# Mutual inductance+ coupling question

Discussion in 'Homework Help' started by nirvanaguy, Dec 8, 2009.

1. ### nirvanaguy Thread Starter New Member

Oct 3, 2009
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So I was looking at this problem and my teacher only went over the case with only a voltage and 1 inductor in the loop, but clearly 1 loop has a voltage source and a resistor with an inductor whereas the other is just a resistor + inductor.. so I don't really know how to account for the inductor. I tried combining their impedances and doing V/Z=I but then you dont get a function dependent on time.. Part b is easy, but I was stuck on how I could find an expression for current for part A and C because I know if i can find that I can find part c by just integrating power (IV)... but I only know that when one of the currents enter and other exits from dot, the expressions would bee V1=L1(di1/dt) - M(di2/dt) and V2=-M(di1/dt)+L2 (di2/dt)...

knowing that the coupling constant is .5 from M=k(sqrt(L1L2)), is it ok to assume that the voltage V1 (on the left ) is Vg and on the right V2=0? If thats the case then I guess you can just use a system of equations (with 2 equations) and solve for one expression, but then youre left with like a differential equation since you have Vg= L1(dig/dt)-M(dil/dt)+5Ig and Vl= -m(dig/dt)+L2(dil/dt)+15Il

how are you supposed to go about isolating for Ig and Il even with 2 equations..? (if thats the right approach)

can anyone help me out on how to approach this problem. I'd appreciate it. thanks!

Last edited: Dec 8, 2009
2. ### nirvanaguy Thread Starter New Member

Oct 3, 2009
15
0
bump?

I have no idea how to take the resistor into account since all of the examples he did in our class were just only inductors

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Are you obliged to solve the problem in the time domain using differential equations?

It is a far easier task to do this in the frequency domain using complex numbers.

Handling resistors in differential equation form is as one would expect in typical circuit analysis. For the left hand loop the voltage drop across the 5Ω is simply 5*ig and for the right hand loop the 15Ω resistor voltage drop is 15*iL.

4. ### nirvanaguy Thread Starter New Member

Oct 3, 2009
15
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well they want us to find the currents as a function of time.. you're saying if i use it just using complex numbers your way I owuld get the same answer using the diff eq method? because as long as we get the right answer thats all they care about.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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That's correct - the complex number approach will yield the steady-state solution, which is what you were asked to provide.

Do you know how to approach a solution?

6. ### nirvanaguy Thread Starter New Member

Oct 3, 2009
15
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i found somewhere else that you can substitute L(di/dt)=jwLI, so I did it all using complex numbers like that but then I get a phasor for an answer.. does that make sense??

Last edited: Dec 8, 2009
7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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This might help ....

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8. ### nirvanaguy Thread Starter New Member

Oct 3, 2009
15
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im getting the M terms to be negative when you're getting them to be positive.. my teacher told me when they're on opposite sides of the conductor (left dot is on top and right top is on bottom), the coefficients for M is -1 instead of +1?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I use the convention that assumed current flowing into a 'dot' creates a positive mutual voltage at the corresponding 'dot' on the magnetically coupled winding.

Hence the assumed ig flows into the LHS dot which will induce a positive voltage polarity at the RHS dot. Similar reasoning applies for the assumed direction of IL.

You should at least try and come up with a solution - and check your answers - if you are given any.