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Power Being Absorbed

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the power being absorbed by element X in Figure if it is a (a) 100 Ω resistor; (b) 40 V independent voltage source, + reference on top; (c) dependent voltage source labeled 25ix ,+ reference on top; (d) dependent voltage source labeled 0.8v1,+ reference on top; (e) 2 A independent current source, arrow directed upward.

    14tably.png

    2. Relevant equations

    [tex]P = vi[/tex]

    3. The attempt at a solution

    a)

    KVL:

    [tex]20v + R_{1}i - 120v + R_{2}i + R_{3}i + R_{4}i + R_{5}i[/tex]

    [tex]i = \frac{20v - 120v}{R_{1} + R_{2} + R_{3} + R_{4} + R_{5}}[/tex]

    [tex]i = \frac{-100v}{200\Omega} = -0.5A[/tex]

    [tex]v = Ri = (100\Omega)(-0.5A) = -50v[/tex]

    [tex]P = (-50v)(-0.5A) = 25w[/tex]

    b)

    KVL:

    [tex]20v + R_{1}i - 120v + R_{2}i + R_{3}i + R_{4}i + 40v[/tex]

    [tex]i = \frac{20v + 40v - 120v}{R_{1} + R_{2} + R_{3} + R_{4}}[/tex]

    [tex]i = \frac{-60v}{100\Omega} = -0.6A[/tex]

    [tex]P = (40v)(-0.6A) = -24w[/tex]

    c)

    KVL:

    [tex]20v + R_{1}i - 120v + R_{2}i + R_{3}i + R_{4}i[/tex]

    [tex]i = \frac{20v - 120v}{R_{1} + R_{2} + R_{3} + R_{4}}[/tex]

    [tex]i = \frac{-100v}{100\Omega} = -1A[/tex]

    [tex]P = ((25)(-1A))(-1A) = 25w[/tex]

    d)

    current is same as c)

    [tex]i = -1A[/tex]

    [tex]v_{1} = (40\Omega)(-1A) = -40v[/tex]

    [tex]P = ((0.8)(-40v))(-1A) = 32w[/tex]

    e)

    I am not sure how to do e) because I don't have a voltage and the current is 2A.
     
  2. jcsd
  3. Oct 7, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I did (a) and agree.
    My take on (e) is that it is a current regulated supply that puts out whatever voltage it needs to push 2 A into the circuit. Assuming that the 120 V source has no internal resistance to having 2 A pushed backwards into it, you could calculate the 2 A source's potential V:
    V = IR
    20 + V - 120 = 2*100
     
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