# Inductance and Transient Circuits

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1 Chaper H Inducance and Transien Circuis Blinn College - Physics Terry Honan As a consequence of Faraday's law a changing curren hrough one coil induces an EMF in anoher coil; his is known as muual inducance. Similarly, a changing flux in a coil induces an EMF in he same coil; his is self inducance and a circui componen wih inducance is called an inducor. We will also discuss simple circuis wih inducors combined wih our oher linear circui elemens: resisors, capaciors and DC volage sources. H. - Muual and Self Inducance Muual Inducance Consider a pair of coils, one called he primary coil and he oher called he secondary. A curren I hrough he primary creaes a magneic field B which in urn creaes a magneic flux hrough he secondary coil F 2. I ï B ï F 2 A changing curren creaes a changing flux which induces an EMF in he secondary coil. I ï F 2 ï E 2 The above relaionships are proporionaliies. Define he consan of proporionaliy as he muual inducance M 2. E 2 = -M 2 I is beyond he scope of his class o prove ha he muual inducance of coil on coil 2 is he same as ha of 2 on, and his will jus be called M. I E 2 = -M I where M = M 2 = M 2 We will revisi muual inducance in he nex chaper in he conex of a ransformer. A ransformer is he case where, ideally, all he flux from one coil passes hrough he oher. Self Inducance There is also inducance of a coil on iself. This is called self inducance. When we use he erm inducance by iself self inducance is implied. A curren hrough a coil creaes a field and ha causes a flux hrough he coil iself. I ï B ï F A changing curren creaes a changing flux wih induces an EMF in he coil. I ï F ï E The above relaionships are proporionaliies. The consan of proporionaliy is defined as he inducance L. E = -L I The sign in he above expression is due o Lenz's law. Take DV o be he change in he volage when moving hrough he inducor in he direcion of he curren. A simple Lenz's law analysis shows ha if he curren is increasing he volage change is negaive. If we wrie V as he volage drop we ge

2 2 Chaper H - Inducance and Transien Circuis DV = -L I The sign convenions for inducors is he same as ha for resisors and capaciors and V = L I. DV = -I R and V = I R DV = - Q C and V = Q C Inducance of a Long Solenoid Consider a long solenoid of lengh {, wih N urns and a cross-secional area A. As before we define n as he number of urns per lengh n = N ê{. Passing from he curren o he field o he flux gives and using Faraday's law we ge I ï B = m 0 n I ï F = B A = m 0 n I A E = -N F = -N m 0 n A I. We can read he inducance from his expression. Wriing L in erms of boh N and n gives H.2 - Energy Consideraions L = m 0 n 2 A { = m 0 N 2 { A. Energy in an Inducor Inducors, like capaciors, sore energy, while resisors dissipae energy. Use U o denoe he energy in an inducor. The rae ha energy is being sored in an inducor is Inegraing he above expression gives U = P = I V = I L I. U = 2 L I2 + consan. Choosing U = 0 when I = 0 fixes he consan and gives he desired expression. U = 2 L I2 Energy in a Magneic Field In he capaciance chaper we derived an expression for he energy densiy (Energy/Volume) in an elecric field. u = 2 ε 0 E 2 To derive his we used he fac ha he elecric field is uniform inside a parallel plae capacior. Combining expressions for he energy in a capacior and for he capaciance gave he above expression for u. A similar analysis will give he energy densiy in a magneic field. The magneic field is uniform inside a long solenoid. Combining he expression for he inducance of a long solenoid wih he energy in an inducor gives U = 2 L I2 and L = m 0 n 2 A { ï U = 2 m 0 n 2 A { I 2 Using B = m 0 n I and u = U êvolume = U êha { L gives he energy densiy in a magneic field.

3 Chaper H - Inducance and Transien Circuis 3 u = 2 m 0 B 2 H.3 - Transien Circuis and Differenial Equaions The volage drops across a resisor, capacior and inducor are, respecively, V = I R, V = Q C and V = L I. Remember ha hese are volage drops, so D V = -V. If we build a circui ou of hese and dc volage sources, where D V = E, we hen ge an equaion for I and Q. Since I is he ime derivaive of Q his will give a differenial equaion for Q or for I. I = Q Commens on ODEs (Ordinary Differenial Equaions) A differenial equaion (DE) is some equaion involving a funcion and is derivaives. The differenial equaion is solved o find he funcion. The order of a DE is he highes number of derivaives. If here is a mos a second derivaive i is a second order equaion. If i is a funcion of one variable i is an ordinary differenial equaion (ODE). For funcions of several variables here are parial differenial equaions (PDE). For funcions of more han one variable we ake parial derivaives insead of ordinary ones. The differenial equaions course (aken afer Cal. III) is on ODEs. The general soluion of a p h order ODE is any soluion involving p independen arbirary consans. I is easy o verify ha somehing is a soluion o a differenial equaion; i is jus a maer of aking derivaives and plugging ino an equaion. If a soluion has he correc number of arbirary consans hen we can conclude ha his is he general soluion. H.4 - RC Circuis When he swich is hrown o he Charging posiion he curren flows from he baery o charge he capacior. In he Discharging posiion he charge flows from he capacior and is energy is dissipaed in he resisor. The charge on he capacior is relaed o he curren in he wire by I = Q. Noe ha when he capacior is discharging he charge is decreasing and hus he curren is negaive. C R Discharging E Charging Discharging For he discharging case, applying he loop rule o he circui gives:

