Mutually coupled inductors. Coupling coefficient. Power ...

1 Lecture notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris EvstatievMutually coupled inductors . Coupling coefficient. Power and energyof Mutually coupled inductors . Analysis of circuits with mutuallycoupled Equivalent circuits of Mutually coupled inductorsAs was already mentioned in the second topic, when the magnetic field of one coil reaches a secondone the two inductors are Mutually coupled and are characterized by a coefficient of mutualinductance M. Depending on the connection between inductors there are a number of equivalentcircuits which could be used to simplify the circuit Mutually coupled inductors in seriesConsider there are two inductors L1 and L2 in series, which are magnetically coupled andhave a mutual inductance M. The magnetic field of the two inductors could be aiding oropposing each other, depending on their orientation (fig ). a)b)Fig. Mutually coupled inductors and dot convention: a) series aiding inductors ; b) seriesopposing inductors in series as well as the dot convention are presented in fig.

2 Since theirmagnetic fields aid each other, the KVL for this situation is:v=v1+vM+v2+vM=L1diLdt+MdiLdt+L2diLdt+ MdiLdt=(L1+L2+2M)diLdtIn the above equation vM is the voltage drop caused by the mutual inductance M. There aretwo voltage drops vM: one is produced by the additional magnetic flux coming from L1 toL2 and the second one by the magnetic flux coming from L2 to L1. Then the equivalentinductance is:LE=L1+L2+2 MThe second situation is when the magnetic field of the two coils oppose each other (fig. ). Then1 Lecture notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris Evstatievthe KVL law is:v=v1 vM+v2 vM=L1diLdt MdiLdt+L2diLdt MdiLdt=(L1+L2 2M)diLdtAnd the equivalent inductance is:LE=L1+L2 Mutually coupled inductors in parallelConsider two inductors L1 and L2 are connected in parallel and are Mutually coupled withmutual inductance M. The voltage applied on the inductors is vS. Once again there are twopossibilities: the magnetic fields of the two inductors could be aiding or opposing each other.

3 A)b)Fig. Mutually coupled inductors in parallel: a) aiding inductors ; b) opposing , we accept that the inductors are aiding each other (fig. ) and write down two KVLequations:|vS=L1di1dt+Mdi2dtvS=L2di2d t+Mdi1dtNext, the above system is written in matrix form:|vSvS|=|L1 MML2|.|di1dtdi2dt|The determinants are: =|L1 MML2|= M2 1=|vSMvSL2|=vS(L2 M) 2=|L1vSMvS|=vS(L1 M)The solutions for the current derivatives are:di1dt= 1 di2dt= 2 The KVL for the circuit is:i=i1+i22 Lecture notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris EvstatievThe above equation is differentiated:didt=di1dt+di2dt= 1+ 2 =vS(L2 M)+vS(L1 M) M2=vSL1+L2 M2 Then the equivalent inductance is:vS(t)=LEdi(t)dt= M2L1+L2 (t)dtorLE= M2L1+L2 a similar manner can be proven that the equivalent inductance of parallel inductors whosemagnetic fields are oppose each other, is:LE= M2L1+L2+ Elimination of mutual inductanceConsider the case when two coils are Mutually coupled (fig.

4 They could be replaced with anequivalent circuit without mutual inductance and three coils as shown in the figure. To prove thiswe write the system of equations for the original circuit:|v1=L1di1dt+Mdi2dtv2=L2di2dt+Mdi 1dtThen we write the system of equations for the equivalent circuit:|v1=(L1 M)di1dt+Mdi0dtv2=(L2 M)di2dt+Mdi0dtConsidering di0dt=di1dt+di2dt we obtain the same system of equations as for the original circuit:|v1=(L1 M)di1dt+Mdi1dt+Mdi2dt=L1di1dt+Mdi2dtv2=( L2 M)di2dt+Mdi1dt+Mdi2dt=L2di2dt+Mdi1dtIn a similar manner we can prove the circuit for the Mutually opposing )3 Lecture notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris Evstatievb)Fig. Equivalent circuit for: a) aiding inductors ; b) opposing that the nodes 1 and 2 of the original circuit may or may not be connected and that doesn tchange the equivalent General caseConsider two inductors with mutual inductance which may or may not have an electric the two inductors are aiding each other, the equivalent circuit is presented in fig.

5 The mutualinductance is replaced by two dependent sources Mdi2dt and Mdi1dt, which are opposing thedirection of the KVL eqations for both the original and the equivallent circuit is:|v1(t)=L1di1dt+Mdi2dtv2(t)=L2di2dt+Md i1dtNext consider the two inductors are opposing each other (fig. ). The equivallent circuit is thesame however this time dependent sources are with a negative sign: Mdi2dt and notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris EvstatievFig. KVL eqations for both the original and the equivallent circuit is:|v1(t)=L1di1dt Mdi2dtv2(t)=L2di2dt Mdi1dtIf the above time domain variables are replaced with phasors, the dependent sources in theequivallent circuits becomes j MI 1 and j MI 2 for aiding inductors and j MI 1 and j MI 2 for opposing inductors (fig. ).Fig. KVL eqations in complex form for the circuits are:|V 1=j L1I 1 j MI 2V 2=j L2I 2 j MI 1where the plus sign corresponds to aiding inductors and the minus sign to opposing Energy in Mutually coupled inductorsIt was already demonstrated in the second topic that the energy stored in an inductor is:WL= s consider two Mutually coupled inductors (fig.)

