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Inductance Help

  1. Jun 3, 2005 #1
    Inductance Help!!!!

    https://www.physicsforums.com/attachment.php?attachmentid=3779&stc=1


    Hello,

    I was wondering if I could get some help on the following questions,

    First of all lets me explain....I am a high school student who is trying to learn the basics of electrical engernerring on the the side, some day I hope to attend college,and study this field, but at the moment I have an old engineering text book I got from local libary, and I try to solve the questions.....BUT I DO NOT KNOW HOW TO SOLVE THESE EQUATIONS, I have looked the internet through for some input on the formulas to use with Inductance, but find theres too much Algebra involved,without actually giving me the formulas I need!!!

    I would appreciate if someone could look at these questions, and give me some advice on how to solve for each one, and what formulas to use to find the answers, Again I do not attend college, still in grade 10, if possible the answers to each question would be great with the equations worked out so I can follow it would be great!!!

    But I am hoping someone will give me some time on the following questions.

    I look forward to your reply.

    Thanks

    Wolf
     

    Attached Files:

  2. jcsd
  3. Jun 4, 2005 #2
    To solve a circuit you only need two things: The constiutive relations for each element in it, and some conservation law.

    The consitutive relations relate the current flowing through a device to the voltage drop across it.

    [tex]V=IR[/tex] for a resistor

    [tex]V = L\frac{di}{dt}[/tex] for an inductor

    [tex]I = C\frac{dv}{dt}[/tex] for a capacitor

    The conservation laws to use are either Kirchoff's voltage law or Kirchoff's current law. They say that the sum of the voltage drops across all the elements in a loop is zero, and the sum of all the currents entering and exiting a point is zero.

    For your problem, you would want to say that the voltage drop across E, plus the voltage drop across the resistor, plus the voltage drop across the inductor is zero. Then use the consitutive relations to find the terms for this equation. Then you can solve for v(t) (or i(t)), and answer all the questions.

    I'm not sure if you can do these problems without calculus or differential equations though. I'm assuming you don't know it because you said you're only in 10th grade right now. I guess one way would be to use the impedances of the devices, and then look up the inverse Laplace transforms in a table. That only requires solving a quadratic equation for cases like this.

    So is that your problem? That you aren't sure how to solve the differential equations you get?
     
  4. Jun 4, 2005 #3
    Test

    R=11Ohms
    L=6200 H
    E=810 V

    The switch is intially in position A and the inductor is fully charged. Calculate the time after the switch is set toposition B for the inductor Voltage to drop to 230V

    Voltage Loss=E*e^(-t/Tau)
    230V=810V*e^(-t/(6200H/11Ohms)
    0.283950617=e^(-t/563.6363636)
    t=-563.6363636*ln(o.283950617)
    t=709.59 or rounded off=709.60 Seconds

    Any commets would be appreciated

    Wolf
     
    Last edited: Jun 4, 2005
  5. Jun 4, 2005 #4
    Ok, it looks like you were given some formulas for the solutions of simple RLC circuits so you can avoid solving the differential equations. That means if I solve it the long way and get the same thing as you, then your answer is right.

    First off, the equation. Sum of the voltages around the loop equals zero.

    [tex]E=V_R+V_L=i(t)R+L\frac{di(t)}{dt}[/tex]

    The time constant is basically the root of the characteristic equation of the homogenous version here, or if we guess a solution of
    [tex]i(t)=Ae^{rt}[/tex]
    and plug it in to the homogenous equation, I get

    [tex]0=RAe^{rt}+LAre^{rt} \rightarrow r=-\frac{R}{L} [/tex]
    This is the time constant (maybe without the minus sign, I'm not sure).

    Now with some more magic we can solve the nonhomogenous version and after finding some constants get


    [tex]i(t)=-\frac{E}{R}e^{-Rt/L}+\frac{E}{R}[/tex]

    Now that we have an equation for the current through the circuit at any time, you can find the voltage across any element at any time by just using the constitutive relations.

    You want to find the time it takes for the voltage across the inductor to become something, so we need to use

    [tex]v(t)=L\frac{di(t)}{dt}[/tex]

    Since you're given a v(t), plug that in and solve for time.

    [tex]V_{given}=-L\frac{E}{R}\frac{(-R)}{L}e^{-Rt/L}=Ee^{-Rt/L}[/tex]

    [tex]t = -\ln(\frac{V_{given}}{E})\frac{L}{R}[/tex]

    And that's exactly the same equation you were using!

    Only problem is I think you messed up when you found R/L, I think you're off by a factor of 10.

