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Current and resistance in series and parallel circuits

  1. Sep 1, 2015 #1
    1. The problem statement, all variables and given/known data
    List the order of brightness of each bulb (while the switch is open) assuming the bulbs are all of the same resistance.
    image00654b9b5a1.jpg
    2. Relevant equations
    Ohm's Law: Change in voltage = current divided by resistance

    3. The attempt at a solution
    I know that bulbs A,C,D,F are all equal in brightness because they are both simple series circuits (while the switch is open). I know that B and E are also equal in brightness as they are a series circuit.

    My difficulty begins in that the B and E bulb are in a series and also a parallel circuit. Does this make these two bulbs brighter than bulbs A,C,D,F or dimmer?

    Any help with some sort of reference would be very helpful.

    (this is for my Physics Lab homework)
     
  2. jcsd
  3. Sep 1, 2015 #2

    berkeman

    User Avatar

    Staff: Mentor

    There are three strings of two series bulbs connected in parallel across the battery, so you are correct (until the switch is closed). Don't let the drawing fool you with a little extra wire length leading to the B-E bulbs. Each of the three strings of two series bulbs is connected the same way across the battery. :smile:
     
  4. Sep 1, 2015 #3
    Redraw the circuit with all bulbs on the same side of the battery and all series pairs in vertical legs.
     
  5. Sep 2, 2015 #4
    Ah! Thank you for your help, your answer was correct (I double checked using a program at our physics lab). :smile:
     
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