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2 problems

  1. Mar 2, 2004 #1
    A length of metal wire has a radius of 0:002 m and a resistance of 0.1. When the potential difference across the wire is 14 V, the electron drift speed is found to be 0:000229 m/s. Based on these data, calculate the density of free electron in the wire. Answer in units of m^-3.

    r=.002m R=.1 Vd=.000229

    V/R=i
    14/.1=140

    Vd= drift velocity.

    i/(Vd*A)=ne
    140/(.000229*pi*.002squared)=4.86499826e10
    j=neVd
    11140846.02 (I tried plugging in this answer, it's wrong)
    11140846.02/1e-3 = 1114084602e10 (I did this because it says answer is
    supposed to be in 1e-3)

    The answer is wrong I don't know why.

    A 72:5 W, 191.2 V light bulb is plugged into a 61:6 V outlet. If energy costs 4.7 cents/kWh, how much does it cost per month (30 days) to leave the light bulb turned on? Answer in units of
    cents.

    light bulb = 72.5W, 191.2V
    outlet = 61.6V
    Energy = 4.7 cents/kWh
    Days = 30

    P/V = i
    72.5/191.2 = .3791841
    i*V=P
    .3791841*61.6 = 23.35574059W
    23.35574059/1000 = .02335574059kW
    Pt = P*t
    .02335574059*30*24 = 16.81757322
    Cost = P*cents
    16.81757322* 4.7 = 79.04259414 cents
    This is also wrong, I don't know why.

    Any help is appreciated.
     
  2. jcsd
  3. Mar 2, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What happened to "e", the electron charge? Seems like you left it out. (Also, don't carry so many significant digits; three is plenty.)
    You forgot that the bulb is plugged into a lower voltage than it is rated for, so it will consume less power. The bulb's wattage depends on the voltage. First find the resistance of the light bulb (assume it is constant). Then find the power consumption at the new voltage.
     
  4. Mar 2, 2004 #3
    The e wouldn't have mattered by the end because I am taking the value "ne" together

    I don't understand what you're telling me to do here.
     
  5. Mar 2, 2004 #4

    Doc Al

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    Staff: Mentor

    I thought the question asked for electron density, which is "n", not "ne". "ne" would have units of C/m3, not 1/m3.
    Treat the light bulb as a resistor. Find its resistance, based on the rated numbers (watts at a given voltage). Now find out how much power is used when that bulb (a resistor) is given a different (lower) voltage.
     
  6. Mar 2, 2004 #5
    How sure are you that this is right? Because there is one try left for me to get this right. If you think this is right, can you please double check my work at tell me if it's right.

    r=.002m R=.1 Vd=.000229 e=1.6e-19

    V/R=i
    14/.1=140

    Vd= drift velocity.

    i/(Vd*A)=ne
    140/(.000229*pi*.002squared)=4.86499826e10

    ans/e= n
    4.86499826e10/1.6e-19 = 3.0406e29
    n/1e-3 = 3.0406e32
     
  7. Mar 2, 2004 #6

    Doc Al

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    Staff: Mentor

    Of course I think it's right. But it's your homework. Sounds like you've been doing more "plugging and praying" than understanding. :smile:
    I have no idea what you mean by "n/1e-3", but the answer seems correct; the units are 1/M3 (which means the number of electrons per cubic meter... that's the definition of "n").
     
  8. Mar 2, 2004 #7
    the question said the answer should be in 1e-3. So I divided the answer by 1e-3
     
  9. Mar 2, 2004 #8
    ****, the answer was 3.0406e29. Anyway, thanks for the help
     
    Last edited: Mar 2, 2004
  10. Mar 2, 2004 #9

    Doc Al

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    Staff: Mentor

    The question asked for units of m^-3, which are the proper units for "n", not "1e-3". (At least that's what you said: "Answer in units of m^-3.")

    Perhaps I should have been clearer. I meant that the answer you gave for "n" was correct. (I have no idea why you divided by 1e-3, which is why I questioned it.) Oh, well... sorry.
     
  11. Mar 2, 2004 #10
    Whoa! I just realized that the question says that the answer is supposed to be in 1e-3 but the correct answer was in 1e3, maybe I can ask my prof. to give me credit for the answer!
     
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