    Also, you say that the switch is intially closed so the inductor is fully charged and you want to find the voltage drop after the switch is opened. I think this is a typo because an inductor acts as an open initially, which means if the switch starts open, right after it's closed there will be no current flowing through the inductor, and the entire 810 volts will be dropped across the inductor. Then you use the equation to find the drop in voltage across the inductor. If the switch starts closed, then after a while the inductor will act as a short, so when the switch is opened there will be no voltage drop across the inductor, so all 810 volts will be dropped across the resistor. So if the switch is opened, you want to measure the voltage drop across the resistor, or the voltage increase across the inductor. (Or maybe they just want a negative voltage drop, but that may confuse the young EE student.)

    This may seem like a stupid point, but it's important to understand what's actually going on in the circuit. You can test it by plugging in t=0 and t=inf to find the voltage across both the resistor and the inductor, and hopefullly it matches what I said above.
     
  6. Jun 4, 2005 #5

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  7. Jun 4, 2005 #6
  8. Jun 4, 2005 #7
  9. Jun 4, 2005 #8
  10. Jun 4, 2005 #9
    Ok, forget what I said about your R/L being off by ten. I read your R=10Ohms as R=100.

    That's what I was saying. Eventually the inductor acts like a short circuit, so you can just draw a wire through it. Then you only need to worry about the resistor.
    The equation for current matches with the one I got.

    I'm confused. Are these the solutions? Or your work? What exactly are you looking for now? I'm sorry for using calculus, but I just wanted to show you where it all comes from.
     
  11. Jun 4, 2005 #10
    These are my workings, but again I don't want to know how calculus plays a part in these question, too me that is irrelevant, I need to know, what is the final answer if you look closer at both answers they are off, this is because of the rounding off past the decimal place, so if this were an exam, what would be consider correct, if the questions states nothing about rounding off!!

    I pretty much know how to solve these questions, but like I mentioned before I just want someones opinion on if they are correct or not, I do not want to think I am doing this correct only to find out down the road that this is wrong!!

    So once again, If you can post your workings, so I can compare my workings to yours, I want to see how you play out the process with concern towards the decimal place!!

    And again thanks for your post, I enjoy it!!!
     
  12. Jun 4, 2005 #11
    So that's your workings using my formula? Now I'm not so sure of my work... I think I may have used the wrong initial conditions to find the constant. Although it matches the equation you used in the earlier post, so I don't know. I should probably mention that I don't actually know much about EE, and I should probably step down and let somebody else solve the actual circuit.

    But I can say a few things about the math.

    When you have

    [tex]V_1 = V_2 e^{st}[/tex]

    There's no need to replace e with 2.71... and do all that other fancy stuff. You said your calculator can do log base e, so use

    [tex] log_e(\frac{V_1}{V_2}) = log_e(e^{st}) = st[/tex]

    As for digits of precision, it's best to carry the symbols through as far as you can. If you can get an expression for t in terms of all the given parameters, you juts put in those nice round numbers and the calculator will keep all the extra digits for the intermediate compuations and your answer should be more accurate.

    For example, if I use

    [tex]t = -\ln(\frac{V_{given}}{E})\frac{L}{R}[/tex]

    I get t = 709.59278....

    Just by plugging 709.59 into this

    [tex]V(t)=Ee^{-Rt/L}[/tex]

    I get V = 230.001

    So by using symbols as long as possible, you preserve accuray.

    Symbols also help you from making mistakes by dimensional analysis. If I get an equation that says i(t) = (V0 + V1/R)*exp(t), I know something is wrong because V1/R is current, but V0 is voltage. And I can't add a current and a voltage. But if you just had numbers there, you would never know you had a mistake.
     
  13. Jun 4, 2005 #12
    Thanks

    Hey.....Thanks for your input, like I said before I am just looking for someone else to bounce these question off, and see what someone else thinks, I really don't have anybody else to ask if they are correct or not, but I do appreciate you opinion, and thats what I am looking for, so please if I post another question please enlighten me, I value your opinion!!!

    Wolf
     
  14. Jun 4, 2005 #13

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    Last edited: Jun 4, 2005
  15. Jun 5, 2005 #14
    I think I know what I did wrong in the first problem. I shouldn't have had the E in my differential equation because the switch was opened, so E was zero. That would give me something of the form

    [tex]i(t)=I_0e^{-Rt/L}[/tex]

    Now since the switch had been closed for a long time, there is no voltage drop across the inductor, so all of the voltage drop is on the resistor, so Io = E/R.

    [tex]i(t)=\frac{E}{R}e^{-Rt/L}[/tex]

    Now as time increases, the current in the inductor goes to zero which makes sense because the switch is open, so the inductor will discharge and eventually act like a short circuit. So I think that's the equation you should have used in the first problem I tried to do. But if I was wrong before, maybe you shouldn't trust me now either.

