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Help with circuits (Current, Voltage Drop, and Power Dissipation)

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Basically, I need to know the power dissipation (W) for three resistors using aw 12V battery: 120, 80, and 50 ohms.

    https://webwork2.uncc.edu/webwork2_files/tmp/Fall2014-Engr1201-Common/img/2ff2f0ce-cb10-3786-8109-c42346130316___1a2e5eeb-8657-3983-941f-b7d8a9123afa.gif [Broken]

    2. Relevant equations
    I think the formula for voltage drop is Voltage Drop = Resistance * Current, and I thought the formula for power dissipation was V^2/R.


    3. The attempt at a solution

    It also asks for the current, which I think is 0.05A and voltage drop across all three resistors.

    I think the answers for voltage drop are 6, 4, and 2.5 V, and the current was 0.05A.

    I thought the power dissipation for the three resistors were 1.2, 1.8, and 2.9 W.

    The website says I got "at least one wrong." Thanks, WeBWork.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 2, 2014 #2
    I just got the power dissipation problems. All I need is the voltage drop. Apparently, V=IR doesn't apply here?
     
  4. Sep 2, 2014 #3

    berkeman

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    Staff: Mentor

    Adding your voltage drops gives 12.5V, which is more than your voltage source.

    I also got a different current when I divided 12V/Rtotal. Can you show your work in those two steps?
     
    Last edited by a moderator: May 6, 2017
  5. Sep 2, 2014 #4
    12V/250(Rtotal) = 0.048A which I rounded to 0.05A.

    Vdrop1 = 120 * .05 = 6
    Vdrop2 = 80 * .05 = 4
    Vdrop3 = 50* .05 = 2.5

    Do you think I just rounded too early?
     
  6. Sep 2, 2014 #5
    Yea, I reentered them without rounding, and they were correct. Thank you, berkeman.
     
  7. Sep 2, 2014 #6

    berkeman

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    Staff: Mentor

    Glad to help. Yeah, definitely don't round anything until the end. :smile:
     
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