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

[gneill]

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.

 

[The_daddy_2012]

I knew it would be something ridiculously simple. Thank you very much gneill. I'm laughing at myself at the moment.

 

[The_daddy_2012]

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.

 

[earthloop]

Thanks gneill and the-daddy. Really appreciate the help.

 

[Gremlin]

(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.

 

[Gremlin]

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.

 

[Gremlin]

(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,

 

[earthloop]

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!
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

 

[MrBondx]

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 :

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?

 

[earthloop]

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.

 

[Gremlin]

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:

I²_rms 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?

 

[MrBondx]

I am not sure what you mean, show what you have worked out and I can help.

Got this question sorted, thanks anyway.

 

[Birchyuk]

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?

 

[MattSiemens]

I'm really struggling with part C of this question to, I'm going around in circles can anyone help?

 

[gneill]

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.

 

[MattSiemens]

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 :-)

 

[gneill]

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:

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:

$\frac{V}{jωL_{eq}} = I_1 + I_2$

 

[Jerremy_S]

Hi All,

I am a bit confused about part d):
I calculated L using:

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.

 

[gneill]

Too many L's

Hi All,

I am a bit confused about part d):
I calculated L using:

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.

 

[Jerremy_S]

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

 

[gneill]

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.

 

[MattSiemens]

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

 

[gneill]

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)

 

[MattSiemens]

So I am on the right track though gneill?

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

Thanks for your help!

 

[gneill]

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

 

[js3]

Hi, I'm a bit stuck on part d).
I am struggling to see how others have achieved 0.75L for Leq.

So far i have used the Leq equation and simplified down to Leq = L² - (0.5L)² / L

I can't seem to get 0.75 from this, any help would be appreciated.

 

[gneill]

Hi, I'm a bit stuck on part d).
I am struggling to see how others have achieved 0.75L for Leq.

So far i have used the Leq equation and simplified down to Leq = L² - (0.5L)² / L

I can't seem to get 0.75 from this, any help would be appreciated.

Can you show your simplification steps in detail?

 

[js3]

Yes sure.
M = k√L1L2

L1=L2

M = k√L²

M = kL

k = 0.5 therefore M = 0.5L

Substituting into: Leq = (L1L2 - M²) / (L1+L2 - 2M)

Leq = L1L2 - (0.5L)² / L1+L2 - 2(0.5L)

Where L1=L2=L

Leq = L² - (0.5L)² / L + L - L

= L² - (0.5L)²/ L

 

[gneill]

M = k√L1L2

L1=L2

M = k√L²

M = kL

k = 0.5 therefore M = 0.5L

Substituting into: Leq = (L1L2 - M²) / (L1+L2 - 2M)

Leq = L1L2 - (0.5L)² / L1+L2 - 2(0.5L)

How did the numerator and denominator end up being broken into separate terms? By convention what you've written is parsed as:

$L_{eq} = L_1 L_2 - \frac{(0.5 L)^2}{L1} + L2 - 2(0.5 L)$

You need to make more use of parentheses to keep the terms grouped and make the implied order of operations clear.

Try again being sure to keep the numerator and denominator terms of the original equation together.

 

[js3]

How did the numerator and denominator end up being broken into separate terms? By convention what you've written is parsed as:

$L_{eq} = L_1 L_2 - \frac{(0.5 L)^2}{L1} + L2 - 2(0.5 L)$

You need to make more use of parentheses to keep the terms grouped and make the implied order of operations clear.

Try again being sure to keep the numerator and denominator terms of the original equation together.

Apologies for that gneill, i'll try again.

$L_{eq} = \frac{L_1 L_2 - (0.5L)^2}{L_1 + L_2 - 2 (0.5L)}$

$L_{eq} = \frac{L^2 - (0.5L)^2}{L + L - L}$

$L_{eq} = \frac{L^2 - (0.5L)^2}{L}$

Hopefully that makes for better reading.

 

[gneill]

Much better. Can you continue to simplify the equation? Hint: Let 0.5L be L/2.

 

[js3]

Much better. Can you continue to simplify the equation? Hint: Let 0.5L be L/2.

$L_{eq} = \frac{L^2 - 0.25L^2}{L}$

 

[gneill]

$L_{eq} = \frac{L^2 - 0.25L^2}{L}$

Okay, continue.

 

[js3]

Okay, continue.

Would i then divide the L, leaving:

$L_{eq} = L - 0.25L$

No idea why I'm struggling with this!

 

[gneill]

Would i then divide the L, leaving:

$L_{eq} = L - 0.25L$

No idea why I'm struggling with this!

You just need to practice your algebra.

Continue. Factor out the L on the right hand side.