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Ohm's Law, Resistance, and Power

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Please look at the attachment.


    2. Relevant equations

    ΔV=I*R
    R=ΔV/I
    Power=I*ΔV


    3. The attempt at a solution

    I said that ΔV of A and ΔV of B are the same (each 3/2 V). Using this, I got that the current at A = 3/20 Amps. However, this means that the current at B and C is each 3/40 Amps. However, then the 1.5 V * 3/40 does not equal 10 ohms, which is given. How am I supposed to go about this problem to find the power dissipated in each bulb and find what changes in terms of bulb brightnesses in A and B if bulb C is unscrewed?
     

    Attached Files:

  2. jcsd
  3. Feb 28, 2013 #2

    SteamKing

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    You have one resistor in series with two others which are in parallel. You have not found the correct current in the circuit.
     
  4. Feb 28, 2013 #3

    Simon Bridge

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    The resistance of B and C in parallel is less than the resistance of A, so the volt-drop will be less.
     
  5. Mar 1, 2013 #4
    How so? They're each 10 ohms.
     
  6. Mar 1, 2013 #5
    J, they are in parallel. There are 2 possible paths for current so there is less total resistance for electrons being pushed through this path. You can think of this sort of like having 2 checkout lanes open at a grocery store. More shoppers can move through faster than if there were one lane open.

    Resistors in series (in a row) are added as R1+R2. Resistors in parallel (multiple paths) are not added this way. Hint: If there are two parallel paths, both must have the same voltage across them.
     
  7. Mar 1, 2013 #6

    CWatters

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    Apply KCL to the point where A is connected to B and C. You will find the current through B (or C) will be half the current through A. So the voltage drop oved B will be half that of A.
     
  8. Mar 1, 2013 #7

    Simon Bridge

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    Thanks folks - jperentosh: yes they are correct. By now you should have seen the equations for resistors in series and parallel. You can always check with an ohmmeter if you don't believe it.
     
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