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# of elecron/sec in a 3 bulb circuit

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    When a single round bulb of a particular kind and two batteries are connected in series, 5* 10^18 electrons pass through the bulb every second. When a single long bulb of a particular kind and two batteries are connected in series, only 2.5*10^18 electrons pass through the bulb every second, because the filament has a smaller cross section.


    Figure 18.68
    In the circuit shown in Figure 18.68, how many electrons per second flow through the long bulb?

    E is the electric field in the filament
    A is the cross-sectional area of the filament
    i is electron current
    l is the length of the filament
    n is the # of electrons per unit volume (unknown)
    u is the electron mobility (unknown)

    2. Relevant equations

    Circuit 1 is the main circuit, with long bulb A, and two identical round bulbs, both labeled B.
    Circuit 2 is the circuit with 1 long bulb, C.
    Circuit 3 is the circuit with 1 round bulb, D

    [tex]i_1=nA_auE_a=n2A_buE_b[/tex]
    [tex]i_2=nA_auE_c=2.5e18[/tex]
    [tex]i_3=nA_buE_d=5e18[/tex]
    [tex]2emf=l(E_a+2E_b)=E_c l=E_d l[/tex]

    3. The attempt at a solution

    note: [itex]A_b>A_a[/itex]
    [tex]E_c=E_d[/tex]
    I am assuming that all the filaments have a length 'l.'

    I have created fractions of current to produce some hopefully useful formulae.

    From [itex]i_1[/itex], [itex]A_a E_a=2A_b E_b[/itex]

    From [itex]\frac{i_1}{i_2}[/itex], [itex]E_a i_2 =2 E_b i_3[/itex]

    Also, [itex]i_1=\frac{E_a i_2}{E_c}=\frac{2 E_b i_3}{E_c}[/itex]

    And [itex]\frac{i_2}{i_3} = \frac{A_a}{A_b}[/itex]

    [tex]E_a+2E_b=2E_b+E_b \frac{i_3}{i_2}=E_c[/tex]


    Please just give me a nudge in the right direction. Thanks!!!
     

    Attached Files:

    Last edited: Mar 9, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    Isn't it just about voltage, current and combining resistances?
     
  4. Mar 10, 2012 #3
    We haven't talked about resistance or Ohm's law yet. We can only use Kirchhoff's node rule,
    "In the steady state, the electron current entering a node in a circuit is equal to the electron current leaving the node."
    We can also use "i=nAuE" and the Loop Rule, where voltage around a closed path is zero in a circuit.

    Is there an easier way to do it with resistance? Please help, but I would also like to know how to do it without Ohm's Law.
     
  5. Mar 10, 2012 #4
    Wow, how did I not see it? In formula 4

    Then [itex]A_b=2A_a[/itex] !

    So now formula 1 can be written as [itex]E_a=4E_b[/itex]

    There is an error in the 4th relevant equation. It should be [itex]2emf=l(E_a+E_b)=lE_c[/itex] because the closed path only travels through 1 round bulb, not 2.

    and the rest is algebra
     
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