RL Circuit

Discussion in 'Homework Help' started by ihaveaquestion, May 9, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    This one I'm not sure how to do...

    Thankfully it should be easier being that it's first-order.

    http://img7.imageshack.us/img7/5259/93582031.jpg

    So the switch is initially open and no current is flowing
    After a long amount of time all of the current is going to flow through the inductor

    Final current through the indcutor 8V/4Ohm = 2 A?

    I don't know how to setup the differential equation either
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You are correct so far.

    At t=0 the (switch first closes) the inductor current will be zero. So all of the source current is flowing in the two resistors at this instant.

    You could draw the Thevenin equivalent at the inductor terminals at t=0.

    Looking back towards the source in this case, you have the source and the two resistors configured as a voltage divider, with the inductor sitting across the lower arm of the voltage divider.

    So you should be able to find Rth = 4Ω//4Ω and Vth=4V.

    You should then be able to derive the inductor current i(t) with the Thevenin equivalent sitting across the inductor.

    You should get something of the form i(t)=Imax*(1-exp(-at))
     
  3. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Yup.. that makes sense and sounds right tnk, thanks.

    Would anybody know how to setup the differential equation by any chance also?


    Lucky my exam got pushed back until Wednesday... but I don't think the differential equation would be required for this type of problem anyway
     
    Last edited: May 10, 2009
  4. t_n_k

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    See attached
     
  5. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Ah of course..

    Making the differential equation starting with KVL from the THEVENIN equivalent would make much more sense... don't know why I was thinking of doing it from the original circuit

    thanks
     
  6. steinar96

    Active Member

    Apr 18, 2009
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    An easy way to calculate simple first order circuits like this is to calculate the effective resistance the inductor sees in this circuit, and once u've got that you can calculate the time constant L/R.

    In this case the resistance is the 2 4 ohm resistors in parallel which gives 2. So the time constant is 4 / 2.

    You know the natural responce of the circuit which is of the form X(t) = Ae^(-t/TimeConst).

    Since there is a source in this example we have to solve for the particular integral. There is however an easy way to bypass that calculation by looking at the circuit and determining what the current trough the inductor will be when time aproaches infinity.
    Since the initial response will have faded at when time aproaches infinity the inductor is again a open circuit. Which means current will bypass the second resistor and flow entirely trough the wire the inductor is placed at. The particular integral is simply the current when the inductor's response to the changes is over. In this case it's just the source voltage devided by the first 4 ohm resistor (since the inductor is short circuit).

    That would give us I(t -> infinity) = 2A

    The initial condition is I(0-) = 0. So solving for A in the response would give us

    I(0+) = Ae^(-t/2) + 2 = I(0-) = 0. So A = -2.

    The total response would be I(0+) = -2^e(-t/2) + 2 for t > 0
     
    Last edited: May 10, 2009
  7. t_n_k

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    Steinar96 gives a good approach - especially when you are comfortable with the underlying concepts and can bypass a lot of the cumbersome maths.

    My post with the approach similar to your teacher's method was an attempt to align her method with the solution I gave. To avoid confusion.

    As always, I'm as prone to errors as others (possibly more so!) and my analysis has an error which carries through the solution but which remarkably ends up with the right answer - go figure!

    Anyway I've attached my update to the solution which hopefully is correct.
     
  8. steinar96

    Active Member

    Apr 18, 2009
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    Keep in mind that T_N_K's approach assumes u've found the thevenin equavelant circuit for the circuit around the inductor. The R in his solution is Rth and E = Vth, Vth is 4V, or 4*u(t) V if we are to be exact.

    u(t) is the unit step function which has value 0 for t<0, and 1 for t>0.

    If you would like to set up the differential equation without using thevenin you would use KCL. The initial equation would be for the voltage but by inserting proper voltage/current relations you would aquire a differential equation for i(t)

    This method however requires more calculation then by simply finding the thevenin circuit and solving directly for i. There are many ways to solve proplems regarding first order circuits and realising which one is the easiest can make your life alot easier and save you alot of calculations and paper. Writing KCL relations which produces an equation for V is propably the longest since you have to insert the proper relations to relate the voltage to the current function you are trying to solve for.
     
