Equivalent inductance of two coupled inductors in parallel

[Triopas]

Ah okay, so like this?

Rearrange equation (1) for I₁.
$U = L_1 I_1 + M I_2$
$I_1 = \frac {U-MI_2} {L_1}$

Replace I₁ in equation (2) with the above.
$U = L_2 I_2 + M(\frac {U-MI_2} {L_1})$

Rearrange for I₂, then replace I₂ in equation (1) with the above?

 

[gneill]

That's the idea, yes.

 

[Spongecake]

I'm sure i am being silly but what does Jw stand for?

 

[gneill]

I'm sure i am being silly but what does Jw stand for?

$\sqrt{-1} \cdot \omega$

 

[Spongecake]

Thank you gneill

 

[Spongecake]

For part d) i have this (done on mathCAD) is this correct?

 

[gneill]

Looks okay to me.

 

[Spongecake]

Brilliant Thank you

 

[cablecutter]

hi bit of further guidance on part c would be much appreciated.
so far following on from post #91

U=L2i2+M((U-(Mi2))/L1

i am unsure if I'm correct in how I'm going about rearranging for i2 here's what I've got so far.

i2= ((U-M)/L2) +M((U-M)/L1) ? i have a feeling its way out.

 

[gneill]

i2= ((U-M)/L2) +M((U-M)/L1) ? i have a feeling its way out.

Yeah, looks like something's gone wrong with your algebra. Try again, perhaps showing us your work step-by step.

 

[cablecutter]

I'm unsure do i need to try and rearrange U=L2i2+M((U-Mi2)/L1
for i2?

or the original simultaneous equation (2) for i2?...
U = L2i2+Mi1
U-Mi1 = L2i2+Mi1-(Mi1)
(U-Mi1)/L2 = (L2i2)/L2
(U-(Mi1)/L2 = i2

 

[gneill]

I'm unsure do i need to try and rearrange U=L2i2+M((U-Mi2)/L1
for i2?

Yes.

 

[cablecutter]

so from U = L2i2+M(U-(Mi2))/L1

UL1 = L1L2i2+MU-M²i2

UL1-MU = L1L2i2-M²i2

UL1-MU = i2(L1L2-M²)

i2 = (UL1-MU)/(L1L2-M²)

am i right so far ?

 

[gneill]

Looks fine.

 

[cablecutter]

UL1 = L1L2i2+MU-M²i2

UL1-MU = L1L2i2-M²i2

UL1-MU = i2(L1L2-M²)

i2 = (UL1-MU)/(L1L2-M²)

am i right so far ?[/QUOTE]

now replace i2 in equation (1) with above? do i replace both i2 and i1 in equation 1 or just i2?
can end up with:

U = L1i1+M((UL1-MU)/(L1L2-M²))

or

U = L1(U-(M((UL1-MU)/(L1L2-M²))/L1) + M((UL1-MU)/(L1L2-M²)) ?

you can then minus the L1 to give

U = U-(M((UL1-MU)/(L1L2-M²)) + M((UL1-MU)/(L1L2-M²)) which will end up canceling out to 0=0?

 

[gneill]

I think we're losing the plot a bit here. Your goal is to find expressions for $I_1$ and $I_2$ starting with the equations

$U = L_1 I_1 + M I_2~~~~~~~$ (1)
$U = L_2 I_2 + M I_1~~~~~~~$ (2)

Two equations in two unknowns. Everything else is treated as known constants. This is a typical pair of simultaneous equations which you wish to solve for the variables $I_1$ and $I_2$.

These equations have been solved in this thread already (granted the thread is getting rather large due to it being continuously "reawakened" by students with the same question and issues...), so I'm able to recap here without really giving anything away that hasn't already been presented.

Isolate $I_1$ equation (1):

$I_1 = \frac{(U - M I_2)}{L_1}$

Plug that expression for $I_1$ into equation (2):

$U = L_2 I_2 + M \frac{(U - M I_2)}{L_1}$

Solve for $I_2$:

$I_2 = \frac{L_1 - M}{L_1 L_2 - M^2}$

Now you have an expression for $I_2$ that only involves the known values.

Do a similar thing to find the expression for $I_1$, or use this $I_2$ result in equation (1) to eliminate $I_2$ there and solve for $I_1$, or simply look at the symmetry of the two equations and write the result for $I_2$ by inspection from the result for $I_2$.

 

[cablecutter]

okay so from

U = L1i1+M((L1-M)/(L1L2-M²))

L1i1 = U-M ((L1-M)/(L1L2-M²)

i1 = (U-M((L1-M)/(L1L2-M²)))/L1

Thanks for your Help gneill

 

[osykeo]

on part c i follow it up to solving for i2 but what i can't understand is how the U has been dropped;

by my workings i get:

I2= (U(L1-M)/L2L1-M^2)

where as the final for your i2 has dropped the U ?

am i missing something obvious here, seems to happen when youve be staring at a question for so long.

 

[Jason-Li]

on part c i follow it up to solving for i2 but what i can't understand is how the U has been dropped;

by my workings i get:

I2= (U(L1-M)/L2L1-M^2)

where as the final for your i2 has dropped the U ?

am i missing something obvious here, seems to happen when youve be staring at a question for so long.

I think we're losing the plot a bit here. Your goal is to find expressions for $I_1$ and $I_2$ starting with the equations

$U = L_1 I_1 + M I_2~~~~~~~$ (1)
$U = L_2 I_2 + M I_1~~~~~~~$ (2)

Two equations in two unknowns. Everything else is treated as known constants. This is a typical pair of simultaneous equations which you wish to solve for the variables $I_1$ and $I_2$.

These equations have been solved in this thread already (granted the thread is getting rather large due to it being continuously "reawakened" by students with the same question and issues...), so I'm able to recap here without really giving anything away that hasn't already been presented.

Isolate $I_1$ equation (1):

$I_1 = \frac{(U - M I_2)}{L_1}$

Plug that expression for $I_1$ into equation (2):

$U = L_2 I_2 + M \frac{(U - M I_2)}{L_1}$

Solve for $I_2$:

$I_2 = \frac{L_1 - M}{L_1 L_2 - M^2}$

Now you have an expression for $I_2$ that only involves the known values.

Do a similar thing to find the expression for $I_1$, or use this $I_2$ result in equation (1) to eliminate $I_2$ there and solve for $I_1$, or simply look at the symmetry of the two equations and write the result for $I_2$ by inspection from the result for $I_2$.

osykeo, I have the exact same query, although I think I have got the answer. You have to do it as follows:

I1=(U(L2-M))/(L1*L2-M^2)
I2=(U(L1-M))/(L1*L2-M^2)
Hence:
U=Leq (I1+I2)
U = Leq (U(L2-M))/(L1*L2-M^2)+(U(L1-M))/(L1*L2-M^2)

U = Leq ((UL1+UL2-2MU)/(L1*L2-M^2)

Then rearrange this for Leq