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Homework Help: Simple DC Circuit

  1. Feb 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Rank the brightness of five bulbs (3 different circuits) considering all the batteries and light bulbs are identical.

    1st circuit - 1 battery hooked to bulb A.
    2nd circuit - 1 battery with positive end hooks to bulb B then bulb B connects to bulb C then comes back to negative side of battery. (series)
    3rd circuit - 2 batteries (series) with positive end of first battery hooks to bulb D then bulb D connects to bulb E then comes back to negative side of other battery.

    2. Relevant equations

    V=IR

    3. The attempt at a solution

    I think it's this...A=D>E>B>C...

    I think I am missing something really important...question seems to be so...simple!
     
    Last edited: Feb 26, 2007
  2. jcsd
  3. Feb 26, 2007 #2
    Have you just solved it intuitively or have you also set up some equations?
    And why is A=D=E > B=C?
     
  4. Feb 26, 2007 #3
    Update, D>A=E>B>C.

    I can't relate this to anything. I am clueless of this kind of stuff.
    I simply reordered based on what I experimented previously.
    :(
     
  5. Feb 26, 2007 #4
    Your problem asks for the brightness.
    On what does the brightness depend?
    After you answered that question, can you relate it to your relevant equation?
     
  6. Feb 26, 2007 #5
    Brightness depends on current, I, right?
    so D should be the brightest due to number of batteries then A=E>B>C...(?)
     
  7. Feb 26, 2007 #6

    ranger

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    The brightness of each bulb depends on the power across the bulb. Do you know how to find power?
     
  8. Feb 26, 2007 #7
    Hm...is this necessary steps?
    I have not learned it...
    It seems like it can be solved really easily but...:confused: :frown:

    By the way, this isn't suppose to be workout problems...
    It is prelab that I am suppose to do...
     
  9. Feb 26, 2007 #8

    ranger

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    Well, yea I guess you can solve it by making educated guesses. But that is not the way the how these things should be done. Like I said before, brightness is related to power. The general equation for power is:
    P = IV

    Since all of your circuits are series circuits and each bulb has equal resistance. What do you know about voltage and current in a series circuit?
     
  10. Feb 26, 2007 #9
    indirectly proportional(?)
    But that means currents gets smaller as voltage increase...
     
  11. Feb 26, 2007 #10

    ranger

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    Voltage and current are directly proportional (V=IR). What I was asking you is how is the voltage across a bulb related to the current through it? You just need to remember the simple rules of a series circuit.

    The question is not that hard. Its just relating several concepts that may cause confusion - power, ohms law, and series circuits.
     
  12. Feb 26, 2007 #11
    ok so...back to the top...

    assuming that each battery is 9 volt and since all resistor inside of bulbs is same...
    current must be same as long as there are same amount of batteries, which tells me that A>B>C.

    Now the circuit 3...since there are two batteries, do i simply add 9+9=18 so that D>E?

    I have no idea what's keeping me from understanding this simple concept.
     
  13. Feb 26, 2007 #12
    D=E > A > B=C...
    My final answer :)
     
  14. Feb 26, 2007 #13
    I am leaning towards on that one now...
     
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