4 Chaper H - Inducance and Transien Circuis 0 = I R + Q C. Using he fac ha he curren is he derivaive of he charge we can rewrie his as a firs order differenial equaion. where is defined as he ime consan 0 = R Q + C Q or Q = - Q = R C. Since he derivaive of an exponenial is iself we can guess a soluion of e -ê and i is easy o verify ha his is a soluion. If we muliply his by a consan a hen i sill is a soluion QHL = a e -ê. Since a is arbirary and we are beginning wih a firs order ODE we can conclude ha his is he general soluion. If we define he iniial charge as Q 0 hen Q 0 = QH0L = a and he soluion becomes QHL = Q 0 e -ê. Q Q 0 - Q 0 Ineracive Figure - Discharging an RC Circui Charging The loop rule for he charging case gives: 0 = E - I R - Q C and we can rewrie his as a firs order differenial equaion. E = R Q + C Q Using he same ime consan we can guess a soluion of he form QHL = a e -ê + b. Insering his ino our differenial equaion gives E = -R a e-ê + C Ia e-ê + bm The erms involving a cancel and he equaion requires ha b has he value b = E C for i o be a soluion. This means ha QHL = a e -ê + E C is a soluion o he differenial equaion. Since i is a firs order equaion and a is an arbirary consan we can conclude ha i is he general soluion. The soluion we desire is where a = 0 he capacior is uncharged QH0L = 0, giving QHL = Q I - e -ê M where Q = E C.

5 Chaper H - Inducance and Transien Circuis 5 Q = E C Q H- - LQ Ineracive Figure - Charging an RC Circui H.5 - RL Circuis L R B E A Decaying Curren Begin wih swich A closed and B opened. This creaes a curren hrough he inducor and resisor. Close swich B and hen open A. This causes he curren o flow hrough he op branch of he above circui. Applying he loop rule around he circui gives 0 = L I + R I. This is a firs order ODE (ordinary differenial equaion) similar o ha of a discharging capacior where he ime consan is defined by I = - I, = L R. The general soluion of his differenial equaion for he curren IHL is IHL = I 0 e -ê, where he arbirary consan is labelled I 0 because i is he curren a ime zero. This is a simple exponenial decay, analogous o he decay of he charge for a discharging capacior.

6 Chaper H - Inducance and Transien Circuis The iniial energy in he inducor U = 2 L I 0 2 is convered o hea in he resisor. I I 0 - I 0 Ineracive Figure - Curren Decay in an RL Circui Growing Curren Wih boh swiches opened giving zero curren, close swich A a = 0. This causes he curren o gradually build up o a seady-sae value. Apply he loop rule o he circui gives he firs order ODE. Guess a soluion of he form Insering his guess ino he differenial equaion gives E = L I + R I. IHL = a e -ê + b. E = L - a e-ê + R Ia e -ê + bm. The definiion of he ime consan gives a cancellaion of he e -ê erms for any value of a, making a our arbirary consan. Our guess will only be a soluion when b has he value b = E R. Since we have a soluion wih one arbirary consan and i is a firs order ODE we can conclude ha IHL = a e -ê + E R is he general soluion. We are looking for a soluion where he iniial curren is zero IH0L = 0, giving The growh of he curren can be wrien as a = - E R. is he seady-sae curren. IHL = I I - e -ê M where I = E R

7 Chaper H - Inducance and Transien Circuis 7 I = E êr I H- - LI Ineracive Figure - Curren Growh in an RL Circui H.6 - LC Circuis, RLC Circuis and Their Mechanical Equivalens There is an imporan analogy beween he LC Circui and he mass-spring sysem. We will see ha he soluion o boh equaions is oscillaory. We can add damping o hese cases by insering a resisor ino he elecrical sysem and adding fricion o he mechanical sysem. The LC Circui C L Consider a simple loop circui conaining jus a capacior C and inducor L. The loop rule gives 0 = L I + Q C. Wrie he curren I as he ime derivaive he charge on he capacior and define he angular frequency by I = Q This gives he second order ODE w 0 = L C. 0 = 2 Q 2 + w 0 2 Q. The Mass-Spring Sysem Ineracive Figure

