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Homework Help: Two quick questions - circuits

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data

    #1 A power source produces a "sawtooth" wave of current, reversing direction at 1s. After 2s (The current starts at 0, reaches 1 amp at 1 second, reverses direction to -1 amp, and then reaches 0 again at 2 seconds), calculate the total energy supplied to a 5000 ohm resistor.

    #2 Different measurements are taken by attaching various resistors R across a cell terminal and recording the voltages and corresponding currents I. From the graph, find

    (a) the open circuit voltage of the cell.
    (b) the short-circuit current
    (c) the maximum power that the cell can deliver to an external resistance.

    The graph is a negative linear line. V on the x-axis, I on the y-axis. The starting point is at () V, 3 amps) and the end point is at (1 V, 0 amps).

    2. The attempt at a solution

    #1 - Energy = C (charge) x V, so I'm tempted to say that energy is zero because the current is positive for the first second and negative for the second, thus they should cancel.

    #2 - I'm not sure what the open/short circuits are referring to... could anyone clarify? As for c, power P = V*I, but at what point is this maximum?
    Last edited: Jan 6, 2010
  2. jcsd
  3. Jan 6, 2010 #2


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    For the first one, use E=I2Rt to find the energy for the two directions.
  4. Jan 6, 2010 #3
    The current is not constant, so how could I use that equation? Here are the graphs I described:

  5. Jan 7, 2010 #4
  6. Jan 7, 2010 #5


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    Integrate: energy = ∫ power dt = ∫ I2R dt.
  7. Jan 7, 2010 #6
    You know [tex]I(t)[/tex] over the relevant intervals. You said it yourself, the current is a function of time. You can't just say that [tex]\int I^2 dt = I^2 t[/tex]
    That's not true when [tex]I[/tex] is a function of time!

    Use the differential form:
    [tex]P=\frac{dE}{dt}[/tex] to rewrite: [tex]dE=I^2 R dt[/tex] and then integrate over the relevant intervals.
    Last edited: Jan 8, 2010
  8. Jan 7, 2010 #7
    So integrating both sides yields:

    energy = ∫ I2R dt

    What function am I integrating though? All I have is this graph :(...something to do with a Sin wave perhaps?
    Last edited: Jan 7, 2010
  9. Jan 7, 2010 #8
    Find the current as a function of time over the two relevant intervals. You drew a perfect diagram of a linearly rising current, now just derive its equation!
  10. Jan 7, 2010 #9
    Can I approximate the sawtooth as two integrals,
    I(t) = t for t=0 to t=1
    I(t) = t - 2 for t=1 to t=2

    Doing this and integrating, I get 0 for energy.
    Last edited: Jan 7, 2010
  11. Jan 7, 2010 #10
    Square of the current!
  12. Jan 7, 2010 #11
    Oops. After squaring I(t) for both integrals, I'm getting 10000/3 for E. Does this make sense? I was expecting zero.
  13. Jan 7, 2010 #12
    It's correct. The power supplied to a resistor is always positive if there is a nonzero current. if the power supplied could become negative, then the resistor would be supplying power to the battery. This power is equal to [itex] I^2R [/itex] wich is always positive as well.
  14. Jan 7, 2010 #13
    Thanks, that makes sense.

    Could someone confirm these answers for the second question?
    a) 1 volt
    b) 3 amp

    for c)... The equation for I=3-3V so the power would be P=IV=3V-3V^2, right?

    This has a maximum when dP/dV =0 = 3-6V or V=1/2.

    so Pmax =(3-3(1/2))*(1/2)= 3/4?
  15. Jan 8, 2010 #14
    Haha I'm in your class! but yes I got the same answers as you did for that problem.
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