Little Circuit Challenge, DC Current In Inductor

Thread Starter

MrAl

Joined Jun 17, 2014
9,626
For the zero coil resistance case, this is only always true if the initial currents in the inductors are both zero when the voltage is applied. Otherwise, if there are initial currents in the inductors when the voltage is applied, and if the sum of those currents is 1 A, then those currents will persist for all time (as with WBahn's example). If the sum of the initial currents is something other than 1 A, the currents will vary exponentially for a while, and then settle down to values that will persist.
Hi,

Yes i think that is right. In my next post i'll quote the final value function where the initial currents can be made to be zero or non zero for quick observation.
 

Thread Starter

MrAl

Joined Jun 17, 2014
9,626
Removing L2 to find the DC current in L1 is completely unjustified. You seem to be trying to use a completely invalid version of superposition.

Furthermore, you go from shorting out the inductor to removing it from the circuit (which is open-circuiting it).

Imagine a circuit in which L1 and L2 are in series. What would the result be of removing L2 to find the DC current in L1?

Or have them in series but also have each have a parallel resistor across it, say R1 and R2. To keep things finite, have a resistor Ro in series between the supply voltage Vo and the inductor/resistor combinations. Please show how removing L2 to find the DC current in L1 and then doing the reverse to find the DC current in L1 yields a correct answer.

Let's put some numbers to this. Make Vo = 10 V, R1 = 4 Ω, R2 = 9 Ω, R3 = 1 Ω, L1 = 1 H, L2 = 2H.

According to you, if we remove L1 then the DC current in L2 is 2 A while when we remove L2 the DC current in L1 is 1 A. Yet if we do it the right way, we get that the DC current in both inductors is 10 A.

Even if this did somehow work, your reasoning for why the actual DC current would be shared equally is specious -- IF the inductors were identical and IF the initial conditions had equal currents in each of them, THEN you could make such an argument based on symmetry. But they aren't identical and we don't know whether the initial conditions were the same for each or not. It comes down to it making "sense" only because that's the result you would like to have.



You either have to specify constraints that make the initial conditions immaterial OR specify the needed initial conditions. You are wanting to do neither. You are imposing the constraint that the initial voltage at t = -∞ is E1 but you won't specify the initial currents in the inductors at that time. This is needed since you also don't specify constraints, such as inductor series resistance, that would make that information immaterial. Can't have it both ways.



You are making a lot of unsupported assertions here. Take a 1 H superconducting coil and get 10 A persistent flowing downward in it. Take another superconducting coil of 2 H and get a 9 A persistent current flowing upward in it. Now connect them in parallel and put them into that circuit with your 1 V, 1 Ω source. At this point you will have 10 A in one coil, 9 A in the other, 1 A in the resistor, and the persistent mode switches will have a zero net current in them. Now open the switches. You will have 10 A flowing downward in the first coil, 9 A flowing upward in the second, and and 1 A flowing in the resistor. Note that the details of whether the DC voltage was somehow "pure" is completely immaterial because the assembly of the circuit is not dependent on that in any way.
Hi,

I cant remember what i was thinking when i wrote that, but rather than take the time to go over it all again i'll just post the final value funciton and a plot of the current in L2 with initial currents in both varying from 0 to 5 amps.
The initial post did not mention initial currents, but it turns out that it's an interesting aspect of this circuit too and credit for that goes to WBahn :)

The function i got was this:
iL2=2*i2/3-i1/3+1/3

where i1 and i2 are initial currents in L1 and L2 respectively.
This function remains to be verified and there may have to be constraints placed on the initial values here.
What is interesting is that the current can be negative, if the initial current in L1 was high enough. This might be because the inductor acts as a current source if there is another inductor to dump it into.
See what you think, and remember it could be wrong.

The plot is shown in the diagram, it's a slanted plane.
 

Attachments

The initial post did not mention initial currents, but it turns out that it's an interesting aspect of this circuit too and credit for that goes to WBahn :)

The function i got was this:
iL2=2*i2/3-i1/3+1/3
I gave numerical examples of what final currents would be for several initial currents way back in posts #7, #8, #12 and #13, and described the behavior, which is qualitatively different depending on whether i1 and i2 sum to 1, or not.

