Determine the voltage across 2 µf capacitor if current through it is i(t = 6e -300t ma Since v = 1/C t 0idt + v(0 and v(0 = 0 v = 1/(2 x 10-6 t 06e -3000t dt.10-3 = (3 x 10 3 /-3000 x e -3000t t 0 =(1 e -3000t V

15 Ex : Find the energy stored in each capacitor under dc condition. under dc i = 3/(3=2+4 x 6mA = 2 ma v 1 = 2000i =4V v 2 = 4000i = 8V w 1 = 1/2C 1 v 12 = 16 mj w 2 = 1/2C 2 V 22 = 128 mj 2mF 2kΩ 6mA 3kΩ v 1 2kΩ 6mA 3kΩ 5kΩ 4mF 5kΩ v 2 4kΩ 4kΩ

16 Ex : Find the energy stored in each capacitor under dc condition. 1kΩ 20uF 10V 10uF 6kΩ

17 4.3 Serial-Parallel Combination Inductors in Series The inductor are forced to carry the same current; thus only one current for series combination. v 1 = L 1 di/dt v 2 = L 2 di/dt v 3 = L 3 di/dt v = v 1 + v 2 + v 3 = (L 1 + L 2 +L 3 di/dt = L eq di/dt L eq = L 1 + L 2 + L L N

18 Inductors in Parallel The inductor in parallel have the same terminal voltage. In the equivalent circuit, the current in each inductor is a function of the terminal voltage and the initial current in the indicator. i 1 = 1/L 1 t זt0vd + i 1 (t 0 i 2 = 1/L 2 t זt0vd + i 2 (t 0 i 3 = 1/L 3 t זt0vd + i 3 (t 0 i = i 1 + i 2 + i 3 So I = (1/L 1 + 1/L 2 + 1/L 3 t זt0vd + i 1 (t 0 + i 2 (t 0 + i 3 (t 0 or I = 1/L eq t זt0vd + i(t 0 where 1/L eq = 1/L 1 + 1/L 2 + 1/L 3 i(t 0 = i 1 (t 0 + i 2 (t 0 + i 3 (t 0

19 Ex : Find Leq. Leq = 18H 4H 20H 7H L eq 8H 10H 12H

20 Ex : Find Leq. 20mH 100mH 40mH Leq = 25mH 50mH 40mH 30mH 20mH L eq

21 Ex : Find v c and the energy stored in capacitor and inductor under dc condition. Under dc condition, capacitor is Open and inductor is short. i = i L = 12/(1 + 5 = 2A v c equal to voltage across 5Ω v c = 5i = 10V Energy in capacitor, w c = 1/2Cv c 2 =1/2(1(10 2 = 50 J Energy in inductor, w L = 1/2Li L 2 =1/2(2(2 2 = 4 J 12V 12V 1Ω 1Ω 4Ω 1F 4Ω 5Ω 5Ω 2H

22 Ex : If i(t = 4(2 e -10t ma and i 2 (0 = -1 ma. Find v and i. From i(t = 4(2 e -10t ma i(0 = 4(2 1 ma = 4mA Coz i = i Leq = 5H So v(t = Ldi/dt = 5(4(-1(-10e -10t mv = 200e -10t mv v 1 (t = Ldi/dt = 2(-4(-1(-10e -10t mv = 80e -10t mv v 2 (t = v(t V(t = 120e -10t mv i 1 (t = t 0v 2 dt +i 1 (0 = 120/4 t 0e -10t dt + 5 ma =-3e -10t t 0 + 5mA = -3e -10t = 8-3e -10t ma L eq i 2 (t = t 0v 2 dt +i 2 (0 = 120/12 t 0e -10t dt -1mA =-e -10t t 0-1mA = -e -10t = e -10t ma 4H i v 1 i 1 i 2 4H 12H v 2

