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

[js3]

You just need to practice your algebra.

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

The penny has dropped, L minus 0.25xL leaves three quarters of an L!
i.e. 0.75L
Thank you gneill.

 

[David J]

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)

Good Morning

I have started looking at this question now (part a) My notes tell me virtually nothing about this question so I am trying to get an understanding by looking through the posts of this thread. I have noticed the terminology jw used a lot but i could not work out what it was but then someone changed it to the latex form ##jw##. And now in this post it has been replaced by E, E being equivalent to ##\frac{v}{jw}## The E means more sense to me than##jw## but I was just wondering if someone could explain where ##jw## is derived from please?? Thanks

 

[gneill]

Good Morning

I have started looking at this question now (part a) My notes tell me virtually nothing about this question so I am trying to get an understanding by looking through the posts of this thread. I have noticed the terminology jw used a lot but i could not work out what it was but then someone changed it to the latex form ##jw##. And now in this post it has been replaced by E, E being equivalent to ##\frac{v}{jw}## The E means more sense to me than##jw## but I was just wondering if someone could explain where ##jw## is derived from please?? Thanks

##ω## is the angular frequency of the voltage source V which is presumed to be an AC sinusoidal source. ##j## is the square root of negative one, that is, ##j = \sqrt{-1}##. That makes ##jω## an imaginary value.

The mathematics used for AC circuit analysis is based on complex arithmetic. Currents and voltages are represented by complex phasors, and component impedances are complex values. The complex arithmetic automatically takes care of the phase relationships between currents and voltages that arise due to reactive components.

 

[David J]

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?

Thanks

Hello, I am looking at this question and struggling to understand how ##I1## and ##I2## were obtained. Do calculators offer this function? Not necessarily to solve for ##I1## and ##I2## but to re arrange for ##I1## and ##I2## as is required here ??

 

[ShortCircuit]

Hi all,

I'm new to writing on the forum but hopefully get this presented right.

Here's my problem b) - obtaining the current ratio I1/I2.

Here is where I am:

First of I have the equations from question A)

ABEF : V= jw*L1*I1+jw*M*I2

ABCDEF : V=jw*L2*I2+jw*M*I1

Now on to question B)

V=V

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

Factorise jw

jw(L1*I1+M*I2) = jw(L2*I2+M*I1)

Divide by JW

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

let V/jw = E

Eq (1) E = L1*I1+M*I2

Eq (2) E = L2*I2+M*I1

find I1 and I2

Eq (1) E = L1*I1+M*I2

Divide by L1-M*I2

E/L1-MI2 = I1

Substitute into Eq2

Eq (2) E = L2*I2+M(E/L1-MI2)

remove the brackets

E = L2*I2+ME/L1-MI2

Divide by I2 and multiply by E

I2 = E*L2+ME/L1-M

Substituting I1 and I2 back into Eq 2

E = L1(E/(L1-M(E*L2+ME/L1-M))) + M (E*L2 + ME/L1-M)

this is why I think I've got this wrong! after day 3 of head scratching.

Many thanks in advance

 

[gneill]

Hi ShortCircuit, Welcome to Physics Forums.

First of I have the equations from question A)

ABEF : V= jw*L1*I1+jw*M*I2

ABCDEF : V=jw*L2*I2+jw*M*I1

Now on to question B)

V=V

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

Factorise jw

jw(L1*I1+M*I2) = jw(L2*I2+M*I1)

Divide by JW

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

At this point you should be able to find the ratio I1/I2 with a bit of algebra on the last expression. Just gather the I1 terms on one side and the I2 terms on the other...

 

[ShortCircuit]

Hi Gneil,Thanks for the prompt reply clearly I was going off at a tangent and over complicating things.Ok so following on fromL1*I1+M*I2 = L2*I2+M*I1Dividing by I2 and I1(L1+M)/I1 = (L2+M)/I2Subtract ML1/I1 = L2/I2Then multiple by I1, Divide L2I have the RatioL1/L2 = I1/I2

 

[gneill]

Your algebra does not look correct. Can you show more detail in the steps you took? You said, "dividing by I2 and I1", but I don't see how you could arrive at:

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

 

[ShortCircuit]

It doesn't work ...

Try again, sorry

 

[ShortCircuit]

Ok so here's were I am and I'm lostL1*I1+M*I2 = L2*I2+M*I1

Dividing by L2

(L1*I1+M*I2)/L2 = I2+M*I1

Divide L1

(I1+M*I2)/L2 = (I2+M*I1)/L1

Subtract M

(I1+I2)/L2 = (I2+I1)/L1

?