    Anyway, now to the current problem. Since the other equation I got was for the switch open initially, that should work here. So,

    [tex]i(t)=\frac{E}{R}(1-e^{-Rt/L})[/tex]

    I think the equation you have there is for the voltage across the resistor.
    According to my equation,

    [tex]V_L=Ee^{-Rt/L}[/tex], and you have [tex]V_R=E(1-e^{-Rt/L})[/tex], and we know that VR+VL = E, so adding those two together does indeed give E.

    You could also say that the voltage across the resistor must be the voltage from the source minus the drop across the inductor, and just by looking at the equation you can see that those are the two terms in it.

    But that's ok, because we know that the sum of all the voltages around the circuit must be zero, and E is constant, so if the inductor voltage drops by V0 volts, then the resistor voltage must increase by V0 volts. So you could use either equation to find the time it takes for a change in voltage of some amount to happen on either element.

    So do this problem exactly as the one before. Plug the given voltage value into the equation for VL, then solve for t, and convert that to taus.

    Alternatively, if you want to use the VR equation, since the inductor voltage starts at E and drops to 120, the change in voltage is (E-120), and as I said before, a change on the inductor will have the same change on the resistor, so plug VR = (E-120) into your equation for resistor voltage and see if you get the same answer. That will be a good way to check if everything is making sense.

    (To see how to get the tex right, click on one of my equations and it will show you the tex code.)
     
  16. Jun 5, 2005 #15
    Peavey_Wolfgang_200,
    There are only two forms of the solutions to the differential equations for these first order circuits.

    First order circuits are those that contain only one capacitance or one inductance, and are called first order because the differential equation that describes their operation is "first order". You don't have to worry about that at this stage (although it will be relevant when you come to study these circuits in an engineering degree).

    What is important is that the current, or the voltage, with either go up or go down, and that tells you what the solution looks like. Those that go up look like [tex]v(t)=V(1-e^{-t/\tau})[/tex] and those that go down look like [tex]i(t)=I.e^{-t/\tau}[/tex]. These equations don't allow for the cases where the levels don't start of finish at zero, so it is easier to use the general form of the solution, which covers all cases. The general form of the solution is

    [tex]x(t) = x_{final} + (x_{initial} - x_{final} ) e^{(-t/\tau)} [/tex]

    where [tex] x [/tex] is the variable that you are interested in (voltage or current). The value of [tex] \tau [/tex] depends on the capacitance or inductance in the circuit and the total resistance seen by that C or L during the change (sometimes some resistors will not be involved after a switch changes position). For an inductive circuit [tex]\tau=L/R[/tex] and for a capacitive circuit [tex]\tau=RC[/tex]. (You should be able to work out why the L and C are on the top in these equations, and why the R changes position).

    In the problem posed,
    [tex]\tau=L/R=860/16=53.57 seconds[/tex]
    [tex]v_L(t) = 170 = v_L_{final} + (v_L_{initial} - v_L_{final} ) e^{(-t/53.57)}[/tex]
    [tex]\Rightarrow 170 = 0 + (850 - 0) e^{(-t/53.57)}[/tex]
    [tex]\Rightarrow 170 = 850 e^{(-t/53.57)}[/tex]
    [tex]\Rightarrow 170/850 = 0.2 = e^{(-t/53.57)}[/tex]
    [tex]\Rightarrow ln(0.2) = -t/53.57[/tex]
    [tex]\Rightarrow (-1.609)\times(-53.57) = t[/tex]
    [tex]\Rightarrow t=86.22 seconds[/tex], or [tex]t=1.61 \tau[/tex]

    If you use this equation all the time, all you have to do is work out the time constant and the initial and final values (which you have been doing anyway) and plug them in.

    On the number of decimal places ... the resolution of your answer is bound by the number of significant digits in the information you start with. In the case you have been given, the resistor is only given as two digits, which would suggest that there is error in the third digit (it was not given as 16.0 ohms). This means that the third digit of your answer will have error in it, and it is not useful to record any extra digits. Thus I gave the answer as [tex]1.61 \tau[/tex] rather than [tex]1.609 \tau[/tex] or anything else. (Of course, if your examiner says "calculate to the fifth decimal place" then you have no choice).

    Hope this helps. (When you do the engineering program you will have to solve the differential equations so that you know how to handle second order circuits, with both L and C ... but leave that until you have to do it.)
     
  17. Jun 5, 2005 #16
    Inductance QUestion Thank you

    Hey First Of all I want to thank LeBrad & Xodar, you guys have helped me out!!

    And secondly Xodar I like how you showed me your workings and explanation of how to solve the equation!!....and your easier formula expression is also great,

    What do you think of the first question? does your answer come close the the answer I posted, your opinion would be great, especially if you show me how you came to your conclusion.

    I look forward to conversing with you guys and many more in the days to come!!

