1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simple circuit

  1. Jun 3, 2010 #1
    Quick question, if i had a circuit of 5 light bulbs whereby 4 were running parellel in a sequence and one was running seperatly, would one be brighter than the others?
    sorry if it is hard to visulise :)
     
  2. jcsd
  3. Jun 3, 2010 #2
    I assume from your question that 4 bulb connected in parallel is in series with 5th bulb to a voltage source.
    Now assuming all bulbs are identical, i.e., they have same resistances (say x) you can find the total resistance. And if you find voltage drop across these bulbs, you can find which bulb will be brighter.
     
  4. Jun 3, 2010 #3
    so shall i just substitute x into where the values of resistance would be. Im not given any values for anything
     
  5. Jun 3, 2010 #4
    Assume each bulb has resistance R and the whole circuit is driven by a source that produces V volts between its two terminals. Remember that the brightness is directly proportional to the power delivered to the bulb (assuming no or identical power loss).

    Using these, you should be able to compare their brightnesses.
     
  6. Jun 3, 2010 #5
    No need to know the value of resistance, because you have to find which bulb is brighter and not how much right?
    By simple common sense we know that the bulb with more voltage across it glows brighter. So you may check across which bulb more voltage drop will occur.
     
  7. Jun 4, 2010 #6
    Even if you knew the value of the resistance, you will need to know much more to actually calculate a value for luminous flux or intensity. Comparison is the only sensible thing that I can see being done given only these pieces of information.

    That is only true if the bulbs are identical. As I said the brightness is directly proportional to the power dissipated by the resistor inside the bulb. That's not a function of only the voltage across it. If you aren't convinced, you can easily try this by using two different light bulbs and a 9V battery.
     
  8. Jun 4, 2010 #7
    so am i right in believing it to be bulb p? the one on its own but in parallel with the others
     
  9. Jun 7, 2010 #8
    Which one are you referring to as bulb p?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook