Leq = (L1*L2-M^2)/(L1*L2-M^2)Hope this helps

Click For Summary
The discussion revolves around the analysis of a circuit with two inductors connected in parallel and the application of Kirchhoff’s voltage law. Participants derive the current ratio I1/I2 in terms of the inductances L1 and L2, finding I1/I2 = (L2-M)/(L1-M). They also confirm that the equivalent inductance Leq can be expressed as Leq = (L1*L2-M^2)/(L1+L2-2*M). A capacitor is introduced, leading to calculations for the required inductance L when the circuit operates at a frequency of 1 MHz, with participants arriving at values around 33.33 μH. The conversation emphasizes the importance of understanding the relationships between inductance, mutual inductance, and circuit behavior at resonance.
  • #31
I am suffering on (c).

I need to find I which is I1+I2

Therefore I need to find I1 and I2

From (b) working backwards by one step.

i1 (L1-m) = i2(L2-m)

Therefore:

i1 = (i2 (L2-m))/ (L1-m)
i2 = (i1 (L1-m))/ (L2-m)

Adding them together I still have two values for I

If I use substitution

I end up with.

i1(L1-m)(L2-m) = i1 (L1-m)(L2-m)

Which I is true, but not at all helpful!

How on Earth do I start this question!
 
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  • #32
earthloop said:
Thanks for quick reply, I tried this earlier with no luck... maybe its how I am inputting it into wolfram, I am not sure. I input :
a*b*c+a*d*e=v, a*f*e+a*d*c=v
All I get is a bunch of zero solutions.
For (b) use the two equations you obtained in (a).

Quoting the legend that is gneill. V=V right?
 
  • #33
Ok so far I have a few values, I have tried following some of the earlier instructions but I'm stuck for part (d)

As everyone L=1/((2pif)^2)C

L=25.33 microhenrys

M=k sqrt L1L2

Or M=kL

K=0.5

M=0.5L

With rearranging

Leq = (L^2 - m^2)/(2L -2m)

Replace m with kL

Leq=(L^2-0.5L^2)/(2L-(2x0.5L)

Leq=(L^2-0.5L^2)/L

Leq=(L-0.5L)

Leq=0.5L

Leq=25.33 microhenrys

25.33 microhenrys = 0.5L

L=50.66 micro-henrys

is this right or have I done it wrong?
 
  • #34
Yeah I have tried using the equations from a), but am struggling with solving them simultaneously.

V=jw*L1*I1+jw*M*I2
V=jw*L2*I2+jw*M*I1

I have input it into my calculator to solve for I1 and I2 and get the correct answers. I can't get the same answer in Wolfram. I would really like to know how to solve these two for I1 and I2 without using my calculator. Any more help would be hugely appreciated. Maybe just the first step to solving, to get the ball rolling?

The_daddy_2012 said:
Ok so far I have a few values, I have tried following some of the earlier instructions but I'm stuck for part (d)

As everyone L=1/((2pif)^2)C

L=25.33 microhenrys

M=k sqrt L1L2

Or M=kL

K=0.5

M=0.5L

With rearranging

Leq = (L^2 - m^2)/(2L -2m)

Replace m with kL

Leq=(L^2-0.5L^2)/(2L-(2x0.5L)

Leq=(L^2-0.5L^2)/L

Leq=(L-0.5L)

Leq=0.5L

Leq=25.33 microhenrys

25.33 microhenrys = 0.5L

L=50.66 micro-henrys

is this right or have I done it wrong?
I don't think that's the correct answer, I think its around 33. I have it worked out at home on paper, I will have to check tomorrow unless somebody else beats me to it.

Thanks
 
  • #35
The_daddy_2012 said:
M=0.5L

With rearranging

Leq = (L^2 - m^2)/(2L -2m)

Replace m with kL

Leq=(L^2-0.5L^2)/(2L-(2x0.5L)
M is 0.5L, but you've replaced M2 with 0.5 L2. The 0.5 should not have escaped the squaring!
 
  • #36
earthloop said:
Yeah I have tried using the equations from a), but am struggling with solving them simultaneously.

V=jw*L1*I1+jw*M*I2
V=jw*L2*I2+jw*M*I1

I have input it into my calculator to solve for I1 and I2 and get the correct answers. I can't get the same answer in Wolfram. I would really like to know how to solve these two for I1 and I2 without using my calculator. Any more help would be hugely appreciated. Maybe just the first step to solving, to get the ball rolling?
This is basic simultaneous equations.

Rearrange the first equation so you have I1 = <stuff>.

Substitute that <stuff> into the second equation where I1 appears. Solve for I2. Use this result to replace I2 in the <stuff> in order to obtain a value for I1.
 