8 8 Chaper H - Inducance and Transien Circuis The mechanical analog of his is a mass-spring sysem. The force of a spring is given by Hooke's law F = -k x. Applying Newon's second law gives a second order ODE Defining he angular frequency by F ne = m a ï -k x = m 2 x 2 w 0 = k m, gives he second order ODE 0 = 2 x 2 + w 0 2 x. The Analogy wihou Damping Summarizing he analogy as saed above: The charge, curren and derivaive of he curren are he analogs of he posiion, velociy and acceleraion. The energy in he inducor U L = 2 L I2 and he kineic energy of he mass K = 2 m v2 are analogous, as are he energy in he capacior U C = 2 C Q2 and he poenial energy of he spring U = 2 k x2. LC Circui Q I = Q I L C Mass Spring x v = x a = 2 x 2 m k w 0 = L C w 0 = k m U L = 2 L I2 K = 2 m v2 U C = 2 C Q2 U = 2 k x2 Soluion of he Undamped ODE Wih he analogy clearly saed, le us solve he second order ordinary differenial equaion 0 = 2 Q 2 + w 0 2 Q. Since he second derivaives of boh sine and cosine are he negaives of hemselves, i follows ha cos w 0 and sin w 0 are soluions o our differenial equaion. Because he ODE has he simple properies (a homogeneous linear equaion) ha: i follows ha HiL a consan imes a soluion is a soluion and HiiL he sum of wo soluions is a soluion, QHL = B cos w 0 + C sin w 0 is a soluion, where B and C are arbirary consans. Since we have soluion o a second order ODE wih wo arbirary consans, we can conclude ha his is he general soluion. Seing QH0L = Q 0 and IH0L = I 0 where I = Qê gives B = Q 0 and C = I 0 ê w. Anoher way of presening his soluion is QHL = Q max coshw 0 + fl where he arbirary consans are Q max, he ampliude, and f, he phase angle.

9 + 6 p 2 Chaper H - Inducance and Transien Circuis 9 Q Q max I 0 Q 0 2 p T = w Ineracive Figure - Charge as a Funcion of Time for an LC Circui LCR Circui C R L We can add damping o he LC circui by adding a resisor R o he series circui. The loop rule gives This gives us he second order ODE where w 0 is he same as in he undamped case and 0 = L I + I R + Q C. 0 = 2 Q + b Q + w 2 0 Q 2 b = R L. The Damped Mass-Spring Sysem The mechanical damping is achieved by adding viscous fricion, which is he force -bv, o he Hooke's law force. Newon's second law gives This becomes he second order ODE F ne = m a ï -k x - b v = m 2 x 2. where w 0 is he same as in he undamped case and now 0 = 2 x + b x 2 + w 0 2 x b = b m.

10 0 Chaper H - Inducance and Transien Circuis The Analogy wih Damping We can add o he previous able he values of he damping consans b. LCR Circui R b = R L Damped Mass Spring b b = b m Noe he similariies beween resisance and fricion. Boh remove energy from he sysem, fricion removes mechanical energy and resisance removes elecrical energy. The energy los o each goes o hea. Soluion wih Damping 0 = 2 Q + b Q + w 2 0 Q 2 To solve he above differenial equaion we will guess a soluion of he form QHL = A e -g coshw + fl. This will be a soluion only when he consans g and w have specific values, which we will wrie in erms of he consans in he equaion, b and w 0. The consans A and f will be our arbirary consans. To verify his is a soluion and find g and w, we mus plug our guess ino he differenial equaion. Firs we mus evaluae he derivaives 2 Q Q = A e -g coshw + fl. Q = A e coshw + fl - w sinhw + fld. 2 = A e-g AIg 2 - w 2 M coshw + fl + 2 g w sinhw + fle. 0 = 2 Q + b Q + w 2 0 Q ï 2 0 = A e -g AIg 2 - w 2 M coshw + fl + 2 g w sinhw + fle +A e - b g coshw + fl - b w sinhw + fld +A e -g A w 0 2 coshw + fl + 0 E For his equaliy o be rue a all imes, he erms muliplying coshw + fl and sinhw + fl mus separaely vanish, giving The second expression gives 0 = g 2 - w 2 - b g + w 0 2 and 0 = 2 g w - b w. and plugging his ino he firs expression gives g = b 2 w = w b2 4.

11 Chaper H - Inducance and Transien Circuis A Q 2 p w 0 -A Ineracive Figure - Charge as a Funcion of Time for an LCR Circui

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