 

[gneill]

You should do as I previously suggested and first gather all the "like" current terms together. So move all the I1 terms to the left hand side and all the I2 terms to the right hand side. I think that'll clear your path.

 

[ShortCircuit]

Got it!

(L1*I1)+(M*I2) = (L2*I2)+(M*I1)

Subtract (M*I1)

(L1*I1)+(M*I2)-(M*I1) = (L2*I2)+(M*I1)-(M*I1)

Factorize

I1(L1-M)+(M*I2) = (L2*I2)

Subtract (M*I2)

I1(L1-M)+(M*I2)-(M*I2) = (L2*I2)-(M*I2)

Factorize

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

Divide by I2

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

Divide by (L1-M)

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

 

[ShortCircuit]

Hi gneill,

I'm looking at c now - showing that L1 and L2 can be replaced by the equivalent inductor.

Following on from previous advise let

E=L1*I1+M*I2 (eq 1)
E=L2*I2+M*I1 (eq 2)

I'll broken down eq 1 as an example.

E=L1*I1+M*I2

Subtract (M-I2)

E-(M*I2) = L1*I1+M*I2 -(M*I2)

Divide by L1

E-(M*I2)/L1 = L1*I1/(L1)

E-(M*I2)/L1 = I1

Likewise for I2 I end up with

E-(M*I1)/L2 = I2

I then substituted these values back to form a simultaneous equation using my answer from A) and get

V=jw*L1*(E-(M*I2)/L1)+jw*M*(E-(M*I1)/L2)

V=jw*L2*(E-(M*I1)/L2)+jw*M*(E-(M*I2)/L1)I'm I right so far?

 

[gneill]

It looks okay. You should be sure to use parentheses to ensure that there is no ambiguity in the order of operations. Thus:

(E-(M*I2))/L1 = I1 ##~~~~##-and-##~~~~## (E-(M*I1))/L2 = I2

Treat these as two equations in two unknowns and solve for I1 and I2 (The expression for I1 should not contain I2, and similarly the expression for I2 should not contain I1).

As a suggestion, you won't have to return to the part A result if you define the equivalent inductance to be

V = jω L_eq(I1 + I2)

or in terms of your E:

E = L_eq(I1 + I2)

Use your "solved" expressions for I1 and I2 from above to proceed.

 

[Student12345]

Hi I'm just starting this question and and maybe this is a little stupid but what does M represent in this circuit? At first I assumed both L1and L2 had an inductance of M but it can't be. Is it XL?

 

[gneill]

Hi I'm just starting this question and and maybe this is a little stupid but what does M represent in this circuit? At first I assumed both L1and L2 had an inductance of M but it can't be. Is it XL?

M is the mutual inductance that links the two inductors.

 

[Triopas]

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.

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:
View attachment 109129
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##

Hello gneill,

I have tried to follow your instruction above for part c) and was wondering if I'm on the right lines?

We have the Kirchoff's loop equations from part a) right? i.e.

##V=(jwL_{1}I_1{})+(jwMI_{2})##
##V=(jwL_{2}I_2{})+(jwMI_{1})##

So transposing for I, I get:

##\frac {V-(jwMI_{2})} {jwL_{1}} =I_{1}##
##\frac {V-(jwMI_{1})} {jwL_{2}} =I_{2}##

Cancelling jw and filling in the formula you gave:

##\frac{V}{jωL_{eq}} = I_1 + I_2##
##\frac{V}{jωL_{eq}} = \frac {V-(MI_{2})} {L_{1}} + \frac {V-(MI_{1})} {L_{2}}##

Feel like I've gone very wrong somewhere!

Thanks

 

[gneill]

Hi Triopas.

What you have is three equations to work with, one being the definition introduced for ##L_{eq}##, and the other two being the expressions obtained from your KVL. I'd suggest first making the substitution:

##U = \frac{V}{j ω}##

in your three equations so that you can work without complex values. The three equations become:

##U = L_{eq}(I_1 + I_2)##
##U = L_1 I_1 + M I_2##
##U = L_2 I_2 + M I_1##

Start by solving the last two equations for ##I_1## and ##I_2## in terms of U, L's and M. There should be no ##I##'s in the solutions, just U, L's, and M.

 

[Triopas]

Thank you gneill.

Should I solve the second two equations for I1 and I2 as simultaneous equations?

##L_1 I_1 + M I_2 = L_2 I_2 + M I_1##

 

[gneill]

Thank you gneill.

Should I solve the second two equations for I1 and I2 as simultaneous equations?

Yes.

##L_1 I_1 + M I_2 = L_2 I_2 + M I_1##

Don't start by equating them; The resulting solutions for the ##I##'s should involve ##U##, ##L_1##, ##L_2##, and ##M##.

 

[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?