6 The Power transferred from the first to thesecond coil is:pM12(t)= notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris EvstatievFig. we integrate the Power from t1 to t2 the energy is:WM12= t1t2pM12(t)dt= Power transferred form the second to the first coil is:pM21(t)= the transferred energy is:WM12= t1t2pM21(t)dt= energy stored in the two coils due to their self-inductance is:WL1= ,WL2= the total energy stored in two Mutually coupled inductors is:W= + plus sign corresponds to aiding inductors and the minus to opposing energy usage is always positive we can rewrite the above equation (use equation for opposinginductors) + 0or12( L2i2)2+i1i2( L1L2 M) 0 Considering ( L2i2)2 is always positive or zero, then the second term is also greater orequal to zero: L1L2 M 0 This way is defined the Coupling coefficient k:k=M L1L2 The Coupling coefficient takes values in the range 0 k 1 and shows how good the couplingbetween the two coils is.

7 For coils which are not coupled , k=0 and in case of ideal Coupling (only possible in theory) k= Analysis of circuits with magnetically coupled Kirchhoff s laws analysisThe analysis of circuits with magnetically coupled inductors could be achieved using an equivallentcircuit without magnetic couples. The analysis using the Kirchhoff s laws includes creating theequivallent circuit and analyzing it by writing a system of equation whose number is equal to thenumber of unknown notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris EvstatievExample: Estimate the currents for the circuit in fig. using the Kirchhoff s can be seen that the mutual inductance j M is given as mutual resistance:j M= First we are going to create an equivallent circuit by replacing the mutual inductance withdependent source (fig. ). Since both currents I 1 and I 2 enterthe dots, the dependentsources are with plus we write the system of equations:|I 1=I 2+I 31 3=10I 1+(1+j2)I 3 2= (1+j2)I 2+0I 3 Fig.

8 Matrix form it becomes:[I 1I 2I 3][ 111101+ + ]=[010]7 Lecture notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris EvstatievThe determinants are: =[ 111101+ + ]= (1+j2)+10(1+ )+ (1+ ) 10== +1+10+j15+ j5= +j10 1=[01111+ + ]=1+ +j 2=[ ]= 3=[ 110101+j2101+ ]=1+ the currents are:I 1= 1 =1+ +j10= + 2= 2 = +j10= 3= 2 =1+ +j10= + Nodal analysisThe nodal analysis cannot be applied if dependent sources are used. However in certain situationswhere the mutual inductance could be eliminated, the nodal analysis can be : Estimate the currents for the circuit in fig. using nodal the two coils have a common end, the mutual inductance could be replaced with anequivallent circuit with three coils (fig. ). Their resistances are:j L1=j2 j L2=j3 j L3= Fig. we can apply nodal voltage analysis. The KCL is:8 Lecture notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris EvstatievI 1=I 2+I 3= 1 V 110+ 11+ +V 1 node voltage then becomes:V 1=110+ + +110+ + the currents are:I 1= + + + 2= + 3= + Mesh analysisThe mesh analysis method can be applied for analysis of circuits with mutual indcutance.

9 Howeverin order to do that the dependent sources should be expressed with the mesh currents. Once this isdone the equations are written according to : Estimate the currents for the circuit in fig. using mesh we should estimate the branch currents with the mesh currents:I 1=I 1(k)I 2=I 1(k) I 2(k)I 3=I 2(k)Then the equivallent dependent sources become (fig. ) 2= (I 1(k) I 2(k)) 3= 2(k)Fig. two KVL equations are:9 Lecture notes in Theory of electrical engineering. Assoc. Prof. Dr. Boris Evstatiev1 2(k)=I 1(k)(10+1+j2) I 2(k)(1+j2)1=I 1(k)(11+j2) I 2(k)(1+ ) 2(k) 1(k)+ 2(k)=I 2(k)(1+j2+j3 j3) I 1(k)(1+j2)0= I 1(k)(1+ )+I 2(k)(1+j)In matrix form the equations are:[I 1(k)I 2(k)][11+j2 (1+ ) (1+ )1+j]=[10]The determinants are: =[11+j2 (1+ ) (1+ )1+j]=11+j11+j2 2 1 + +j10 1=[1 (1+ )01+j]=1+j 2=[11+j21 (1+ )0]=1+ the mesh currents are:I 1(k)= 1 =1+ +j10= + 2(k)= 2 =1+ +j10= + the branch currents are:I 1=I 1(k)= + 2= + 3= + Alexander Ch.

10 , Sadiku M. Fundamentals of electric circuits. Fifth edition. McGraw-Hill. Hayt W., Buck J. Engineering electromagnetics. Sixth edition. McGraw-Hill. 2001. Chapters 5and