Power dissipated in RC & RL circuits

Discussion in 'Homework Help' started by The knight, Jan 12, 2011.

  1. The knight

    Thread Starter New Member

    Jan 12, 2011
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    Greetings,

    I hope you are well, and I'd like to ask you about the power dissipated due to capacitor and inductor,

    For example, a capacitor, say 2μF connected to the independent votage source 20V, the the source is removed and replaced by a resistor in such a way that the voltage does not change instantenously, the resistor has a resistance of 5Ω. and the qusetion is find the time when the power dissiapted is 75% of the initial power?

    And what will be the time if we use inductor 2mH and do the process?
     
  2. Georacer

    Moderator

    Nov 25, 2009
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    On a resistor, the power dissipaded is \frac{V^2}{R}. Also, the voltage on the resistor will be described by the equation V_0 e^{-\frac{t}{RC}}.
    Therefore, the total power dissipated will be \frac{V_0 \cdot RC}{R} (someone correct me if I 'm wrong).
    Then you need to solve the integral from 0 to t and equate that to 3/4 of \frac{V_0 \cdot RC}{R}. Solving for t will give the answer.
     
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  3. The knight

    Thread Starter New Member

    Jan 12, 2011
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    0
    ^^

    Thank you,

    but when I saw solution in the the book, the author uses 1/3 rather than what you used. I don't why he solves like.

    My question know, why he uses the remaining presntage of energy to calculate the time needed to dissiapate 75% ?

    and In case of inductor, he uses the presnage itself not the remaining
     
  4. Georacer

    Moderator

    Nov 25, 2009
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    As I posted, it is possible that I am wrong. I have never seen an exercise like yours so far. I put my logic down and solved it. Maybe if you write down what your book asks and suggests we will sort it out.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Perhaps the question is really asking at what time the stored energy in the capacitor has reduced to 75% of its original value - rather than when the power has fallen to 75%.

    The stored energy at time t will be given by

    W=\frac{1}{2}CV_c^2(t)

    For the energy to have fallen by 75% would require the voltage to have fallen to √0.75 of the original value or 86.6% at time t.
     
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