23 Capacitors in Series Capacitor connected in series can be reduced to a single equivalent capacitor. If each capacitor carries its own initial voltage, the initial voltage on the individual capacitors is the algebraic sum of the initial voltages on the individual capacitors. 1/C eq = 1/C 1 + 1/C /C N v(t 0 = v 1 (t 0 + v 2 (t v n (t 0 זd (ז i - v(t0 = i/ceq t

24 Capacitor in Parallel The equivalent capacitance of capacitor connected in parallel is simply the sum of the capacitance of the individual capacitors; C eq = C 1 + C 2 +.+C N Capacitor connected in parallel must carry the same voltage. Therefore if there is an initial voltage appear across the original parallel capacitor, this same initial voltage appears across the equivalent capacitance, C eq.

25 Ex : Find the C eq 20µF series with 5µF = 4µF 5uF 60uF 4µF 6µF 20µF = 30µF 30µF series with 60µF = 20µF 20uF 6uF 20uF C eq Ex : Find the C eq 50uF 70uF 60uF C eq 20uF 120uF

26 Ex : Find the voltage across each capacitor. C eq = 10mF The total charge, q = vc eq = 10 x 10-3 x 30 = 0.3C Since i = dq/dt, so charge on 20mF and 30mF is 0.3C So v 1 = q/c 1 = 0.3/(20 x 10-3 = 15V v 2 = q/c 2 = 0.3/(20 x 10-3 = 15V v 3 = 30 v 1 v 2 = 5V v 3 = q/c = 0.3/60 = 5V 30V 20mF 30mF v 1 v 2 40mF v 3 20mF Ex : Find the voltage across each capacitor. 40F 60µF v 1 v 3 60V 20µF v 2 v 4 30µF

27 Summary Relation Resistor(R Capacitor (C Inductor (L v-i v = ir v = 1/C t t0idt + v(t 0 v = 1/2Ldi/dt i-v i = v/r i = Cdv/dt i = 1/L t t0idt + i(t 0 p or w p = i 2 R = v 2 /R w = 1/2Cv 2 w = 1/2Li 2 Series R eq = R 1 + R 2 Ceq = (C 1 C 2 /(C 1 +C 2 L eq = L 1 + L 2 Parallel R eq = (R 1 R 2 /(R 1 +R 2 C eq = C 1 + C 2 Leq = (L 1 L 2 /(L 1 +L 2 At dc Same Open circuit Short circuit Circuit variable cannot change abruptly Not applicable v i

28 4.4 Mutual Inductance When two circuit are linked by a magnetic field, voltage induced in second circuit can be related to time-varying current in first circuit by parameter known as mutual inductance. The self inductance of two coil is L 1 and L 2 and the mutual inductance is M and the easier to analyze is using mesh current. Choose for direction for i 1 and i 2, sum the voltage around each closed path. Coz of M, there will be two voltage across each coil, namely self-induced voltage and a mutually induced voltage. Self induced voltage is the product of the self-inductance of the coil and the first derivatative current in other coil. Mutual induced voltage is the product of mutual inductance of coil and first derivatative current in other coil. V g R 1 L M i L 1 i R 2

29 From the circuit the self-induced voltage across this coil is L 1 (di 1 /dt and the mutually induced voltage across this coil is M(di 2 /dt. When the ref direction for a current enters the dotted terminal of a coil, the ref polarity of the voltage that is induces in the other coil is positive at its dotted terminal and the other way around. The dot convention rule indicates that the reference polarity for the voltage induced in coil 1 by the current i 2 is negative at the dotted terminal of coil 1. This voltage (Mdi 2 /dt is a voltage rise with respect to i1. The voltage induced in coil2 by the current i 1 is Mdi 2 /dt and its ref polarity is positive at the dotted terminal of coil 2. This voltage is a voltage rise in direction of i2. The sum voltage around each closed loop ; -v g + i 1 R 1 + L 1 di 1 /dt Mdi 2 /dt = 0 i 2 R 2 + L 2 di 2 /dt Mdi 1 /dt = 0