Am I making this more complicated than I need? Is spice failing me?

Thread Starter

Electronics117

Joined Sep 9, 2017
27
Im trying to solve for part b) and am just getting hung up on finding the current through an individual inductor it asks for. Im thinking of using a current divider but if i solve for L-equivalent to use the current divider equation I will not be able to extract the current asked for, unless its just that current divided accordingly. But its not matching my spice.

Thanks in advance.
 

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crutschow

Joined Mar 14, 2008
23,544
If you connected the circuit properly, with the correct component values and voltage simulation waveform, then Spice is quite likely giving you the correct answer.
 

Thread Starter

Electronics117

Joined Sep 9, 2017
27
If you connected the circuit properly, with the correct component values and voltage simulation waveform, then Spice is quite likely giving you the correct answer.
What about solving for the theoretical equation?
Just find the series equation with Leq and multiply by 1/3?
 

crutschow

Joined Mar 14, 2008
23,544
The inductors have different values, so you can't just divide the current by 3.
But you'll need to find other help solving the math, as I'm semi-allergic to it. ;)
 

MrAl

Joined Jun 17, 2014
6,649
Hello,

I dont see how you can read that diagram as it is way too light. Here is an enhanced version so everyone here can read it better.

One of the main points here i think is how do you calculate the initial DC current in two inductors with different values that are in parallel. I will pose this question in another section so we can take a closer look at this. In the mean time, here is an enhanced version of the original problem...
 

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The Electrician

Joined Oct 9, 2007
2,750
One of the main points here i think is how do you calculate the initial DC current in two inductors with different values that are in parallel.
An easy solution to that problem is to establish an additional variable, a partition coefficient if you will, that is the ratio of the two DC currents. Include that variable in the total solution.
 

WBahn

Joined Mar 31, 2012
24,857
What about solving for the theoretical equation?
Just find the series equation with Leq and multiply by 1/3?
There's an assumption that you need to make about the initial currents in the two parallel inductors. Normally, you assume that the system starts at rest (zero current). But that doesn't work here because your voltage source is non-zero for all time prior to t=0.

So you have a couple of options. The purest is to assume that there is an arbitrary current circulating in those two inductors and then work the problem and see if that circulating current affects the answers. If it doesn't, then you are set.

If it does, then you need to make further assumptions. Probably the one I would lean toward quickly would be to do the analysis with a waveform that starts with the system at rest and then ramps the voltage to -10 V and let's the system stabilize. You then use this as the initial conditions (at t = -∞) for your actual problem.

You can also determine the rule for current division in parallel inductors that applies under the assumption of zero initial current -- but do be aware that you do have to make that assumption for the rule to be valid.
 

MrAl

Joined Jun 17, 2014
6,649
An easy solution to that problem is to establish an additional variable, a partition coefficient if you will, that is the ratio of the two DC currents. Include that variable in the total solution.
Hi,

Well maybe you could show that.

See what happened was i looked at it as a step response and then came up with a ratio of the two inductors, but when i consider a pure DC i get equal currents in both inductors. See what you can make of it.
 

MrAl

Joined Jun 17, 2014
6,649
What did spice give you?
Hi,

I cant even get a result as MicroCap gives an error that there is an inductive voltage loop, which normally means that an inductor is being driven directly by a voltage source which is not allowed. The only way to get it to work is to add some series resistance to at least one of the two inductors.

I set up a simple circuit with a DC source, a small resistance of 1 ohm, and the two inductors in parallel to try this.
Theoretically i can get an answer if i consider the DC source to be a unit step. So this seems to bring up more questions. As time progresses, the voltage drops to zero, yet there is current in the inductors, which makes sense. With zero voltage though, that means that the current can not change in either inductor. That means that if one inductor has 2/3 of the total current and the other has 1/3 of the total current, it's going to stay that way forever. A strange situation which of course is theoretical only, but when these kinds of questions come up for students the only answer seems to be that more information has to be provided unless there has been a precedent set in the classroom as to how these situations should be handled.
 
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crutschow

Joined Mar 14, 2008
23,544
That means that if one inductor has 2/3 of the total current and the other has 1/3 of the total current, it's going to stay that way forever. A strange situation....
That's not really strange.
To change the current in an ideal inductor you must supply a voltage and since that voltage is zero, there is no change.
 

