CAPACITACE, IDUCTACE, AD MUTUA IDUCTACE 6. The Capacitor 6. The Inductor 6.3 Series-Parallel Combinations of Capacitance and Inductance
6. The Capacitor In this chapter, two new and important passive linear components are introduced. They are ideal models. Resistors dissipate energy but capacitors and inductors are energy storage components. 3 6. The Capacitor Circuit symbol and component model. q@ C v, q(t)= i ( τ) dτ C q:charge C:capacitance, in F(Farad) t C 4
6. The Capacitor t0 vc() t = ic( τ) dτ = ic( τ) dτ + ic( τ) dτ C C C t0 t vc() t @ vc( t0) + ic( ) d... (A) C τ τ t0 dq dvc ic() t = = C... (B) dt dt farad = Coulomb/Volt The unit of capacitance is chosen to be farad in honor of the English physicist, Michael Faraday(79-867). 867). t 5 6. The Capacitor Example : A parallel-plate plate capacitor A C= ε d ε : the permittivity of the dielectric 6 ε material between the plates
Example : (cont.) 6. The Capacitor If v >0 and i >0, or v <0 and i <0 c c c c the capacitor is being charged. If v i <0, the capacitor is discharging. c c dvc p(t)=vc() t ic() t = vc() t C dt () t ε Energy in acapacitor t q w= p d = Cv t = C ( τ) τ c () 0 7 ε 0 6. The Capacitor ( a) When v is constant, then i =0. ( b) v () t is a continuous function C if there is only finite strength energy sources inside the circuit. t+ ε QvC( t+ ε) = vc() t + ic( ) d C τ τ t + lim v ( t+ ε) = v () t, v (0 ) = v (0 ) C C ie., equivalent to open circuit C C C C i.e. v(t) can not change abruptly for finite i (t) 8 c C
6. The Capacitor (c) An ideal capacitor does not dissipate energy. It stores energy in the electrical field. (d) A nonideal capacitor has a leakage resistance ESR : equivalent series resistance 9 6. The Inductor Circuit symbol and component model. λ 0 i t () ( ) λ @ i, λ t v τ dτ = λ : flux linkage, in web-turns : inductance, in H (Henry) 0
6. The Inductor t t0 t i = v( ) d v( ) d v( ) d τ τ = τ τ τ τ + t0 t i() t = i( t0) + v( ) d ( A) τ τ t0 dλ di() t v() t = = ( B) dt dt di() t p() t = vi = i() t dt Energy in an inductor t wt () = p( τ) dτ = i () t 0 Example : An Inductor 6. The Inductor i l A H : magnetic field intensity B : flux density ur ur φ: B da flux uuv r Hdl = i i Hl = i, H = l µ i B = µ H =, µ permeability of the core l
Example : (cont.) 6. The Inductor φ = BA = µ Ai l λ = φ = µ A i l = λ µ A = i l,flux linkage i l A 3 6. The Inductor The unit of inductance is the henry (H), named in honor of the American inventor Joseph Henry (797-878) 878). H = volt-second / ampere (a) when i is constant, then v =0. i.e., equivalent to short circuit (b) i (t) is a continuous function if there is only finite strength source inside the circuit. i t i t v dt t + ( + ) = () + ( τ ) t ( + ) = () ( + ) = ( ) lim i t i t, or i 0 i 0 0 () () ie.. i t can not change abruptly for finite v t. 4
6. The Inductor (c) An ideal inductor does not dissipate energy. It stores energy in the magnetic field. (d) A nonideal inductor contains winding resistance and parasitic capacitance. 5 6. The Inductor Example 3 : Under dc and steady state conditions, find (a) I, V C & I, (b) W C and W I = I = = A + 5 V = 5I = 0V C W C 0 50 = = J W 4 = = J 6
6. The Inductor (a) (b) The capacity of C and to store energy makes them useful as temporary voltage or current sources, i.e., they can be used for generating a large amount of voltage or current for a short period of time. The continuity property of V C (t) and i (t) makes inductors useful for spark or arc suppression and for converting pulsating voltage into relatively smooth dc voltage. 7 6. The Inductor (c) The frequency sensitive property of and C makes them useful for frequency discrimination. (eg. P, HP, BP filters) 8
