Mutual Inductance with Pulsating Current

  • #1
seanhbailey
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I'm not really sure where to put this question.

I'm trying to project current output of a ciruit I am building. Suppose that an initial current which pulsates according to time, t, is given as I = sin(wt)+1. I have two inductors which share the same core in close proximity, the first having N turns, an inductance of L and resistence R1, and the second having M turns, an inductance of P and resistence R2. As a function of t, what would be the current in the second coil if the pulsating current, I, runs through the first inductor?
 

Answers and Replies

  • #2
gneill
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Start by listing what you know about the inductors and their relationships.
 
  • #3
seanhbailey
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If I had a Dc source it is clear that the voltage in the second inductor will be M*(dI/dt) where the mutual inductance M = k*sqrt(L1*L2) where L1 and L2 are the inductances of the first and second coils and k is the coefficient of coupling. However, I don't have a DC source. I know what an AC source would produce as well, but I'm not sure how a pulsating current will differ since I do not have a capacitor present.
 
  • #4
gneill
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If you had a DC source, dI/dt would be zero, and you'd get no current out!
In this case you have an input current function that you can easily differentiate...

What can you say about the mutual inductance and k in this case, given the description of the setup? What can you do with the turns ratio that has been given?
 
  • #5
seanhbailey
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Firstly, even if I had a DC source, I would still get current in the second coil during the first few moments when I turned the power source on, and right after I turned the source off.

Secondly, if I simply took the derivative of the function sin(wt)+1 and multiplied it by the mutual induction, would this give me the voltage in the second coil as a function of time?
 
  • #6
gneill
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Firstly, even if I had a DC source, I would still get current in the second coil during the first few moments when I turned the power source on, and right after I turned the source off.

Secondly, if I simply took the derivative of the function sin(wt)+1 and multiplied it by the mutual induction, would this give me the voltage in the second coil as a function of time?

Are you looking for the transient response or the steady state response?

Since it would appear that you're looking for the current output, it may be simpler to stick with what is known about how current behaves with coupled coils. What do you know about the relationships between the mutual inductance, the coupling constant, the turns ratio, and the currents in the coils? You should have formulae that relate these things.
 
  • #7
seanhbailey
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I'm looking for the transient response. I'm assuming that the current will just be replicated in the second inductor since V2 = k*(N2/N1)*V1 but I'm not 100% sure.
 
  • #8
seanhbailey
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So im looking for answer similar to V2 = V1*k*(N1/N2)*exp(-R1*t/L1) whcih is the voltage as a function of time but im not sure how a pulsating current affects this result, but I'm assuming Im going to end up having to solve a differential equation.
 
  • #9
gneill
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Ah. Well the transient response is going to be a bit more tricky, since the mutual inductance creates a sort of feedback effect from what's taking place in the second coil to the first coil, and vice-versa. There are effective voltage sources that get stuck in series with each coil,

[tex] M \frac{dI_2}{dt}[/tex]

and

[tex] M \frac{dI_1}{dt}[/tex]

These have to be incorporated into your differential equations. The first one goes in series with the first inductor. These voltage supplies can "look like" additional inductance in series with each inductor if you consider that the turns ratio determines the ratio of currents.
 
  • #10
gneill
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You'll have a transient response and a steady state response, each of which can be determined separately. The overall solution will be the sum of the two. Yes, you'll have a bit of slogging to do with differential equations for the transient response. The steady state response should be fairly easy to determine given the form of the steady-state part of the input signal.
 

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