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Circuits: Power, Voltage, Resistance

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data

    The circuit shown is used to produce a current-voltage graph for a 12V, 24W lamp:

    http://img338.imageshack.us/img338/9139/circuitlu6.jpg [Broken]
    http://img338.imageshack.us/img338/9139/circuitlu6.jpg [Broken]
    a) Calculate the resistance of the lamp in normal operation.
    b) Calculate the value for R which wold enable the voltage across the lamp to be varied between 0V and 12V.

    2. Relevant equations

    P = IV
    V = IR

    3. The attempt at a solution

    a) P=IV
    24 = 12I
    I = 2A

    V = IR
    12 = 2R
    R = 6 Ohms

    b) Absolutely no idea how to go about doing this... it seems that the voltage across the resistor R ought to be 8V, but I don't know, since we haven't been told what the component with the arrow pointing into the 24 ohm resistor is...

    Thanks for your help.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 30, 2008 #2
    This is a "potentiometer" A resistor with a sliding contact, forming a variable voltage divider.
    you have to determine such that you get 12 V across the lamp if the contact is at the top and 0 V if the contact is at the bottom.
  4. Mar 30, 2008 #3
    So if the contact is at the top, you get what is in effect two resistors in parallel - one at 24 ohms and the other at 6 ohms? I tried solving it like that... and got nowhere - calculated that the total resistance of the two strands of the circuit was 4.8 ohms... don't know if that's helpful at all, though.
    Last edited: Mar 30, 2008
  5. Mar 30, 2008 #4
    That's certainly helpful. Now you have to find the R that makes the voltage divider formed by R and the 4.8 Ohm resistance produce 12 V
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