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Homework Help: Physics 117 problem about circuits and bulbs

  1. Jul 5, 2012 #1
    1. In the circuit shown below, describe what happens to the brightness of bulbs A and B when the switch is closed. Explain your reasoning. (no other information is given, just asking about the relative brightness of each bulb)

    In the circuit, with the switch open, you have a loop with a battery and bulbs A and B in series. When the switch closes, B becomes parallel with the electrical line that holds the switch. In this case, the electrical line that B is on is inside of the larger loop of the battery, A, and the switch.

    2. Relevant equations V = IR

    3. The attempt at a solution

    I thought at first that the bulbs would both slightly dim because the added switch and line would add a small amount of resistance to the line, but I see that A gets brighter and B goes out. Is this because the current takes the path of least resistance and skips over the B bulb?
  2. jcsd
  3. Jul 5, 2012 #2


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    Staff: Mentor

    Welcome to the PF.

    I don't see your attachment...
  4. Jul 5, 2012 #3
    I do not have a attachment for it, but i think i described it well enough.

    Maybe this will help, tried to type it it best as possible. Ignore the asterisks, I just had to put them in there for spacing. The other lines are the wires.

    | ******* | ********|
    emf **** B ***** switch
    | ******* | ******* |
  5. Jul 5, 2012 #4
    Does the diagram look like the image in the link?
    http://img684.imageshack.us/img684/404/37304acd6959474e821ad8c.png [Broken]
    Last edited by a moderator: May 6, 2017
  6. Jul 6, 2012 #5
    yes, it does.
  7. Jul 6, 2012 #6


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    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, this is the reason why B goes out. The wire that has the switch in line with it acts as a "short circuit." It's called that because it gives the current a shorter path back to ground. Instead of having to go through both A and B, the current can now just pass through A and then directly to ground. So B has no current flowing through it: it has been "shorted out" of the circuit (i.e. bypassed).

    Now all that is left for you to explain is why A gets brighter. Hint: what happens to the voltage across A when you close the switch?
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