    Last edited: May 11, 2009
  9. t_n_k

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    Hi steinar86,

    Agreed with your useful comments.

    I think the Thevenin equivalent looking back from the inductor terminals would be ...

    Vth=4V and Rth=2Ω

    Where Vth is found as the open circuit voltage you would see across the 4Ω resistor in the lower half of the voltage divider arm with the switch closed and the inductor removed.

    Not sure if the OP has moved on from this question anyway .....
     
  10. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Thanks for your input steinar, I appreciate it.

    Tnk's approach is what I am more used to though... but thanks

    I have a question:

    What exactly is i(0+)? what purpose does the plus sign function as? As far as I know it means right when time starts to elapse, i.e. when a switch is first closed for example... is i(0) basically the same as i(0+)? i(0) being the current value right at zero and i(0+) being the current value right when time starts.. seems kind of redundant to me like they would be the same thing
     
    Last edited: May 12, 2009
  11. steinar96

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    Apr 18, 2009
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    It's a convention i was tought to keep track of which side of the switching time you are working with. In this case since we are dealing with an inductor i(0+) = i(0-) which just means the current after switching is initially the same as before switching.
    That does however not apply to all components and that is why some values are different before and after switching, and "noting" down where the value applies can prevent mistakes.

    If we look at the same example, the voltage around the inductor just before switching (as we aproach zero from the left) is 0. But after switching the voltage has risen to some value which is the derivative of the current change.

    that is
    V(0-) = 0
    V(0+) = L * di/dt (i(0+))

    So as you see they arent always the same thing, depending on which side of zero you are at :)
     
  12. ihaveaquestion

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    May 1, 2009
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    so if I had a fully charged capacitor at 20 volts in a circuit and closed the switch to complete the circuit, the voltage across the capacitor is just 20 volts at time (0+)
     
  13. t_n_k

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    Mar 6, 2009
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    This probably relates to the way in which transient conditions are typically defined around time zero - i.e. the time at which some disturbance occurs.

    If you consider the Heavyside or step function, it has the definition

    uo(t) = 0 for t<0 and uo(t) =1 for t>0

    So uo(t) is undefined at t=0 [:confused:], but at some infinitesimal value greater than time zero or t= (0+), uo(t) has a known value of 1. This is more of a mathematical construct rather than a physical reality - like a lot of the theoretical problems we deal with! Nobody actually believes in the Heavyside function as a physically realisable signal.
     
  14. steinar96

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    Apr 18, 2009
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    Yes, since for a capacitor V(0-) = V(0+), similar as the relations i(0-) = i(0+) apply for the inductor. The capacitor does not allow a sudden change in voltage, the inductor does not allow a sudden change in current.

    yeah that stems t_n_k. u(t) is undefined for t = 0 if i remember correctly. The only purpose of functions such as the unit step function or it's derivative the impulse function is to "model" behaviour or signals.

    The - and + refers to left and right side limits as the voltage/current function aproaches zero. Because the response of inductors and capacitors is either a continous or a discontinous function moving from 0- to 0+ depending on whether you are looking at the voltage or current at a location.
    It is also nice to use the +- signs to remind you of where inital values are valid.

    You are propably familiar with the power calculations for capacitors and inductors. The formula for the inductor is the

    integral of w = 1/2 * L * I^2 from minus infinity to t. (minus infinity is replaced with the time when the capacitor starts charging)

    You might not be familiar with this but if current trough a inductor is discontinous it would mean that the function describing the power in the inductor over time would take a huge leap in zero time. Which happens to require infinite amount of energy which is impossible. This means that the current trough the inductor MUST be a continous function and thus i(0+) = i(0-) for the inductor.

    Similar for the capacitor which has the equation
    integral of w = 1/2 * C * V^2 from minus infinity to t

    A discontinous leap in voltage over time would mean a leap in the power stored in zero time which requires infinite energy which as mentioned before is impossible. So the voltage at the terminals of the capacitor MUST be a continous function and thus V(0+) = V(0-) for the capacitor.

    Using 0+ and 0- makes it easier to keep track of initial values before and after switching as long as you remember the relations. And they are also an important attribute describing the behaviour of capacitors and inductors which must apply at all times.
     
    Last edited: May 12, 2009
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