6.3 Series-Parallel Combinations of Capacitance and Inductance capacitors in parallel i i i i v c c c v (0) = v (0) = = v (0) v = v = = v = v Qi = i + i + + i eq dv dv dv = c + c + + c dt dt dt dv dv = ck = Ceq k = dt dt C = c + c + + c v(0) = v (0) k 9 6.3 Series-Parallel Combinations of Capacitance and Inductance capacitors in series i = i = = i Qvk t i d vk t c () = ( τ) τ + ( ) k t0 t v= i( τ) dτ + vk( to) c k= k t k= 0 t = i d + v t C eq t0 t ( τ) τ ( ) = + + + C C C C eq ( ) = ( ) + ( ) + + ( ) v t v t v t v t 0 0 0 0 0 0 0
6.3 Series-Parallel Combinations of Capacitance and Inductance inductors in series i (0) = i (0) = = i (0) i () t = i (t)= = i () t = it () Qv= v + v + + v eq di di di = + + + dt dt dt di di = k = eq k= dt dt = + + + i(0) = i(0) = i (0) = i (0) 6.3 Series-Parallel Combinations of Capacitance and Inductance inductors in parallel Qv = v = = v = v i = i + i + + i t t t i= it + vdτ+ i t + vdτ+ + i t + vdτ ( ) ( ) ( ) 0 0 0 t0 t0 t0 t = vdτ + t = vdτ + it ( 0) t0 t k 0 k= k k= eq = + + + eq ( ) ( ) = ( ) + ( ) + + ( ) 0 0 0 0 i t it it i t i t 0
6.3 Series-Parallel Combinations of Capacitance and Inductance In summary V-I I-V P or W series parallel dc case Resistor Capacitor Inductor V = RI I = V R V P = = I R R R = R + R R eq eq RR = R + R same t v = vt ( 0 ) + idt C t 0 dv i = C dt W = Cv CC Ceq = C + C C = C + C eq open circuit di v = dt t i = it ( 0 ) + v d τ t0 W = i = + 3 eq eq = + short circuit Circuit symbol and model of coupling inductors + v i M i di dt di v = M + dt v = + M + v di dt di dt, : self inductances M : mutual inductance unit : in H 4
Example 4 : Mutual inductance Apply I, with i =0 Hdl= ur v I 5 Assume uniform magnetic field intensity H I µ I H =, B = µ H = l l µ A I φ = B da = l µ A I µ A I λ = φ = ; λ = φ = l l λ µ A λ µ A @ =, M @ = I l I l 6
Dot convention for mutually coupled inductors: When the reference direction for a current enters the dotted terminal of a coil, the polarity of the induced voltage in the other coil is positive at its corresponding dotted terminal. 7 V >0 V i, v >0, φ, ( i =0 ) Another dot of coil should be placed in terminal c. 8
V >0 V Conceptually, one can connect a resistor at cd terminals. Then i will be negative. The generated flux of i will oppose the increasing of φ due to increasing i ( entz law ). Hence, another dot should be placed at c terminal. 9 V >0 V In case, the other dot is placed at d terminal, then i will be positive. Hence, the generated flux of i will be added to the increasing φ due to i. 30
V >0 V Then the induced voltage at coil two will increase and so will i. This will violate the conservation of energy. 3 The procedure for determining dot markings Step Step Step3 Assign current direction references for the coils. Arbitrarily select one terminal of one coil and mark it with a dot. Use the right-hand hand rule to determine the direction of the magnetic flux due to the current of the other coil. 3
Step4 If this flux direction has the same direction as that of the first dot terminal current, then the second dot is placed at the terminal where the second current enters. Otherwise, the second dot should be placed at the terminal where the second current leaves. 33 Example 5 : Determining dot markings Step : Assign i, i, i 3 directions. 34
