# Mutual inductance with a AC voltage source at Steady State Condition with 3 meshes

1. Dec 2, 2012

### ae4jm

When I apply the equations in phasor domain V1=jω(L1)(I1)±jωM(I2) and
V2=jω(L2)(I2)±jωM(I1) and then mesh analysis should I consider the other mesh on the far right side with the capacitor? In other words, this is the first problem that I've seen or had such as this and since I2 is a combination of that mesh and I3 then I'm tempted to add that mesh, but I'm unsure if doing that is the correct way to go, I don't have the answer to the problem to go back and check my work against it. I thought about going into Multisim and trying to get an approximate answer to check my work against. I'm stuck...

Thanks for your help and any advisement...
1. The problem statement, all variables and given/known dataSee the attached file for problem statement, given, and what to find.

2. Relevant equations V1=jω(L1)(I1)±jωM(I2) and
V2=jω(L2)(I2)±jωM(I1)

3. The attempt at a solution I've found the maximum mutual inductance to be 2√2 H and I've found the coupling coefficient to be 0.707.

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2. Dec 2, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Yes, you need to write mesh equations for all three loops. Consider that the all of the components on the right hand side form the load for a transformer.

Although, if you wanted to you could combine the resistors and capacitor into a single impedance. That way you'd reduce the circuit to two loops.

3. Dec 2, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Thanks for your help. Okay, so combining the loops I2 and I3 I have the 2Ω resistor and -j2Ω in series and then in parallel with the 2Ω resistor which totals (1.2-j0.4)Ω in place of the original 2Ω resistor and now that give me the following mesh equations to solve:
For I1: I1(1+j4)-j2I2=100 then
For I2: I1J2+I2(-1.2+j2.4)=0

Then solve for I1 and I2, correct?

4. Dec 2, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Your impedance looks good, but I think you need to pay attention to dot notation on the inductors when you go to write your loop equations; it affects the polarities of the equivalent voltage sources representing the mutual inductance.

5. Dec 2, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Ok, thanks. I'm going to grab some dinner and then I'm coming back to work on this some, as well as cover everything else for this course' final. Thanks a lot, I played around with the transformer equations a little, letting everything in the secondary equal ZL, minus the inductor equations, and with 0=R2 in the secondary. Then I did the same for solving for I1 and I just about have it from there. I was going to use this to check my answer that I got from using mesh analysis and see if they were the same.

Thanks again (gneill)! I'll post back with my answer...

6. Dec 3, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Okay,
I got
I1=24.5∠-75.96°
I2=0.5∠-90°

I double checked what I had with dot notation. I would have +M for the primary because my current goes into j2 in the secondary at the dot, and goes from + to - for a voltage drop across the inductor j4 for its self voltage drop.

Then for the secondary, current I1 goes into the dot at the j4 inductor in the primary which gives me a +M and then for the self voltage I get a + value because current I2 goes into the j2 inductor.

Does this look good? Did you get approximately the same answer as I did?

Thanks for your help...Please let me know if it needs some work, I'm hoping that I have it nailed down though.

7. Dec 3, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

I'm seeing values closer to 30 A in magnitude for both currents. Can you write out your two mesh equations (where the resistors and capacitor have been reduced to a single impedance)?

8. Dec 3, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

(1+j4)I1-j2I2=100 In the Primary
j2I1+(-1.2+j2.4)I2=0 In the Secondary

9. Dec 3, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Ah, you see, I'm finding +j2I2 in the first equation. The induced voltage in the primary due to i2 represents a potential drop for the "KVL walk" in the direction of i1.

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• ###### Fig1.gif
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10. Dec 3, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

I initially have 100-1I1-j4I1+j2I2=0 and then I consolidate and move everything on one side except for the 100 and solve simultaneously...should I rearrange my equation?

11. Dec 3, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

The potential change across the mutual inductance source goes from + to -, so it's a drop just like the change across the resistor is a drop. They should have the same sign.

12. Dec 3, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Is the correct answer:
I1=37.21ang-82.875 deg A
I2=27.735ang-70.56 deg A

Look correct?

13. Dec 3, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Again, that's not what I'm seeing. Let's see your two mesh equations again.

14. Dec 3, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

After changing that induced value in the primary I have:
(1+j4)I1+j2I2=100 and
J2I1+(-1.2+j2.4)I2=0

Thanks again for all of the work that you have put into helping me.

15. Dec 3, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Okay, I like the first equation, not so much the second. For the secondary circuit, you've combined the two resistors and the capacitor into a single impedance. So (assuming units of Ohms), that's 2 || (2 - 2j). That yields (6/5) - (2/5)j. So the total impedance in the loop seen by I2 is then

$2j + \frac{6}{5} - \frac{2}{5}j = \frac{6}{5} + \frac{8}{5}j$

That should make your second equation look like:

$2 I_1 j + \left(\frac{6}{5} + \frac{8}{5} j \right) I_2 = 0$

No problem. Happy to help.

16. Dec 3, 2012

### ae4jm

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

Okay, for the grand finale:

I1=30.71∠-47.49° A
I2=30.71∠169.38° A

Thanks a ton for all of your help and for not letting me drown out in the open ocean, LOL. I can see that I would have gotten the incorrect answer anyways even if my equations were correct to begin with. I had errors in my equation that you had pointed aut and errors in adding my rectangular coordinates by adding j2 to (1.2-j0.4) and just added the j2 to a positive j0.4, wrong no matter how I look at it. Thanks a lot!

17. Dec 3, 2012

### Staff: Mentor

Re: Mutual inductance with a AC voltage source at Steady State Condition with 3 meshe

That looks much better!

Glad it all worked out. Happy to help.

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