Maximum stored energy of an inductance

Discussion in 'Homework Help' started by divisortheory, Sep 26, 2016.

  1. divisortheory

    Thread Starter New Member

    Sep 4, 2015
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    I'm using Schaum's Outline of Electric Circuits in conjunction with Art of Electronics and Practical Electronics for Inventors. This is entirely self-study, and not actually part of any coursework.

    The books I'm using though do not progress at the same pace as the schaum's outlines, so when I get to the section on Inductance and Capacitance I'm a little lost since I haven't encountered these concepts in the main books I'm reading yet. That said, I decided to give the problems a shot anyway. Here is one:

    An inductance of 2.0mH has a current i = 5.0(1 - e^(-5000t)). Find the corresponding voltage and maximum stored energy. So I read the quick summary of inductance, and it says V = L di/dt. Great, so I just plug the numbers in and V = 50.0 e^(-5000t). Seems I have that part correct.

    For the next part, the "worked solution" says "Since the maximum current is 5A, the maximum stored energy is ½L I_max^2 = 25.0 mJ.

    I don't follow this logic.

    The way I did this is to say P = VI, so P = 50e^(-5000t) × 5(1 - e^(-5000t)). And since P(t) = dW/dt, it should be the case that W is maximized when P(t) = 0. Solving the aforementioned P for 0, I get t=0. Sure enough, at t=0 the current is 5A as the solution suggests, and the voltage is 50V.

    What next? Do I have to integrate this? Ugh. If I do it and set t=0 in the result I do get 25.0mJ, but I feel like I'm missing something, because the solution jumped straight to ½L I_max^2. It sure seems like it's integrating L I with respect to I, but why?

    (BTW, is there a way to format superscripts and subscripts on this forum?)
     
  2. RBR1317

    Active Member

    Nov 13, 2010
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    The way I see it is at t=0 the current i=0, and rises to its maximum value within one second. Also, the energy stored by the inductor is found in the magnetic field generated by the current in the inductor. The voltage across the inductor is related to how fast the current through the inductor is changing and is not directly related to the energy stored in the field.
     
  3. crutschow

    Expert

    Mar 14, 2008
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    The energy applied is equal to the voltage times the current (which is constantly changing) at any instant in time.
    Thus to get the total applied energy (which is stored in the inductive magnetic field) you need to integrate voltage and current over the time the voltage is applied
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello,

    Cant you just use (1/2)*L*I^2 or do you want to prove that somehow?
     
  5. WBahn

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    Mar 31, 2012
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    You don't have to integrate each time because that is where the

    E = ½ L·I²

    comes from.

    Start with a discharged inductor (I = 0) and then determine the total work applied to it in order to bring it to a current I and you will discover that it is ½ L·I². Hence the energy stored in an inductor is a matter of state, meaning that if I tell you the inductance and the current as some moment in time, you can tell me the energy stored in the inductor at that same moment in time.
     
  6. divisortheory

    Thread Starter New Member

    Sep 4, 2015
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    I suppose that E = ½ L·I² is a standard result then I don't need to integrate it. As I said though, I hadn't gotten to the part in my texts that discussed this, so it kind of surprised me. I guess it makes sense though. If a discharged conductor has I=0 and a charged conductor has I=I_max, and the power at a particular point in time is L·I, then obviously the total work done is the integral of this with respect to I.

    One more question though. When I actually did all this by hand (work shown in the OP), I ended up finding that at t=0, I(0) = 5.0mA. And yet it would seem to make sense (and you kind of hinted), that at t=0 you would have a discharged inductor (so I=0). Where is this discrepancy coming from?
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Didn't you come up with this for i(t)?

    i = 5.0(1 - e^(-5000t)).

    First off, you are missing units and that will get you in trouble. You yourself are bouncing back and forth between using 5 A and 5 mA.

    But leaving that aside for now, what is i(t = 0)?

    Is not -5000t equal to zero when t = 0?

    What is e^0 ?
     
  8. divisortheory

    Thread Starter New Member

    Sep 4, 2015
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    Ahh, you are right. I did all that work and then somehow missed the 1 - . I was just computing 5 * e^(-5000t).
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Okay, stuff like that happens.

