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Identical bulbs, order them brightest to dimmest

  1. Apr 1, 2016 #1
    1. The problem statement, all variables and given/known data
    There are 5 identical bulbs. All are glowing. Rank from brightest to dimmest. Some may be equal.

    2. Relevant equations
    I= (ΔV)/R
    Req = [(1/R1)+ (1/R2) +...]^-1 (Parallel resistors; same delta V)
    Req = R1 + R2 + .... ( Series resistors; same current I)


    3. The attempt at a solution

    Let all the bulbs have resistance R.

    A. Combine bulbs R and S ( parallel, same ΔV)
    Req = [(1/R)+(1/R)]^-1 = R/2

    B. Combine R&S with P and T ( series, same I )
    Req = (R/2) + R + R = 2.5 R

    C. Combine (R&S, P, T) with Q ( Parallel, same ΔV)
    Req = [(1/R)+(1/2.5R)]^-1 = 0.71R
    ΔV = ε- - > I = ε/Req = 1.4ε/R

    Now decompose to find individual currents:

    A. R(Q) = R; ΔV = ε --> I(Q) = ε/R

    R(R&S, P, T) = 2.5R --> I (R&S, P, T) = 0.4ε/R
    I (P) = I (T) = 0.4ε/R

    R(R&S)= R/2
    ΔV (R&S) =( 0.4ε/R)*(R/2)= 0.2 ε
    ΔV (R) = ΔV (S) = 0.2ε
    R (S) = R (R) = R
    I (S) = I ( R) = 0.2ε/R

    ⇒Q >P=T >S =R

    This is incorrect t and I'd like to know what I did wrong. thanks in advance for any help. tmp_4735-20160401_072217-639066610.jpg
     
  2. jcsd
  3. Apr 1, 2016 #2

    cnh1995

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    They are not in series. What about Q?
     
  4. Apr 1, 2016 #3

    cnh1995

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    You can see that the power consumed is proportional to the square of current through the bulbs, since all have the same resistance. From this logic, the brightest one will have maximum current. Which one do you think will have maximum current, by observation? Which bulbs can you say will glow with equal brightness, just by observation? It is not necessary to find the equivalent resistance of the entire circuit. All you need to do is apply current division principle properly in your mind. For the dimmest bulb(s), you need a little bit simplified circuit.
     
    Last edited: Apr 1, 2016
  5. Apr 1, 2016 #4

    Just by observation, I'd say P will be the brightest and R and S will have the same brightness.
     
  6. Apr 1, 2016 #5

    cnh1995

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    Correct!
     
  7. Apr 1, 2016 #6
    Q seems to be in series with (RS)....Right? (Connected end to end).
     
  8. Apr 1, 2016 #7
    uploadfromtaptalk1459540213304.jpg
     
  9. Apr 1, 2016 #8

    haruspex

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    No, it's not as simple as that. Some current can bypass Q yet flow through R.
    Your diagram with the collapse of R and S into one resistor is wrong. There should not be a direct link from T to P.
     
  10. Apr 1, 2016 #9
    uploadfromtaptalk1459542981023.jpg
    Is this diagram correct? Then P would be in series with Q and (RS) would be on series with T.....?
     
  11. Apr 1, 2016 #10

    haruspex

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    Take it one step at a time. At each step, you can combine resistors:
    - as being in series if connected end to end and all the current that flows through one must flow through the other
    - as being parallel if connected at each end
    RS is certainly in series with T, so combine those. What next?
     
  12. Apr 1, 2016 #11
    Then I combine (RST) with Q (parallel).
    Then I combine (RSTQ) with P ( series).....Correct? uploadfromtaptalk1459547218067.jpg
     
  13. Apr 1, 2016 #12

    haruspex

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    Yes.
     
  14. Apr 1, 2016 #13
    Thank you so much!
     
  15. Apr 1, 2016 #14

    cnh1995

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    < Mentor Note -- @cnh1995 is posting a clarifying post here to be sure that the OP understands what a clean solution is, after the OP has said they have solved the problem. >

    Screenshot_2016-04-01-18-29-22.png
    Here's how it will look like...
    Asuming the resistance of each bulb is r,
    P is the brightest. R and S have same brightness. Current through T is sum of currents through R and S, hence, T will be brighter than R and S. R||S in series with T give a resistance of 3r/2. This 3r/2 is in parallel with Q( resistance r). Since Q<3r/2, current through Q is more than that through 3R/2 i.e current through T.
    Hence,
    P>Q>T>R=S.
     
    Last edited by a moderator: Apr 1, 2016
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