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Lightning flashes

  1. Nov 12, 2015 #1

    MBBphys

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    1. The problem statement, all variables and given/known data
    (a) It is estimated that the average electric charge carried in a lightning flash is 5C. If the p.d. between the cloud and the ground is about 800 MV, approximately how much energy is transferred in a flash?

    (b) In a typical thunderstorm, lightning flashes strike the ground at intervals of about 3 minutes. Over the whole surface of the Earth the total current carried in this way between the atmosphere and the ground averages 1800 A. Estimate the average number of thunderstorms taking place at any instant over the whole Earth.

    2. Relevant equations
    W=VQ
    Q=It
    Anything else?

    3. The attempt at a solution
    For part (a), it was quite simple:
    5 * 800 * 106 = 4 * 109 Joules.

    But for part (b), I am stumped; don't know where to begin. Could you please help?

    Thank you!
     
  2. jcsd
  3. Nov 12, 2015 #2

    MBBphys

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    PS- the answer book gives 11,000 (to 2 significant figures), but I have no clue how it got there.
     
  4. Nov 12, 2015 #3

    haruspex

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    Create some variables to represent the various statistics: rate of lightning flashes in one thunderstorm; charge transferred per flash; total current from all thunderstorms at one time. Another for the thing to be found: number of thunderstorms at one time.
    Then think how you can combine these to obtain other interesting information, such as the average current due to one thunderstorm.
     
  5. Nov 13, 2015 #4

    andrevdh

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    Think electric circuit. Each thunderstorm is another battery in parallel.
     
  6. Nov 13, 2015 #5
    Something is missing from the problem.
    The problem says that a single flash delivers 5 C.
    So if a single storm lasts 3 min it delivers 5 C.
    But if it lasts 30 minutes it delivers 50 C.
    So from the information given one doesn't know the total charge related to a given storm.
    Maybe you also need to make an estimate of the length in time of a single storm.
     
  7. Nov 13, 2015 #6

    haruspex

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    No, there's enough information to answer the question posed.
     
  8. Nov 14, 2015 #7
    Well, I'm still somewhat confused.
    5 C / flash = 5 C / 180 sec = .0278 amps is the current delivered by each storm.
    Now if the total current is 1800 that gives
    N = 1800 / .0278 = 64,800 for the number of storms taking place.
    Now if you take energy / flash = 4 * 10E9 J found in part (a)
    With 180 sec per flash then the average current is 4 * 10E9 J / 180 sec = 2.2 * 10E7 amps for N storms
    Dividing by 1800 gives 2.2 * 10E7 / 1800 = 12,200 storms, approximately the given answer.
    Of course that implies 1800 amps / storm not 1800 amps over the whole surface of the earth.
     
  9. Nov 14, 2015 #8

    MBBphys

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    I am not really understanding any of this.It seems like there are a lot of assumptions to make?
     
  10. Nov 14, 2015 #9
    I agree with your answer for part (a) that the total energy in a single flash is 4 * 10E9 J.
    I think that we can agree I = Q / t and t here is the time between flashes for one storm.
    So I = 5C / 180 sec for the current being delivered by a single storm.
    Then I = .028 amps for a single storm.
    If 1800 amps is the current over the entire surface of the earth then
    N = 1800 / .028 = about 65,000 storms.
    I made an error when dividing E by t to get the current I:
    Error: (With 180 sec per flash then the average current is 4 * 10E9 J / 180 sec = 2.2 * 10E7 amps for N storms)
    Another way to look this is:
    E / t = (Q * V) / t where E = 4 * 10E9 J that you found in part (a)
    Since I = Q / t the definition of average current
    I = (E / t) / V = (4 * 10E9 / 180) / 8 * 10E8 = .028 the average current for a single storm.
    Now, I can't guess how the number 11,000 is obtained from that value.
     