    Now let's consider your lack of using units.

    The final current in the inductor is not 5. It is 5 A. Because you didn't include the units, you messed up and said that the current (though you thought it was the initial current) was 5.0 mA. That's an error of three orders of magnitude!

    Nearly all physical quantities have units and those units are an integral part of the quantity. It is meaningless to say that someone is 72 tall or that they weigh 100. Those give you zero information about the quantities described. You can't even make comparisons with them. Is someone that is 72 taller or shorter than someone that is 32? You have zero idea.

    Beyond that, if you religiously track your units throughout your work, then you will find that you will catch most (not all) of your mistakes shortly after you make them instead of hours later (or not at all). In the academic realm this means you lose points unnecessarily. In the real world it means things get damaged and people get injured and killed unnecessarily.

    Sadly, the people that write text books almost universally come from a purely academic background with little to no industry experience. As a result they have the academic mindset and are very sloppy with units (even when they've gone to lengths to say how important they are -- but they really only mean that it's important on the final answer). This attitude then infects the vast majority of students learning from those texts.

    Consider the equation you started with (and that I'm guessing was given to you as is)

    i = 5.0(1 - e^(-5000t))

    This tells us nothing about the magnitude of the current -- is it 5 nA, 5 mA, 5 A, 5 kA? Any of those might be reasonable depending on the specifics.

    This tells us nothing about the time scale of the decay -- is t in microseconds, milliseconds, seconds, hours? Again, any of those might be reasonable depending on the specifics.

    So we are left with having to guess -- and engineering is NOT about guessing (at least not about this kind of stuff).

    I'm guessing (see -- having to guess, which is major-time not good) that this is 5 A and that the time is in seconds.

    Since the argument to any transcendental function has to be dimensionless, that means that the 5000 has units of inverse-seconds. Thus this equation should be

    <br />
i(t) \; = \; 5.0 \, A \, \(1 \; - \; e^{-5000 \, s^{-1} \; t} \)<br />

    In addition to being correct, by having the units we gain the option to write things in a manner that is easier for us to comprehend, such as:

    <br />
i(t) \; = \; 5.0 \, A \, \(1 \; - \; e^{\frac{-t}{200 \, \mu s}} \)<br />

    Now we can see immediately that we are dealing with a system that has a 200 μs time constant. If someone tells us that t = 1.2 ms we don't have to guess whether, perhaps, the t in the equation was supposed to be in milliseconds or not. Give me a time in picosecond or minutes and I can unambiguously tell you what the value of the exponent is in that equation.

    I can also do the math without having to go to base units, which tend to be either so big or so small that humans are more likely to make mistakes. For instance, if you want the voltage across the inductor, then

    <br />
v(t) \; = \; L \frac{di(t)}{dt}<br />
v(t) \; = \; 2.0 \, mH \frac{d}{dt}\(5.0 \, A \, \(1 \; - \; e^{\frac{-t}{200 \, \mu s}} \) \)<br />
v(t) \; = \; \( 5.0 \cdot 2.0 \) \, mH \cdot A \frac{d}{dt}\(1 \; - \; e^{\frac{-t}{200 \, \mu s}} \)<br />
v(t) \; = \; 10.0 \, mH \cdot A \( - \(\frac{-1}{200 \, \mu s} \) \) e^{\frac{-t}{200 \, \mu s}}<br />
v(t) \; = \; \frac{10.0}{200} \, \frac{mH \cdot A}{\mu s} \, e^{\frac{-t}{200 \, \mu s}}<br />
v(t) \; = \; \frac{10.0}{200} \, \frac{1000 \, m}{1} \, \frac{mH \cdot A}{\mu s} \, e^{\frac{-t}{200 \, \mu s}}<br />
v(t) \; = \; 50.0 \, \frac{m \cdot mH \cdot A}{\mu s} \, e^{\frac{-t}{200 \, \mu s}}<br />
v(t) \; = \; 50.0 \, \frac{H \cdot A}{s} \, e^{\frac{-t}{200 \, \mu s}}<br />

    Note that m·m cancels μ since (1 milli)(1 milli) = (1 micro).

