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Circuit problem: Will the fuse blow out?

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Number 14.


    2. Relevant equations
    A fuse will blow out if there the current going through it is larger than the ratings of the fuses.
    V = IR



    3. The attempt at a solution

    The current from the battery comes out of the negative side (small line) and goes through F1 with almost no resistance. F1 then blows out. The resistor near S3 reduces the current

    V = IR
    6 = I(3)
    2 = I

    Since 2 < 10, F2 does not blow out

    How about F3?

    Does not the current flow through S1 straight into F3 with almost no resistance? Why doesn't F3 blow out?
    Does the current does flow out of the negative end of the battery too?

    Attached are the problem and solution sheets.

    001.jpg
    002.jpg

    Thanks!
     
  2. jcsd
  3. Jun 20, 2011 #2

    I like Serena

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    Hi joej24! :smile:

    For current to flow, it must always be able to flow from a positive pole to a negative pole.
    If there is no connection, or rather no potential difference, current can not flow.

    As it is, the potential on loose en of F3 will change to match the potential on the battery side.
    The effective potential difference will remain zero and no current will flow.
     
  4. Jun 20, 2011 #3
    Actually The current of through the 2 ohm and the 3 ohm resistance would have to go through F2. It's easy to see that the current is still smaller than 10 A, even if F3 does
    not blow.

    There's a closed path through the power supply, F2, S4, the 2 ohm resistance, F3 and S1

    If the current through this path is bigger than the rating of F3 it will blow. (after F1 blows, F1 will short out the power supply, so you get no voltage across the 2 ohm resistor, and therefore no current through F3 until F1 blows.)
     
  5. Jun 20, 2011 #4

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    I believe the problem states that S1, S2, and S3 are closed.
    In particular S4 is not closed.
     
  6. Jun 20, 2011 #5
    Oh, I see. Since S4 is open, there is no current through F3.

    So even though the current passes through the fuse (F2) before going through the resistor, F2 does not blow? In other words, The current does not change ONLY on the part of the wire past the resistor? The presence of that Resistor (3 ohms) in that specific closed path makes the current be the same, or 2 amperes, throughout it?

    Also, how do we know that the current through F1 is greater than 10 A? It's only given that there is no resistance in that path and that the battery has a potential of 6V.
     
  7. Jun 20, 2011 #6

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    Right! :smile:


    Yes, in a wire without branches the current is the same everywhere.
    (And in a branch the current splits in such a way that current-in equals current-out.)


    With zero resistance, you'll get infinite current (with a perfect battery).
    It's called a short-circuit, meaning in practice that any fuses will burn out.
    The problem states that the battery has "negligible internal resistance", meaning it is near perfect.
     
  8. Jun 21, 2011 #7

    Ouabache

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    If you prefer to think of this empirically, you can still solve ohm's law i = v/r .. i = 6/0 (which is undefined), however if you take the limit as r --> 0, i = 6/(small number close to zero) = infinity, or infinite current, so the current (infinite amps) through F1 > 10A.

    extra credit: What would happen if F1 was not there and the conductor was continuous through S2?
     
  9. Jun 21, 2011 #8
    I am not sure if I understand the question exactly.
    If F1 was replaced by F2, F2 would blow out too because
    If F1 is just taken out and everything else is left in place, then the current would go normally through F2 and it would not blow out because there is a resistor that decreases the intensity of the current like we figured out before.

    Thanks PF and everyone!
     
  10. Jun 23, 2011 #9

    Ouabache

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    Extra Credit:
    I wasn't suggesting to just take out F2. I mean, replace it with the same conductor it has above and below it (not leave it an open circuit).
     
  11. Jun 25, 2011 #10
    If F1 was taken out and the conductor continuous through S2, none of the fuses would blow.
     
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