    Thanks Again for your guys help!!!

    WOLF
     
  18. Jun 5, 2005 #17
    Another question

    What about this one!!

    (using same diagram as before)

    R = 16 Ohms, L = 720 H, and E = 880 V. The switch is intially in position B and the inductor is ully discharged. Calculate the number of time constants after the switch is set to position A for the inductor voltage to drop to 110 V ( Calculate to two decimal places)

    Using the below formula

    [tex]x(t) = x_{final} + (x_{initial} - x_{final} ) e^{(-t/\tau)} [/tex]

    [tex]\tau=L/R=720/16=45 seconds[/tex]

    [tex]v_L(t) = 110 = v_L_{final} + (v_L_{initial} - v_L_{final} ) e^{(-t/45)}[/tex]
    [tex]\Rightarrow 110 = 0 + (880 - 0) e^{(-t/45)}[/tex]
    [tex]\Rightarrow 110 = 880 e^{(-t/45)}[/tex]
    [tex]\Rightarrow 110/880 = 0.129411765 = e^{(-t/45)}[/tex]
    [tex]\Rightarrow ln(0.129411765) = -t/45[/tex]
    [tex]\Rightarrow (-2.04475598)\times(-45) = t[/tex]
    [tex]\Rightarrow t=92.01 seconds[/tex] or [tex]t=2.04 \tau[/tex]

    Is this Correct?
     
  19. Jun 5, 2005 #18
    THIS IS FOR Xodar

    I think i undestand how to solve for voltage drop, but when talking about voltage rising, do we use the same formula,

    For example, still using the same digram!!

    R = 29 Ohms, L = 360 H, and E = 760 V. The switch is fully discharged. Calculate the number of time constants after the switch is set to position A for the resistor voltage to rise to 640 V ( Calculate to two decimal places)

    Would you still use this formula [tex]x(t) = x_{final} + (x_{initial} - x_{final} ) e^{(-t/\tau)} [/tex] or something else?

    If you could work this one out so I can follow it, would be great!!

    I look foward to your response

    WOLF
     
  20. Jun 5, 2005 #19
    Hi Wolf,
    The general solution we have been using is for the current series circuit or the voltage across the inductor (it is for the parameters of the reactive element) ... which means it does not generally apply directly to the resistor voltage.

    However, you can use it to solve for [tex]v_R(t)[/tex] by solving for the current in the inductor or the voltage across the inductor (whichever seems easiest at the time).

    Because you have a series circuit,
    [tex]V_s=v_L(t)+v_R(t)[/tex]
    [tex]\Rightarrow v_R(t)=V_s-v_L(t)[/tex]
    [tex]\Rightarrow v_R(t)=V_s-[v_{Lfinal} + (v_{Linitial} - v_{Lfinal} ) e^{(-t/\tau)}][/tex]
    [tex]\Rightarrow v_R(t)=V_s-[0 + (V_s - 0 ) e^{(-tR/L)}][/tex]
    [tex]\Rightarrow v_R(t)=V_s- V_s \times e^{(-tR/L)}[/tex]
    and now solve for [tex]t[/tex] with the parameter you have given.

    Alternatively, you know the
    [tex]v_R(t)=R \times i_L(t)[/tex]
    [tex]\Rightarrow v_R(t)=R \times[i_{Lfinal} + (i_{Linitial} - i_{Lfinal} ) e^{(-t/\tau)}][/tex]
    [tex]\Rightarrow v_R(t)=R \times[\frac{V_s}{R} + (0 - \frac{V_s}{R}) e^{(-tR/L)}][/tex]
    [tex]\Rightarrow v_R(t)=V_s - V_s \times e^{(-tR/L)}[/tex]
    (same equation).

    you should get 1.85[tex]\tau[/tex], unless I made a math error.
    The answer for the earlier question certainly seems correct.

    You can estimate the answer by knowing that after one time constant you will be 63 percent closer to the final answer. So for the case in the current question, you are going from zero volts (on the resistor) toward 760 volts, and want to know how long it will take to get to 640 volts. 63% of 760 is about 480 volts... so it is more than one time constant. 63% of the next bit (760-480) is another 180 volts, giving 480+180=660v after two time constants. So you expect your answer to be a bit less than two time constants, which it is.
     
  21. Jun 5, 2005 #20
    Xodar

    Hey, I really appreciate the time you are spending, in helping me with this subject!!

    But could you show me your working for the prvious question, eventhough I understand what you are saying, I like to have a hardcopy so i can follow,

    I know this takes a little time to do, but i would appreciate it,

    And also, I notice you do not allow PM's, would you consider giving me a e-mail address, so in the future I can e-mail you and get your opinions.

    Again Thanks

    Wolf.
     
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