Expressions for the final currents are part of the solutions to the differential equations. I didn't give general expressions for them, but that's easy enough.

Here are solutions for the differential equations in the zero resistance case, where the initial currents are i1 and i2:

Pic9.png

I've highlighted the final current expressions in red, and we see that:

iL1 = 2/3 + i1/3 - 2 i2/3

iL2 = 1/3 + 2 i2/3 - i1/3

This verifies your result.
 

Thread Starter

MrAl

Joined Jun 17, 2014
9,626
I gave numerical examples of what final currents would be for several initial currents way back in posts #7, #8, #12 and #13, and described the behavior, which is qualitatively different depending on whether i1 and i2 sum to 1, or not.

Expressions for the final currents are part of the solutions to the differential equations. I didn't give general expressions for them, but that's easy enough.

Here are solutions for the differential equations in the zero resistance case, where the initial currents are i1 and i2:

View attachment 180199



I've highlighted the final current expressions in red, and we see that:

iL1 = 2/3 + i1/3 - 2 i2/3

iL2 = 1/3 + 2 i2/3 - i1/3

This verifies your result.

Hi,

Oh ok yes thanks for doing this too.

I think it is most interesting that if one coil has enough current, it forces some current into the other coil too and it just loops around (DC current).
So with no resistance we have a current loop for certain initial current values.
I think that is most interesting also because the voltage across the two tend to zero in the steady state.
 

Thread Starter

MrAl

Joined Jun 17, 2014
9,626
For the zero coil resistance case, this is only always true if the initial currents in the inductors are both zero when the voltage is applied.
Hi,

Well actually, the condition for that to happen is given by:
i1=2*i2

where i1 and i2 are the initial currents in L1 and L2 respectively.
Of course i2=0 and therefore i1=0 is one choice, but then so is i2=1 and therefore i1=2 is another.
I guess these solutions would be the intersection of that slanted plane with a line that lies entirely in the plane.
 
Hi,

Well actually, the condition for that to happen is given by:
i1=2*i2

where i1 and i2 are the initial currents in L1 and L2 respectively.
Of course i2=0 and therefore i1=0 is one choice, but then so is i2=1 and therefore i1=2 is another.
I guess these solutions would be the intersection of that slanted plane with a line that lies entirely in the plane.
I put the word "always" in there on purpose. I didn't want to get into the details of the case where the initial currents are non-zero, but giving a case where the final currents were guaranteed to be iL1 = 2/3 and iL2 = 1/3. I also guessed that this was probably the case crutschow was contemplating.

In fact, the primary constraint on the currents is that the sum of the final currents must be 1 A.

You can pick an arbitrary value for the initial current in L1 (designated i1) and the desired final current in L1 (designated iL1).

Then iL2 = 1 - iL1 and i2 = i1/2 - (3 iL1)/2 + 1

Interestingly, if i1 = 2/3 and i2 = 1/3, then iL1 = 2/3 and iL2 = 1/3, but if i1 =1 and i2 =1 then iL1 = 1/3 and iL2 = 2/3

The case WBahn gave could have started with i1 = 20, i2 = -4 and progressed to final currents of iL1 = 10, iL2 = -9.
 

Thread Starter

MrAl

Joined Jun 17, 2014
9,626
I put the word "always" in there on purpose. I didn't want to get into the details of the case where the initial currents are non-zero, but giving a case where the final currents were guaranteed to be iL1 = 2/3 and iL2 = 1/3. I also guessed that this was probably the case crutschow was contemplating.

In fact, the primary constraint on the currents is that the sum of the final currents must be 1 A.

You can pick an arbitrary value for the initial current in L1 (designated i1) and the desired final current in L1 (designated iL1).

Then iL2 = 1 - iL1 and i2 = i1/2 - (3 iL1)/2 + 1

Interestingly, if i1 = 2/3 and i2 = 1/3, then iL1 = 2/3 and iL2 = 1/3, but if i1 =1 and i2 =1 then iL1 = 1/3 and iL2 = 2/3

The case WBahn gave could have started with i1 = 20, i2 = -4 and progressed to final currents of iL1 = 10, iL2 = -9.

Hi,

I guess we could compare this to the single shorted out inductor with initial DC current case next.
 
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