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Homework Help: What happens to a lamp in series if lamp in parallel is broken?

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm looking through some old IB exams, and I came across this (supposedly fairly easy)question. I'm having some problems, and I feel there's something really basic I'm missing.

    http://img52.imageshack.us/img52/5139/physicsproblem.jpg [Broken] http://g.imageshack.us/img52/physicsproblem.jpg/1/ [Broken]


    2. Relevant equations
    V = IR
    P = VI
    R of parallel = (1/R1) + (1/R2)


    3. The attempt at a solution
    My first instinct was that the correct answer is B is the correct answer; the brightness of Lamp L increases, the brightness of lamp M stays the same. My reasoning there was that since there is no longer a parallel, all the current will be forced to go through lamp L. Since the lamps have equal resistance, the current going through lamp L should double.

    Initially, I didn't think M would change. But since the resistance of what used to be the parallel (lamp L) now is R rather than (2/R), the total resistance of the circuit has increased, and thus the current has decreased. Which would make lamp M less bright.

    I actually thought of the last part while writing this post, so now I'm wondering if there's more to it than that, and if I'm correct. Thanks a lot.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 10, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Don't try to solve it 'in your head'--do it on paper. Compare the current through each lamp before and after lamp N breaks. Hint: Call the voltage V and the resistance of each lamp R. Then actually calculate the currents so you can compare them.
     
  4. Jan 10, 2010 #3
    I seem to be horribly lost in these calculations. Oh, the frustration.

    So, first, if I want to find the total resistance in the parallel:
    1/R(P) = 1/R + 1/R = 2/R = 1/(R/2)
    R(P) = R/2

    Then, to find the current in the parallel:
    I(P) = (2V/3)/(R/2) = 4V/3R

    And since V and R are the same in both lamp L and N:
    I(L) = 2V/3R
    I(N) = 2V/3R

    For M:
    I(M) = (V/3)/R = V/3R

    And then after the loss of N:
    I(L) = (V/2)/R = V/2R
    I(M) = (V/2)/R = V/2R

    Meaning, Lamp L has decreased and lamp M has increased. That sounds terribly wrong to me. Where am I messing up?

    Thanks :)
     
  5. Jan 10, 2010 #4

    Doc Al

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    Staff: Mentor

    You're not messing up at all. :approve:

    (You can't trust your intuition in areas that you have little real experience in.)
     
  6. Jan 10, 2010 #5
    Lovely.

    Went ahead and dug out the markscheme made by my Physics teacher. It claims that C (Lamp L increases, Lamp M decreases) is the correct answer, which is the opposite of my result. Did my teacher make a mistake, then? Very possible, he does that a lot.
     
  7. Jan 10, 2010 #6
    In fact, everyone makes mistakes sometimes, so it might be as well to look at this again.

    You are right to say that, after the break, both lamps get I = V/2R,.

    But what really happens before the break? Are you all sure of that?
     
  8. Jan 10, 2010 #7
    Hint: before the break, it appears that you have calculated that lamp M receives one-third of the total voltage. Not so!
     
  9. Jan 10, 2010 #8
    Hmm. Indeed. Would I be right in assuming that that parallel receives 1/4 the voltage, and lamp M the rest (since V = RI, and Vp = (R/2)(I/2)) ? If so, I get the following calculations

    So, first, if I want to find the total resistance in the parallel:
    1/R(P) = 1/R + 1/R = 2/R = 1/(R/2)
    R(P) = R/2

    Then, to find the current in the parallel:
    I(P) = (V/4)/(R/2) = V/2R

    And since V and R are the same in both lamp L and N:
    I(L) = V/4R
    I(N) = V/4R

    For M:
    I(M) = (3V/4)/R = 3V/4R

    And then after the loss of N:
    I(L) = (V/2)/R = V/2R
    I(M) = (V/2)/R = V/2R


    And that is an increase in L and decrease in M, as required.

    Correct? If no, do help. I am not very strong in electric circuits. Have been neglecting it for too long :)
     
  10. Jan 10, 2010 #9
    Close, but no cigar, as they used to say. This isn't a hard problem really, but a mistake early on led you astray. You have worked on this so I guess it would be OK to give some hints. To find the voltage fraction with all bulbs working:

    The total circuit resistance is the sum of the parallel pair M and N, = R/2 as you correctly calculated earlier, plus the resistance of M; another R. Circuit resistance = R/2 + R =3R/2

    So current through I(M) = V/3R/2 = 2V/3R To get the V(M), multiply by R: V(M) =2V/3
     
  11. Jan 10, 2010 #10
    It occurs to me that if you know a formula for potential dividers, you may have been using it for this problem.

    If so, that would be perfectly OK, but be careful with which resistor you regard as R1, and which as R2.
     
  12. Jan 10, 2010 #11
    Thank you so much for the help :) I get it now.. I think.
     
    Last edited: Jan 10, 2010
  13. Jan 10, 2010 #12
    Final calculations, then.

    1/R(P) = 1/R + 1/R = 2/R --> R(P) = R/2

    R(total) = R/2 + R = 3R/2

    I(M) = V/R = V/(3R/2) = 2V/3R

    V(M) = I(M)R = 2V/3

    Thus:
    V(P) = V/3
    I(P) = V(P)/R = (V/3)/(R/2) = 2V/3R

    I(L) = I(N) = V/3R

    Then after the break:
    I(M) = V/2R
    V(M) = RI(M) = V/2 (Felt the need to confirm that it indeed receives half the voltage)
    I(L) = V/2R

    Which is a decrease in M and an increase in L, as required.

    Everything look correct now?
     
  14. Jan 10, 2010 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Sorry about that. I misread your earlier answer. :redface:

    Yes, looks good now.

    (Thanks to Adjuster for jumping in.)
     
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