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  • #37
I knew it would be something ridiculously simple. Thank you very much gneill. I'm laughing at myself at the moment.
 
  • #38
earthloop said:
Yeah I have tried using the equations from a), but am struggling with solving them simultaneously.

V=jw*L1*I1+jw*M*I2
V=jw*L2*I2+jw*M*I1

I have input it into my calculator to solve for I1 and I2 and get the correct answers. I can't get the same answer in Wolfram. I would really like to know how to solve these two for I1 and I2 without using my calculator. Any more help would be hugely appreciated. Maybe just the first step to solving, to get the ball rolling?

I don't think that's the correct answer, I think its around 33. I have it worked out at home on paper, I will have to check tomorrow unless somebody else beats me to it.

Thanks
I don't know if you've solved it yet, but if you haven't here's some help. Write both of your equations down, then write down v=v underneath and see what you you spot about these three equations. Can you do some substitution?

Then simply follow through using basic algebra.
 
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Likes earthloop
  • #39
Thanks gneill and the-daddy. Really appreciate the help.
 
  • #40
his_tonyness said:
(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

Assuming we're looking at the same course notes - please can you tell me where in the notes you have got the above from? I can't see anything like that in the notes - lesson 12 pg35 has something similar but it's less than half a page and is only the bottom half of the equation.
 
  • #41
Gremlin said:
Assuming we're looking at the same course notes - please can you tell me where in the notes you have got the above from? I can't see anything like that in the notes - lesson 12 pg35 has something similar but it's less than half a page and is only the bottom half of the equation.

And then i saw page 2 of this thread and arrived at the answer.
 
  • #42
his_tonyness said:
(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)

The Attempt at a Solution



(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)

Tony (or anyone else),

Can you tell me why, when answering part (b) you did this I1= (V*(L2-M)) / (jw*(L1*L2-M^2)) , I2=(V*(L1-M)) / (jw*(L1*L2-M^2))?

I answered (b) fairly easily using the equations from part (a), namely V = jωL1I1 + jωMI2 & V = jωL2I2 + jωMI1

jωL1I1 + jωMI2 = jωL2I2 + jωMI1

simplified down to

I1 / I2 = (L2-M) / (L1-M)

which is the same answer that you got. I'm just curious as to why you decided to jump into it the way you did, as i can see nothing in the course notes that would have made me come at it the way you did. But then the way you came at it made made part (c) easier - a part I'm now struggling to answer.

Thanks,
 
  • #43
Hi Gremlin,
I don't think Tony "jumped" into that answer for I1, I certainly didn't, and if he did I'm curious too! o_O
Have you actually tried expressing your answers from a) (v=...) as their currents I1 and I2? And then substituting them back into the answers from a)?
Have a go at it and show some working out and let me know if you need any help!
EL
 
  • #44
his_tonyness said:

Homework Statement


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.

Homework Equations

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

For (a) isn't V supposed to become negative when it goes to the other side of the equal sign. Or what am I missing?
 
  • #45
MrBondx said:
For (a) isn't V supposed to become negative when it goes to the other side of the equal sign. Or what am I missing?

I am not sure what you mean, show what you have worked out and I can help.
 
  • #46
I don't really want to make a new thread for this as it probably doesn't warrant it, but in my textbook i have the equation:

I2rms R= { (I sin (ωt))2 R}avg

The I in the brackets has a little cresent or semi circle above it - does anyone know what this denotes?
 
  • #47
earthloop said:
I am not sure what you mean, show what you have worked out and I can help.

Got this question sorted, thanks anyway.
 
  • #48
Hi all,

I am really struggling with part c of this question and have been stuck for a while, i would really appreciate any help.

I think i have the strategy worked out which is:

Use Loop ABEF and re-arrange for I1
Use I1 in Loop ABCDEF to re-arrange for I2
Use I2 in loop ABEF to calculate I1

I then should add I1 and I2 which should give me Leq.

Is this correct?
 
  • #49
I'm really struggling with part C of this question to, I'm going around in circles can anyone help?
 
  • #50
MattSiemens said:
I'm really struggling with part C of this question to, I'm going around in circles can anyone help?
You have to show what you've tried. No help without first showing your own efforts, as per the forum rules.
 
  • #51
My attempt was that V1=V2,

so jw(L1i1+Mi2) = jw(L2i2 + Mi1) dividing both sides by jw gives;

(L1i1+Mi2) = (L2i2 + Mi1)

Then rearranging the i1/i2 ratio to have an equation for i1 and i2and then substituting into the above equation.

This is where it starts to get a bit messy and I go off the beaten track.