    From the constitutive equation for inductance itself we see that

    <br />
1 \, H \; = \; 1 \, \frac{V \cdot s}{A}<br />

    Hence

    <br />
v(t) \; = \; 50.0 \, \frac{H \cdot A}{s} \, e^{\frac{-t}{200 \, \mu s}}<br />
v(t) \; = \; 50.0 \, \frac{H \cdot A}{s} \, \frac{V \cdot s}{H \cdot A} \, e^{\frac{-t}{200 \, \mu s}}<br />
v(t) \; = \; 50.0 \, V \, e^{\frac{-t}{200 \, \mu s}}<br />

    And if the units hadn't worked out to a voltage at the end, we would have known that we had made a mistake somewhere along the way and gone and tracked it down. This is MUCH better than just taking on a V onto the end of the final answer because that is what we want the units to be.

    Note that I did the above in much more detail than is normally needed just to show how it all works out.
     
  10. divisortheory

    Thread Starter New Member

    Sep 4, 2015
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    Thanks for the detailed response. yes, the lack of units has bothered me in some of these books. I've seen questions like "Given the following equation, calculate the value of I after 2 minutes". That's great, but does that mean t is in minutes? who knows. In any case, you are correct that I should be more careful.

    May I ask how you are formatting the equations? Is there some kind of built-in LaTeX integration or are you rendering them externally and then pasting images?
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    The forum supports LaTeX directly via the MimeTex rendering engine. You put your LaTeX code between TEX tags.
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello again,

    To prove the (1/2) L*I^2 you can do the following.

    The voltage across the inductor is:
    V*e^(-(t*R)/L)

    and the current through the inductor is:
    (V/R)*(1-e^(-(t*R)/L))

    so the instantaneous power is:
    V^2*e^(-(t*R)/L)*(1/R-e^(-(t*R)/L)/R)

    and integrating that we get the total power:
    (L*V^2)/(2*R^2)

    and note that V^2/R^2=(V/R)^2=I^2 so we substitute and get:
    (L/2)*I^2

    which is again (1/2)*L*I^2 where I=V/R.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    That's not a very convincing proof for a couple of reasons.

    First, it starts with an equation for voltage and a separate equation for current that have no link between them. They are just given out of thin air. Second, they apply only to one very specific situation, namely a first-order L-R circuit. Thus the result is only proven for the specific case of a first-order L-R circuit.

    The proper and general way to do it is to use the definition of work and the constitutive equation directly.

    The energy stored in the inductor is equal to the electrical work done on the inductor as the current changes from 0 to some final current I.

    <br />
E(I) \; = \; \int_o^I dW<br />
E(I) \; = \;  \int_0^T{\frac{dW}{dt}dt}<br />
E(I) \; = \; \int_o^T {p(t) \, dt}<br />
E(I) \; = \; \int_o^T {v(t) \cdot i(t) \, dt }<br />
E(I) \; = \; \int_o^T {L \frac{di}{dt} \cdot i(t) \, dt }<br />
E(I) \; = \; \int_o^I {L \cdot i \, di }<br />
E(I) \; = \; L \int_o^I {i \, di }<br />
E(I) \; = \; \frac{1}{2} L \cdot I^2<br />

    This assumes nothing about the circuit that the inductor is part of or the shape of the current waveform. The only thing it assumes is that the inductance is constant (which isn't always a good assumption).
     
  14. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello,

    Wasnt he working with an RL circuit?
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    He was asking about why they could jump straight to using E = L·I²/2.

    If you want to use your explanation, then you have to caveat it with the proviso that you've only shown that they can jump straight to using that expression when they are working with a first order series RL circuit. In fact, that explanation only applies to the final energy in the inductor in response to a step input voltage.

    If he has a second order circuit, he has to start from scratch.

    If he has a ramp input, he has to start from scratch.

    If he has a sinusoidal input and wants to know the energy at an arbitrary point in the cycle he has to start from scratch.

    Doesn't that seem awfully limiting?

    Wouldn't it be better to show that the energy stored in an inductor obeys that relationship regardless the circuit topology and regardless of the waveform or the history of the current in the inductor? That it ONLY depends on the inductance and the current in the inductor at that moment in time?
     
  16. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Sure, that's fine with me.
    No problem :)
     
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