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A quick question about Rheostat

  1. Sep 27, 2008 #1
    Hi all,

    I have an attachment circuit quickly done in paint of what my question is based on. Btw, the LDR can be replaced with any Conductor as I am doing the ohm's law experiment.

    I am doing this experiment in school (not homework) and just need a bit of help. I'm not very good at physics sorry.

    To my understanding, the electrons flow from the negative terminal of the power supply each with a 2V voltage. Some will flow up towards the rheostat and some will flow to the LDR. My concern are the ones that flow to the LDR. I was taught that even if the electrons have split routes, each electron still holds a charge of 2V, so the voltage is not decreased. This would mean that the Voltmeter should read 2V no matter what right?

    I was taught that a component will always use up the voltage whether it is 2V or 100V, so the Voltmeter should read 2V right?

    However, according to my results, the Voltage went from 0.32 to 0.67. These results aren't wrong, its just that I dont understand it.

    Voltage is Directly Proportional to Current and my graph and results show that. what I dont understand is:

    1) How is it possible for the Voltage to change from 2V, (I did not alter the powerpack, it was kept on 2V thorughout the experiment)

    2) What is the point of the rheostat if it is at the end of the circuit? I slide the rheostat up and down and the voltage and current on the voltmeter and ammeter do change even though it is at the end of the experiment!

    Sorry for my lack of physical knowledge. Any help will be totally appreciated.

    Attached Files:

  2. jcsd
  3. Sep 27, 2008 #2
    Ok thanks for all the help people o_0

    I found out that the Long side is negative, so the rheostat is at the start not end, which explains a lot.

    Prob solved.
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