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Power absorbed by multiple elements

  1. Aug 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the power absorbed at t=0,t=0+,and t=200 mS by each of the elements in the circuit of fig 11.25 if us is equal to (a) -10u(-t) V (b) 20 + 5u(-t) V


    2. Relevant equations



    3. The attempt at a solution
    http://www.chegg.com/homework-help/...pter-11-problem-3e-solution-9780073529578-exc

    The link posted above is the solution I'm trying to understand it I can also upload my work but it is more or less a copy of the solution so I saw no point. What I don't understand starting on the first part is how Vs(t)=-10u(-t) V=-10 at negative infinity to 0? First off what is u(-t) is this reactance or impedance? Furthermore if I took the limit if this as t approaches negative infinity would the function go to infinity not -10? So obviously it's not infinity even though it seems that it would, so I would have to assume u(-t)=1 for it to work. Why one because I am assuming that it has to do with current not flowing before time t=0? This also then begs another question as to why from 0 to infinity it is 0 implying that the voltage stops after time t=0 because u(-t)=0.Well anyways this is just two question of many that I have, so I am gonna ask this question first and slowly propagate through my errors in this problem?
     
  2. jcsd
  3. Aug 27, 2013 #2

    rude man

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    First off, I call the resistor R and the inductor L.

    vs(t) is voltage.

    vs = -10u(-t) means vs was -10 Volts from time = -∞ to t = 0, then vs = 0V. So at t=0 what is the current thru R? And 200 ms later what is it?

    PS - L does not 'absorb' power, it stores it and discharges it. So the only element you have to worry about is R.
     
  4. Aug 27, 2013 #3
    Hmm could it be i(t)=Vs/R(1-e^(-Rt/L))u(-t) which gives me -10 V at t=0 but I fear that this isn't even close?
     
  5. Aug 27, 2013 #4

    rude man

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    It's not only close, it's exactly right.

    But you did not need to solve for V(t). Any function of time, like a voltage V(t), when multiplied by U(-t) means the function is V(t) until t = 0, then it's zero. It's the opposite of V(t)*U(t) which is zero until t = 0, then = 1.

    Your function is V(t)*U(-t).

    U(t) = 0 for t < 0, then = 1
    U(-t) = 1 for t < 0, then = 0.
    More generally, U(t-τ) = 0 for t < τ, then = 1
    U[-(t-τ)] = 1 for t < -τ, then = 0
     
  6. Aug 27, 2013 #5
    Ok so thanks for the help on the first part. Let me ask you another question is the solution posted correct? Also at t=0- the voltage across the inductor voltage is Vl(0-)=L(di(0-)/dt) then L(d(-10)/dt)=0 why is this is this just how an inductor works?
     
  7. Aug 27, 2013 #6
    Oh wait the solution can't be correct because no power is absorbed by an inductor
     
  8. Aug 27, 2013 #7

    rude man

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    That's right.

    With V(t) = 20 + 5U(-t) that means the input votage is 25V until t = 0-, the it's 20V for t >= 0+.

    So you need to establish the initial current i(t) at t = 0 and then solve your equation for the current after that. Same for part (a). Then power dissipated in R = [i(t)^2]*R.

    To sum up, (1) find your initial current based on the input voltage at t = 0-
    (2) calculate the total current at t = 0 and t = 200 ms so you need i(t), t >= 0+
    (3) compute (i^2)*R in all four cases.

    When calculating the total current, remember that it consists of two parts:
    1. the time-varying current due to the initial current i(0), which is the current with zero input voltage, and
    2. the time-varying current due to the input voltage as though the initial current were zero.

    Using the principle of superposition you add both parts to determine the current as a function of time t > 0. The two currents have different equations BTW.
     
  9. Aug 27, 2013 #8
    So initial current at t=0- the current is -10 A. Which if I plug into the power equation i^2R I get 100W at t=0-.
     
  10. Aug 27, 2013 #9
    Which is for R
     
  11. Aug 27, 2013 #10

    rude man

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    That's correct. And BTW the current is the same at t = 0+ as at t = 0-. That's because current thru an inductor can't change instantaneously.

    Now the harder part: how do you compute the power at t = 200 ms?
     
  12. Aug 27, 2013 #11
    Ok so then the power at P(0+)=100 W and so is P(0). So I found an equation from the solution that I think might work so P(200 mS)=100e^(-200/250) W = 44.93 W I am goons keep looking for equations because I feel this is incorrect
     
  13. Aug 27, 2013 #12

    rude man

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    'Fraid so. Remember I said you need to consider two separate currents: assuming zero volts input for all t > 0, the current at t=0 and how it behaves for t>0; and the current due to the finite input voltage at t = 0+ assuming no initial current. Then you add the two currents.

    Start with figuring the current for t>0 as if Vin = 0 for all t >= 0+.
     
  14. Aug 27, 2013 #13
    Wait if the input voltage is zero wouldn't that make the current at those values assuming no voltage 0 too?
     
  15. Aug 27, 2013 #14

    rude man

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    The current was not zero at t=0 even if Vin = 0 for t =>0. There is current in the inductor even if you short the input at t=0.
     
  16. Aug 27, 2013 #15
    Well the voltage across the inductor would be -10e^(0)=-10 V so Il=-10 A at t=0
     
  17. Aug 27, 2013 #16

    rude man

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    No, the voltage across the inductor at t = 0- is zero. For an inductor, V = L di/dt and di/dt = 0 at t = 0- since by then the current is a nice constant -10A. At t = 0- the entire input voltage is across the resistor.

    The current has had infinite time (from -∞ to 0-) to settle out.

    You need to solve for the current where R is in series with L, the combo is shorted out, but an initial current runs thru R and L. What is the time response of that current?
     
  18. Aug 27, 2013 #17
    L/R so 250e-3/1= 250e-3 sec
     
  19. Aug 27, 2013 #18
    I looked up an equation for series RL circuits don't know if its helpful but I(t)=I(1-e^(-t/Tao)) which when I solve
    -10(1-e^(-200/250)) which gives -5.506
     
  20. Aug 27, 2013 #19
    Dang I guess you won't be getting back with me tonight oh well I have to turn it in at 9 tomorrow. Still I want to know about it
     
  21. Aug 27, 2013 #20

    rude man

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    ???

    I mean, what is the expression for i(t)?
     
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