# Homework Help: Parallel Mutual Inductances

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1. Feb 1, 2017

### Staff: Mentor

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.

2. Feb 1, 2017

### 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)

3. Feb 8, 2017

### 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?

4. Feb 8, 2017

### Staff: Mentor

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 Leq(I1 + I2)

or in terms of your E:

E = Leq(I1 + I2)

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

5. May 10, 2017

### 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?

6. May 10, 2017

### Staff: Mentor

M is the mutual inductance that links the two inductors.

7. Jan 3, 2018

### Triopas

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

8. Jan 3, 2018

### Staff: Mentor

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.

9. Jan 3, 2018

### Triopas

Thank you gneill.

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

$L_1 I_1 + M I_2 = L_2 I_2 + M I_1$

10. Jan 3, 2018

### Staff: Mentor

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

11. Jan 4, 2018

### 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?

12. Jan 4, 2018

### Staff: Mentor

That's the idea, yes.

13. Jan 19, 2018

### Spongecake

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

14. Jan 19, 2018

### Staff: Mentor

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

15. Jan 19, 2018

### Spongecake

Thank you gneill

16. Jan 19, 2018

### Spongecake

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

#### Attached Files:

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17. Jan 19, 2018

### Staff: Mentor

Looks okay to me.

18. Jan 19, 2018

### Spongecake

Brilliant Thank you

19. Mar 5, 2018

### 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 heres what I've got so far.

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

20. Mar 5, 2018

### Staff: Mentor

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