Parallel Mutual Inductances

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1. Jan 18, 2015

his_tonyness

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 :

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

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2. Jan 18, 2015

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.

3. Jan 18, 2015

his_tonyness

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

1/(2*π*√(L*1*10^-9))=1000000 , L=25.33uH

4. Jan 18, 2015

Staff: Mentor

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)?

5. Jan 18, 2015

his_tonyness

jwC= j0.0063
=25uH

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
6. Jan 18, 2015

Staff: Mentor

Well, the problem states that L1 = L2. So replace them with a single variable.

7. Jan 18, 2015

his_tonyness

Leq = L1 and L2, but L1=L2, so one inductor is = leq/2 which is 25/1= 12.5uH

8. Jan 18, 2015

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.

9. Jan 18, 2015

his_tonyness

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

10. Jan 18, 2015

Staff: Mentor

That looks about right. I'm seeing a slightly higher value, but that could be down to rounding of intermediate values.

11. Jan 18, 2015

his_tonyness

Re-calculated with Lt= 25.33*10^6 , I get 33.77uH

12. Jan 18, 2015

Staff: Mentor

Okay, now we agree :)

13. Jan 18, 2015

his_tonyness

Wonderful, thanks for your help again. :)

14. Jan 18, 2015

Electest

Hi

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

Regards

15. Jan 18, 2015

Staff: Mentor

What isn't you don't follow? Can you show what you've tried / thought about?

16. Jan 18, 2015

Electest

Hi gneill

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

17. Jan 18, 2015

Staff: Mentor

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>.

18. Jan 18, 2015

Electest

Thanks Gneill

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

Regards

19. Jan 19, 2015

Electest

Hi Gneill

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

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

20. Jan 19, 2015

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.