I also tried rearranging the formula Leq = V/I using either V1 or V2 and I = i1 + i2.

I have seen how the Leq formula can be equated using differentiation but I don't think this is the answer that's been looked for?

Thanks for your quick reply :-)
 
  • #52
MattSiemens said:
My attempt was that V1=V2,

so jw(L1i1+Mi2) = jw(L2i2 + Mi1) dividing both sides by jw gives;

(L1i1+Mi2) = (L2i2 + Mi1)

Then rearranging the i1/i2 ratio to have an equation for i1 and i2and then substituting into the above equation.
This gets you as far as a relationship between i1 and i2, but doesn't address the equivalent inductance since it doesn't relate the total current to the driving voltage.
This is where it starts to get a bit messy and I go off the beaten track.

I also tried rearranging the formula Leq = V/I using either V1 or V2 and I = i1 + i2.

I have seen how the Leq formula can be equated using differentiation but I don't think this is the answer that's been looked for?

Thanks for your quick reply :-)
There's only one voltage here, V the source voltage. If you could find an expression for V/I , where I is the total source current, then you would have the impedance of the load.

Here's a starting point I can suggest. Use KVL and write two loop equations per:
upload_2016-11-19_12-30-11.png

Handle the mutual inductances as required, of course. Two equations, two unknowns (##I_1## and ##I_2##). Solve for the two currents. The sum of the two currents is the total current, so that:

##\frac{V}{jωL_{eq}} = I_1 + I_2##
 
  • #53
Hi All,

I am a bit confused about part d):
I calculated L using:
rsrlcc-2.gif

L = 25,33 x 10^-6Hr

Next, knowing that M=kL and k=0,5

I Calculated Lt:

Lt = (L1L2 - M^2)/(L1+L2-M2) = (L^2-M^2)/(2L-2M) = (L^2 - 0,25L^2)/L = 0,75L^2/L = 0,75L

Then: Lt = 0,75 * (25,33 x 10^-6Hr) = 18,9975x 10^-6

I am not sure what supposed to be done next.

Please help.
 
  • #54
Too many L's :smile:

Jerremy_S said:
Hi All,

I am a bit confused about part d):
I calculated L using:
rsrlcc-2.gif

L = 25,33 x 10^-6Hr
That L would be the required equivalent L. So whatever value of inductance is used for L1 and L2 should yield that value when the equivalent inductance is found.
Next, knowing that M=kL and k=0,5

I Calculated Lt:

Lt = (L1L2 - M^2)/(L1+L2-M2) = (L^2-M^2)/(2L-2M) = (L^2 - 0,25L^2)/L = 0,75L^2/L = 0,75L
Lt is then the equivalent inductance when some (other) value L is used for L1 and L2. It's not the same as the L you found for resonance. Lt is in fact what you want for the the resonance inductance.
Then: Lt = 0,75 * (25,33 x 10^-6Hr) = 18,9975x 10^-6
So no, because of the "too many L's" issue you've got the working backwards. Lt is what you want to be equal to your 25.33 μH. You want to find the L that gives you that value in your expression Lt = 0.75 L.
 
  • #55
Thanks for your reply:)

So it looks like Lt should be found first and then used as L in the resonance equation giving 0,75L = 25,33 x 10^-6Hr = 33,77x10^-6 ?

I hope that amount of L's is about right this time...

Cheers
 
  • #56
Yes, that looks right. You can check back through the early posts in this thread to see similar (which L is which) issues being sorted out.
 
  • #57
Hi gniell

I'm still over complicating this I think :-(

I have the two loop equations taken from part (A)

V=JWL1i1 + JWMi2 (1)

V=JWL2i2 + JWMi1 (2)

I have been trying to simplify for i1 and i2 and then substituting in i2 in (1) and i1 in (2) but I'm getting overly complicated equations.

My attempts have been in the order of;

factorising JW and then dividing by this;

V/JW = L1i1 + Mi2

Subtracting Mi2 to isolate L1i1 and dividing by L1;

V/JWL1 - Mi2/L1 = i1

the same approach for i2 but on substituting for either i1 or i2 its starting to get complicated.

Thanks for the help
 
  • #58
Can't do much about the algebra getting messy other than organize the symbols better. To unclutter things a bit let E = V/JW.

Your two equations are then:

E=L1i1 + Mi2 (1)
E=L2i2 + Mi1 (2)
 
  • #59
So I am on the right track though gneill?

That's enough for me to go on :-)

Thanks for your help!
 
  • #60
MattSiemens said:
So I am on the right track though gneill?
Yes. You want to find expressions for I1 and I2 that involve only the L's, M, and V. After that it's not so bad :smile:
 

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