  11. Nov 14, 2015 #10

    haruspex

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    No assumptions need to be made. There is enough information in the question to come up with an answer (though not the given answer).
    @J Hann , I can understand your posting all your working because you are trying to understand how the answer can be 11,000. That baffles me too, but putting that aside, I feel you have spelled it out far too much. We need to help MBBphys figure it out, not lay out the answer.

    @MBBphys :
    We are told that each thunderstorm has one 5C flash each 3 minutes on average. What does that represent as an average current?
     
  12. Nov 20, 2015 #11

    MBBphys

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    5/(3*60) A.
    And we have 1800 A overall.

    So, 1800/ [5/(3*60)] = 64800 lightning flashes.
    Thanks.
     
  13. Nov 23, 2015 #12

    andrevdh

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    The average duration of a lightning strike is 30 microseconds
    http://thunder.nsstc.nasa.gov/primer/primer2.html [Broken]
    during this short time period the average current will then be
    about 170 kA if we assume that 5 coulombs of charge was
    transferred. At the end of the mentioned webpage it states that
    "Typically, more than 2,000 thunderstorms are active throughout
    the world at a given moment, producing on the order of 100
    flashes per second." This suggests that the average current from
    these would then be just over 8 A (again assuming 5 C/flash)
     
    Last edited by a moderator: May 7, 2017
  14. Oct 18, 2016 #13
    Agree with andrevdh. I don't think you can just divide 5C by 3 minutes because 3 minutes is not the amount of time it takes to transfer 5C but the interval between strikes. So we know at least that the average current of a lightning strike is 5C/tau ; tau=duration of lightning strike.

    So you could take 1800/(5C/tau) yielding 360tau as # of strikes. But again the question is not clear. Is the 1800A the average current throughout the day or at any given moment in time? We are not told how long each storm lasts. We know at least, each storm produces on average 1 strike /3 minutes so over a 3 minute period you could say each thunderstorm produces on average 1 strike.

    However since we don't know if the 1800A is a measure of, let's say it's instantenous current. Over a 3 minute period this would deliver 324kA. Then we really should take 324kA/(5C/tau)=64,800tau # of strikes.

    Over a 3 minute period then we can assume 64,800tau storms. andrevdh in the above gives 30us as tau=1.944 storms/3minutes=0.0108 storms per second.

    Obviously that number is absurd, but I don't see the flaw in my logic.

    The problem with doing a physics problem is that it's like watching a David Lynch movie. You're never quite sure if the answer is just complicated or the author just threw a bunch of crap at the wall in the hopes you'd think there's an answer.
     
  15. Oct 18, 2016 #14

    haruspex

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    The question says nothing about the average current in one strike during that strike. It is clearly referring to the average total carried by all strikes everywhere, over continuous time.
    You seem to have become confused by J Hann's having chosen to work out the average current for a single storm, but that's still valid. If I drive 100km at 100km/h, but stop for a break of 15 minutes half way, my average speed for the journey is 80km/h
     
  16. Oct 18, 2016 #15
    I would say there are a number of reasons for confusion here.

    A) The stated answer is 11,000 which no permutation of calculations allows for
    B) Textbooks and professors will often forget a key point in formulating the question and correct it later ("But thanks for working on it for 3 hours")
    C) The actual empirical number of thunderstorms is 1,800 a far cry from the answer we find here. I understand this is physics and not engineering, but it's off by order of magnitude.
    D) I was thinking of this 1800A as the cumulative average of all the discharges not continuous time average as the nomenclature says "carried in this way" in conjunction with part a) priming the reader to think about the discharge itself. I was looking at the amount of current actually delivered/carried during discharge rather than long run average measured current. (thereby reducing the average current/storm significantly if you take into account the non-strike time) "Carried in this way" really tripped me up. I would have assumed the correct way of looking at it without that.
    E)Usually multi-part questions will utilize the preceding calculations, so this further tripped me up.

    Looking over your explanation though, I can't see any issue with it and realize my error. Thanks for your help.

    It's been 15 years since I've looked at a physics problem, so you'll have to forgive me.
     
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