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Parallel Mutual Inductances

  1. Jan 18, 2015 #1
    1. The problem statement, all variables and given/known data
    FIGURE 4(a) shows two inductances connected in parallel across an a.c. supply.
    (a) Apply Kirchhoff’s voltage law to loop abef and to loop abcdef of the circuit.

    (b) Hence or otherwise obtain the current ratio I1/I2 in terms of the circuit inductances.

    (c) See if you can show that L1 and L2 can be replaced by the
    equivalent inductor, Leq, of FIGURE 4(b) where Leq = (L1*L2-M^2)/(L1+L2-2*M)

    (d) A 1 nF capacitor is placed across the two inductors (FIGURE 4 (c)). If L1 = L2 = L and k = 0.5, determine the required value of L if the minimum current I flows from the supply when it is at a frequency of 1 MHz.

    2. Relevant equations


    3. The attempt at a solution
    (a) ABEF : V=jw*L1*I1+jw*M*I2, ABCDEF : V=jw*L2*I2+jw*M*I1

    (b) I1= (V*(L2-M)) / (jw*(L1*L2-M^2)) , I2=(V*(L1-M)) / (jw*(L1*L2-M^2))

    I1/I2= (((V*(L2-M)) / (jw*(L1*L2-M^2))) / (((V*(L1-M)) / (jw*(L1*L2-M^2)))

    I1/I2 =L2-M / L1-M

    (c) Leq = I1+I2 , = (((V*(L2-M)) / (jw*(L1*L2-M^2))) + (((V*(L1-M)) / (jw*(L1*L2-M^2)))

    =(V*(L1+L2-2M))/(jw*(L1*L2-M^2)) = V/jw*Leq, Leq = (L1*L2-M^2)/(L1+L2-2*M)

    (d) Xc= 1/(2*pi*f*C) = j159.155 ohms
    The bit i'm stuck on is whether I can solve for L using :
    rsrlcc-2.gif

    However this is for resonant frequency. A push in the right direction anyone on any of these questions. Can someone explain the origin of why in (c) I1+I2= V/jw*Leq
     

    Attached Files:

  2. jcsd
  3. Jan 18, 2015 #2

    gneill

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    Staff: Mentor

    That should be ##I = I_1 + I_2##, the sum of the currents. Current and inductance are completely different animals.
    The inductor and capacitor are in parallel, so it would make sense to view this load as parallel admittances. No current will flow if the total admittance is zero, right? How do admittances in parallel combine? You should be able to easily show that the total admittance is zero at resonance.
    In general the steady state current through an inductance L due to applied potential V is ##I = V/jωL##. In this problem I1 + I2 is the current total ##I## for the load, and V is the potential across the load. So knowing V and I you can determine an equivalent inductance for the load provided that the current has the same form.
     
  4. Jan 18, 2015 #3
    acp183.gif
    Yt=Y+Y+Y...

    wC=j0.006 , Yt=0 , 0= 1/jwL -j0.006, 1/jwL= j0.006 , hence zero admittance

    Since current is at a minimum at zero admittance, I can solve for 1 of the inductors since L1=L2=L
    rsrlcc-2.gif
    1/(2*π*√(L*1*10^-9))=1000000 , L=25.33uH
     
  5. Jan 18, 2015 #4

    gneill

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    Admittance is the reciprocal of impedance. Both are complex values. Don't confuse them with their magnitudes (the square root of the sum of the squares of their real and imaginary components).

    So for the inductor and capacitor in parallel the total admittance is:
    $$Y = \frac{1}{jωL_{eq}} + jωC$$
    You're given values for C and ω, so it's a snap to find Leq if Y is zero.

    With the net inductance Leq in hand, what's your plan to find the actual inductor value L (where there are two inductors couple by mutual inductance)?
     
  6. Jan 18, 2015 #5
    jwC= j0.0063
    =25uH

    ind43.gif

    ind33.gif

    1/Lt=1/L1+1/L2....

    I cant find or think of an equation to extract the value of one inductor, the net inductance cant be halved can it.
     
    Last edited by a moderator: Apr 19, 2017
  7. Jan 18, 2015 #6

    gneill

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    Well, the problem states that L1 = L2. So replace them with a single variable.
     
  8. Jan 18, 2015 #7
    Leq = L1 and L2, but L1=L2, so one inductor is = leq/2 which is 25/1= 12.5uH
     
  9. Jan 18, 2015 #8

    gneill

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    Staff: Mentor

    No, Leq is the inductance value in parallel with the capacitor that you found for minimum current (resonance). You also found an expression for the equivalent inductance for the two mutually coupled inductors. You want to find the values for those two coupled inductors. The problem states that they have the same value, so simplify the equivalent inductance expression accordingly. Use the expression for mutual inductance with the coupling constant k, too.
     
  10. Jan 18, 2015 #9
    Ok, simplifying the Lt/Leq equation to one inductor variable I get Lt = (L^2-M^2)/(2L-2M), With mutual inductance equation,simplifying, M= k*L
    combining the mutual inductance equation with the Leq/Lt equation (i.e Replacing M in Lt equation with k*L)

    I get L = 33.33uH
     
  11. Jan 18, 2015 #10

    gneill

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    That looks about right. I'm seeing a slightly higher value, but that could be down to rounding of intermediate values.
     
  12. Jan 18, 2015 #11
    Re-calculated with Lt= 25.33*10^6 , I get 33.77uH
     
  13. Jan 18, 2015 #12

    gneill

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    Okay, now we agree :)
     
  14. Jan 18, 2015 #13
    Wonderful, thanks for your help again. :)
     
  15. Jan 18, 2015 #14
    Hi

    I don't quite follow question (b), any pointers would be appreciated.

    Regards
     
  16. Jan 18, 2015 #15

    gneill

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    What isn't you don't follow? Can you show what you've tried / thought about?
     
  17. Jan 18, 2015 #16
    Hi gneill

    Thanks for the swift reply

    Part (a) was straight forward, but part (b) I'm not following, as this is not mentioned in any of my study material. I have looked through various books, but am a little stumped. Can't quite understand what they are asking.

    Regards
     
  18. Jan 18, 2015 #17

    gneill

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    They want you find an expression for I1/I2, presumably starting with the loop equations that you obtained in part (a).

    Hint: Start with the loop equations in the form: V = <stuff>.
     
  19. Jan 18, 2015 #18
    Thanks Gneill

    I'll have a look into that and get back to you

    Regards
     
  20. Jan 19, 2015 #19
    Hi Gneill

    Not quite sure why I'm struggling with this, but I am. :nb)

    I have 2 KVL loops

    ABEF V = jwL1I1 + jwMI2
    ABCDEF V = jwL2I2 + jwMI1

    Which I guess I'm looking at solving for both I1 and I2 using simultaneous equations, either by substitution or elimination. But I cannot find a solution. Can you please point me in the right direction as I want to fully understand this.

    Regards
     
  21. Jan 19, 2015 #20

    gneill

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    Staff: Mentor

    You're looking for a ratio of the two unknowns I1 and I2, so you really don't need to solve for particular values of them. Just equate the two (V = V, right?) and simplify. Show your work.
     
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