RBR1317

Joined Nov 13, 2010
502
The purest is to assume that there is an arbitrary current circulating in those two inductors and then work the problem and see if that circulating current affects the answers.
But if you have an arbitrary current circulating in the inductors, then you end up with an arbitrary amount of energy stored in the inductors. So how could one determine the amount of energy stored in each component?

Going back to my old Linear Circuits text by Scott, chapter 14, Power and Energy in Circuits with Energy-Storage Elements, I found this:

"... The energy delivered by an impulse can be either zero or finite, depending on the circuit element involved, and part of the energy may apparently vanish into thin air. The latter phenomenon occurs in lossless circuits and is a consequence of infinite current flowing in zero resistance. {In any lossless system it is impossible to conserve both energy and momentum. We conserve momentum (charge or flux linkage).} A subtle form of this problem occurs when energy is exchanged between lossless elements of the same sort. ..."

It would seem that the subtle form of the problem occurs with parallel inductors. If one of the inductors has an imperceptible (but non-zero) resistance, then all the current flows through the other perfect inductor in the steady-state condition. But if you replace two parallel inductors with a single equivalent inductance, then the energy calculations will make sense.
 

WBahn

Joined Mar 31, 2012
24,857
But if you have an arbitrary current circulating in the inductors, then you end up with an arbitrary amount of energy stored in the inductors. So how could one determine the amount of energy stored in each component?
You are correct. I had a hard time reading the question parts and I thought it was asking only for the amount of energy supplied by the source after t = 0 (so the amount added to the energy stored in each inductor and not the total stored).

Going back to my old Linear Circuits text by Scott, chapter 14, Power and Energy in Circuits with Energy-Storage Elements, I found this:

"... The energy delivered by an impulse can be either zero or finite, depending on the circuit element involved, and part of the energy may apparently vanish into thin air. The latter phenomenon occurs in lossless circuits and is a consequence of infinite current flowing in zero resistance. {In any lossless system it is impossible to conserve both energy and momentum. We conserve momentum (charge or flux linkage).} A subtle form of this problem occurs when energy is exchanged between lossless elements of the same sort. ..."

It would seem that the subtle form of the problem occurs with parallel inductors. If one of the inductors has an imperceptible (but non-zero) resistance, then all the current flows through the other perfect inductor in the steady-state condition. But if you replace two parallel inductors with a single equivalent inductance, then the energy calculations will make sense.
I don't think that description applies here because I think all voltages and currents are finite at all times. The source is delivering a step change in voltage, not an impulse and there's nothing trying to make a step change in current in the inductors.
 

RBR1317

Joined Nov 13, 2010
502
The source is delivering a step change in voltage, not an impulse and there's nothing trying to make a step change in current in the inductors.
It is true that there are no impulsive driving voltages present; nevertheless, depending upon how one may assume current division in parallel inductors, there may be theoretical impulsive conditions between the parallel inductors as the circuit adjusts to new driving conditions. I assume this is what the author was referring to by "a subtle form of this problem occurs when energy is exchanged between lossless elements of the same sort." Also, if energy can apparently vanish into thin air, can it not also appear from thin air? Should we take this to mean that the problem as given is not susceptible to rigorous mathematical analysis?
 

WBahn

Joined Mar 31, 2012
24,857
It is true that there are no impulsive driving voltages present; nevertheless, depending upon how one may assume current division in parallel inductors, there may be theoretical impulsive conditions between the parallel inductors as the circuit adjusts to new driving conditions. I assume this is what the author was referring to by "a subtle form of this problem occurs when energy is exchanged between lossless elements of the same sort." Also, if energy can apparently vanish into thin air, can it not also appear from thin air? Should we take this to mean that the problem as given is not susceptible to rigorous mathematical analysis?
An example of the apparent loss of energy occurs when you do something like charge an ideal capacitor by connecting it to an ideal voltage source. Only half of the energy delivered by the source winds up in the capacitor. But it is not that hard to account for the rest of the energy -- it is "lost" in the connection between the two since you have a nonzero current flowing through a nonzero voltage drop. So it "vanishes" because it is being converted into some other form -- the problem statement is simply too incomplete to determine what that other form happens to be.
 
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