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Relative brightness

  1. Apr 14, 2004 #1
    Assuming there are five identical light bulbs. Light bulb C and D are in a series connection. C and D are in a parallel connection with E. Light Bulb A and B are in a parallel connection. The two parallel connections are wired in a series connection. What would be their relative brightness? I'm thinking that A=B=E=C+D

    ********************
    |----power---------------|
    |*******************|
    |*******************|
    |**|--A--|***|-C-D-|**|
    |---|****|----|****|---|
    ***|--B--|***|--E--|***
    ********************


    in a parallel connection, It=I1+I2, therefore, A=B. In a seriese connection, The total current equals the individual currents, so C=D. Back to the parallel connection, C+D=E. In a series, they are all equal, so (A+B)=(E+C+D). Therefore...A=B=E=C+D. My friend said that it might be: C^D<A^B<E

    Any suggestions?
     
  2. jcsd
  3. Apr 14, 2004 #2
    EDIT: Ignore this post! I misread your diagram; Chi Meson has it right in the next post.

    Empire,

    Think of it more as an electrical problem first, and then worry about brightness.

    Suppose your power source is a voltage of 100V.

    What's the voltage across the B-E branch?

    So what's the voltage across B? Across E?

    What's the voltage across the A-CD branch?

    So What's the voltage across A (be careful!)? Across CD ? Across C? Across D?
     
    Last edited: Apr 16, 2004
  4. Apr 16, 2004 #3

    Chi Meson

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    Another way to think of it: The brightest bulb will have the most current going through it.

    Obviously,the total current through the entire A/B unit will equal the total current through the C/D/E unit. And the current through A and B is equal, and each gets exactly half the total current.

    So ask yourself, where will more current go, through the C-D branch, or the E branch?
     
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