Fundamental questions on current and waves

Discussion in 'Homework Help' started by Niles, Jan 9, 2009.

  1. Niles

    Thread Starter Active Member

    Nov 23, 2008
    56
    0
    Hi all.

    I have 3 quite fundemental questions, which I think I know the answer of. Please correct me when/where I am wrong:

    1: When looking at a coaxial cable which just ends at point x = L (no termination, so the resistance is infinite), then we have that the current satisfies I(L,t)=0. Does this mean that at the end of the cable, it looks like this? (Look at the attached image)

    2: On an Ohm-meter, I can turn the "wheel" to 2 Ω, 20 Ω, 0.2 kΩ, 2 kΩ, 20 kΩ, 200 kΩ and so forth. Lets say that it is placed at 2 kΩ, and the display of the meter shows 3.02. Does this mean that the resistance is 3020 Ω? And likewise, if the "wheel" is placed at 200 kΩ, and the display shows 4.87; does this mean that the resistance is 4870 Ω?

    3: Why does it say 2 Ω, 20 Ω, ... and not 1 Ω, 10 Ω, ...? (This I don't have any attempt on)

    Thanks in advance. I really appreciate you helping me.

    Best regards,
    Niles.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    1. Not absolutely positive, but it sounds right that the voltage is reflected back 180 deg out of phase.

    2. The device is a range selector. Use the full scale selected to judge the magnitude displayed. The decimal point should move in agreement with the scaling, so the 3.02 value in the 2K scale is not possible unless the meter displays 5 1/2 digits. And, if that were the case, the reading would be 3.02 volts.

    3. A meter usually can display X digits, plus a "1", so a 3 1/2 digit display can show a maximum value of 1.999 (or 19.99, 199.9, or 1999). The 1.999 is close enough to 2 to make essentially no difference.
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    I think it is a good idea to note that

    I_reflected = -I_incoming (note the negative)
     
  4. Niles

    Thread Starter Active Member

    Nov 23, 2008
    56
    0
    Thanks guys for answering. It seems like you know alot about coax.-cables and reflections. So here comes another tricky question:

    Lets say that I do not terminate the cable at point x = L, i.e. it is just left open. Then R = infinity, which gives me that the amplitude of the reflected voltage equals the amplitude of the incoming voltage:

    V_reflected = V_incoming.

    So the voltage gets reflected without a phase change. How does this explain the fact that the current is zero at the point x = L, when the cable is not terminated?
     
    Last edited: Jan 10, 2009
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