EE 05 Dr. A. Ziduri Electric Circuits II Mre abut Mutual Inductance Lecture # - -
EE 05 Dr. A. Ziduri The material t be cvered in this lecture is as fllws: Mutual Inductance in Terms f Self Inductance Mesh Current Analysis f magnetically cupled Circuits Prcedure fr Determining Dt Markings Energy Calculatin After finishing this lecture yu shuld be able t: Analyze Circuits with Mutual Inductance Determine the Cefficient f Cupling Determine the Dt Markings f neighbring Cils Perfrm Energy Calculatins in Magnetically Cupled Circuits - -
EE 05 Dr. A. Ziduri Mutual Inductance in Terms f Self Inductance We have seen that L =Np Therefre But fr linear system, L =Np LL =NNpp =NN (pp )(pp ) p =p p p LL =NNp + + p p p p = M + + p p p p Replacing the psitive quantity + + by gives a mre meaningful expressin p p k M=k LL M =k LL r - 3 -
EE 05 Dr. A. Ziduri k is called the Cefficient f Cupling and must lie between 0 and : 0 k When Φ = Φ =0 p =0 =,r k=0 k If n flux linkage between the cils, bviusly M=0 and k=, when Φ = Φ =0, which presents the Ideal State. - 4 -
EE 05 Dr. A. Ziduri Mesh Current Analysis T slve the fllwing example 3-, Let us reprduce the Mesh Current Equatins that describe the fllwing Circuit seen in Lec3: + + di di di di M L L M dt dt dt dt + + di di v + Ri + = g L M 0 dt d t + di di M = 0 Ri L dt (Mesh ) (Mesh ) d t - 5 -
EE 05 Dr. A. Ziduri Example 3- Write a Set f Mesh-Current Equatins that describe the Circuit in the fllwing Figure: What is the Cefficient f Cupling f the Magnetically Cupled Cils? + 4.5 di dt + ( ) 4.5 di + i dt di ( + i3) di vg + 8( i i) + 9 4.5 = 0 (Mesh ) dt dt di di ( + i3) 6( i + i3) + 8( i i) + 4 4.5 = 0 (Mesh ) dt dt di d( i + i3) di 0i3 + 6( i + i3) + 4 + 9 4.5 = 0 dt dt dt (Mesh 3) M 4.5 4.5 k= = = =0.75 LL 4 9 6-6 -
EE 05 Dr. A. Ziduri Prcedure fr Determining Dt Markings The fllwing six steps applied t the fllwing figure determine a set f Dt Markings. ) Arbitrarily Select ne Terminal, (Say Terminal D) f ne cil and mark it with a Dt ) Assign a current int the dtted terminal and label it i D 3) Use the Right Hand rule t determine the directin f the magnetic field established by i D inside the cupled cils and label this as Φ D 4) Arbitrarily pick ne terminal f the secnd cil (Say Terminal A) and Assign a current int it (Shw this as i A ) 5) Use the Right Hand rule t determine the directin f the magnetic field established by i D inside the cupled cils and label this as Φ A 6) Cmpare the directins f the tw fluxes Φ D and Φ A. If the fluxes have the same reference directin, place a Dt n the terminal f the secnd cil where the test current (i A ) enters. In this example the fluxes Φ D and Φ A have the same reference directin, therefre a Dt ges n terminal A. B A (Step 4) (Step 6) (Step 5) Φ A Φ D (Step 3) C (Step ) D Arbitrarily Dtted Terminal (Step ) Fig.3- Set f Cils Shwing Steps fr Determining a Set f Dt Markings - 7 -
EE 05 Dr. A. Ziduri Example 3- The plarity markings n tw cils are t be determined experimentally. The experimental setup is shwn in Fig.3-3. When the switch is pened, the dc Vltmeter kicks upscale. Accrding t the plarity and dt marking shwn, where shuld we place the ther dt? Answer: The lwer terminal f the unmarked cil has the same instantaneus plarity as the dtted terminal. Therefre, place a dt n the lwer terminal f the unmarked cil. - 8 -