Example 5 : (cont.) Step : For coils and, choose first dot as follows 35 Example 5 : (cont.) For coils and 3 36
Example 5 : (cont.) For coils and 3 37 Example 5 : (cont.) Step 3 : Check the relative flux directions and determine the dot position at the other coil For coil and and φ are in the same direction φ M 38
Example 5 : (cont.) For coil and 3 φ 3 and φ are in opposite direction 3 39 Example 5 : (cont.) For coil and 3 φ and 3 φ are in opposite direction 40
Example 5 : (cont.) In summary di di di3 v = + M M3 dt dt dt di di di3 v = M + M3 dt dt dt di di di3 v3 = M3 M3 + 3 dt dt dt + M i i v v + M 3 M 3 + i 3 3 v 3 _ 4 If the physical arrangement of the coils are not known, the relative polarities of the magnetically coupled coils can be determined experimentally. We need a dc voltage source V S a resistor R : to limit the current a switch S a dc voltmeter 4
Step : Assign current directions and arbitrarily assign one dot at coil one. 43 Step : Connect the setup as follows. 44
Step 3 : Determine the voltmeter deflection when the switch is closed. If the momentary deflection is upscale, then If the momentary deflection is downscale, then 45 The reciprocal property of the mutual inductance can be proved by considering the energy relationship. Step : i =0, i increased from zero to I. + v M i i + v di v = dt di = v M dt input power p = vi + vi input energy w di = ( ) i dt 0 I = idi 46 0 = I = t pdτ
Step : keep i =I, i is increased from zero to I Input power Input energy di di di v = + M = M dt dt dt di di di v = M + = dt dt dt p = vi + vi w di di = M I + ( ) i dt dt = pdτ I I = M Idi + 0 0 = M II + I idi 47 Total energy when i =I, i =I w= w + w = I + I + IIM Similarly, if we reverse the procedure, by first increasing i from zero to I and then increasing i from zero to I, the total energy is w= w+ w = I + I + IIM Hence, M =M =M 48
Definition of coefficient of coupling k k @ M 0 k 0 < k <, loosely coupling k <, closely coupling k =, unity coupling 49 According to dot convention chosen, the total energy stored in the coupled inductors should be w= i + i ± Mii 0 In particular, consider the limiting case i + i Mii = 0 The above equation can be put into the following form ( i i) + ii ( M) = 0 50
Thus, w(t) 0 0 only if M when i and i are either both positive or both negative Hence, k k 5 Example 6 : Finding mesh-current equations for a circuit with magnetically coupled coils. i g i 5Ω i g 0Ω i i 4H 60Ω Three meshes and one current source, only need two unknowns, say i and i i 6H ote : current i 4H = i (t) current i 6H = i g - i 5
Due to existence of mutual inductance M=8H, there are two voltage terms for each coil. One can use dependence source to eliminate the coupling relation as follows. d i dt 8 6H 5Ω i 0Ω i g i g 8 di dt 53 i 60Ω 4H d i dt +- 8 6H 5Ω i 0Ω i g i g 6H + - 8 di dt i 60Ω Hence, for i mesh di d i dt dt i i i i i For i mesh 4 + 8 ( g ) + 0( ) + 5( g) = 0 d d 0( i i) + 60i + 6 ( i ig) 8 i = 0 dt dt 54
Summary Objective : Know and be able to use the component model of an inductor Objective : Know and be able to use the component model of a capacitor. Objective 3 : Be able to find the equivalent inductor (capacitor) together with it equivalent initial condition for inductors (capacitors) connected in series and in parallel. 55 Summary Objective 4 : Understand the component model of coupling inductors and the dot convention as well as be able to write the mesh equations for a circuit containing coupling inductors. Chapter Problems : 6.4 6.9 6. 6.6 6.40 6.4 Due within one week. 56