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Simple Circuits for Stupid People

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the voltage V and the power absorbed by the 12-ohm resistor.
    http://i.imgur.com/xCTEI.png

    2. Relevant equations

    p = vi
    v = ir
    KVL
    KCL

    3. The attempt at a solution

    Tried using KVL and KCL in various ways. This book gives horrid examples.
     
  2. jcsd
  3. Mar 22, 2010 #2
    You can solve using mesh analysis if you wanted, but I'd condense that middle branch into an equivalent resistance first.

    4 + (1/3 + 1/6)^-1 = 6. Next, write the mesh loops, but leave out that first loop with the current source (but include its effects in that the current going through the 2 ohm resistor is (I_1 - 6))
     
    Last edited: Mar 23, 2010
  4. Mar 23, 2010 #3
    I found the current in the last loop to be .4 A. Using V = IR to solve to get the voltage, I get 4.8 V at the 12-ohm resistor. I then used P = VI to get a power of 9.6 W, but the book has 1.92 W. Did I do anything wrong or is the book wrong again?
     
  5. Mar 23, 2010 #4
    The current is [tex].4[/tex] and the voltage is [tex]4.8[/tex]. However, [tex].4 \times 4.8[/tex] is not [tex]9.6[/tex]. It's [tex]1.92[/tex](as the book says). Also, [tex]I^2 r = (.4)^2 12 = 1.92[/tex] as well if you wanted to avoid that extra calculation of voltage.

    To see where the current-dependent equation comes from notice:
    [tex]P=IV[/tex]
    and
    [tex]V=IR[/tex]
    substituting [tex]IR[/tex] for voltage into the power equation brings us to:
    [tex] P = I (IR) = I^2R[/tex]
     
    Last edited: Mar 23, 2010
  6. Mar 24, 2010 #5
    Yes, like xcvxcvvc said: P is equal to I2*R as well as V2/R. These only work when you're finding the power absorbed by the resistor!! Also, I think the easiest way to solve this (because mesh equations can get ridiculous and tedious after a while --> source transformation, if you have learned it!
     
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