EE 05 Dr. A. Ziduri Energy Calculatins Fr linear magnetic cupling M =M M=k LL Cnsider the fllwing Circuit Case : Increase i frm 0 t I, hld i = I and i = 0 w dw = L 0 0 W= LI I di Case : Nw we hld i = I and Increase i frm 0 t I p =IM di +i v dt - 9 -
EE 05 Dr. A. Ziduri Energy Calculatins (cnt) The ttal energy stred in the pair f cils when i = I w I I w 0 0 dw = I M di + L i di W=W+ LI +MII W= LI + LI + MII If we reverse the prcedure That is, if we first increase i frm zer t I and increase i frm zer t I - The ttal energy stred is: W= LI + LI + MII At any instant f time, the ttal energy stred in the cupled cils is: w(t)= L i + L i + Mi i In general: w(t)= Li + L i ± Mi i Assuming that bth cil currents entered plarity marked terminals - 0 -
EE 05 Dr. A. Ziduri Example 3-3 Cnsider the circuit in Fig.3-5. Determine the cupling cefficient. Calculate the energy stred in the v = 60 cs 4t + 30 V. cupled inductrs at time t = s if ( ) F 6 Slutin: 60 cs( 4t + ) The cupling cefficient is M.5 k= = =0.56 LL 0 Means that the inductrs are tightly cupled. T find the energy stred, we need t btain the frequency dmain equivalent f the circuit 30 V 60 30, ω = 4 rad 5H jωl = j0ω.5h jωm = j0ω 4H jωl = j6ω 4 6 F = jωc j Ω The frequency dmain equivalent is shwn in Fig.3-6 belw: s - -
EE 05 Dr. A. Ziduri Slutin (Cnt): We nw apply mesh analysis This yields In the time dmain i i =3.905 cs 4t =3.54 cs 4t (0 + j0) I + j0i = 60 30 j0 I + ( j6 j4) I = 0 I Or =.I I ( j4) = 60 30 I = 3.54 60.6 A ( 9.4 ), ( + 60.6 ) I =.I = 3.905 9.4 A At time t=s, 4t= 4rad=9.º, i = 3.905 cs 9. 9.4 = 3.389 A, i ( ) ( ) = 3.54 cs 9. + 60.6 =.84A - -
EE 05 Dr. A. Ziduri Slutin (Cnt): The ttal energy stred in the cupled inductrs is w= Li + Li + Mii w= Li + Li + Mii = (5)( 3.389) + (4)(.84) +.5( 3.389)(.84) = 0.73J 60 30 V - 3 -
EE 05 Dr. A. Ziduri Self Test: When k equals zer, tw inductrs in series cmbine by a) additin b) subtractin c) prduct-ver-sum d) multiplicatin answer: a Tw parallel inductrs with n mutual inductance cmbine by a) additin b) subtractin c) prduct-ver-sum d) multiplicatin answer: c If N=00 and a rate f change f flux is 0.4 Wb/s. The vltage induced in the inductr is? a) 50 b) 80 c) 00 d) 500 answer: b Tw inductrs are wund n a ferrmagnetic cre. The first Inductr has N =00 and a rate f change f flux is 0.4 Wb/s. If the secnd Inductr has N =5 turns and the cefficient f cupling is unity, hw much vltage is induced in the secnd inductr? a) 50 b) 80 c) 00 d) 500 answer: a The self-inductances f the cils in Fig.3-4 are L = 8 mh and L = 3 mh. If the cefficient f cupling is 0.85, calculate the energy stred in the system in millijules when a) i = 6 A, i = 9 A; b) i = -6 A, i = -9 A; c) i = -6 A, i = 9 A; d) i = 6 A, i = -9 A. What cnclusin d yu draw? answer: a) 7.60 mj; b) 7.60 mj; c) 58.40 mj; d) 58.40 mj. Same answer when bth current are entering r bth leaving, and same answer when ne is leaving and ne is entering. - 4 -