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Circuits problem

  1. May 12, 2005 #1
    The identical lamps in the circuit (attachment) each have a resistance R at their rated voltage of 6V. The lamps are to be run in parallel with each other using a 12V source and a series resistor Z in the circuit as shown. If the lamps are to operate at their rated voltage, what must be the resistance of Z?

    A. R/2
    B. R
    C. 2R
    D. Zero

    Any ideas?

    Attached Files:

  2. jcsd
  3. May 12, 2005 #2
    The set of parallel resistors and resistor Z are in series. The net voltage in the circuit is 12V. If each lamp operates at 6V, how does their equivalent resistor's voltage compare? How do series resistors handle voltage in circuits?
  4. May 13, 2005 #3


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    I can't fully guess how many lamps are in nobb's circuit until attachment is approved. Given the solution possibities, I am guessing two lamps.

    It sounds like a voltage divider arrangement, with the output voltage already known (6V). After combining all the parallel resistances (bulbs), you just need to do some algebra to solve for Z.
  5. May 13, 2005 #4
    I know I have to do some algebra to find some sort of ratio. My problem is that I have no idea where to start.
  6. May 13, 2005 #5
    When the electrons leave the negative terminal, they have to go through a 12V total drop. Which means once theyve gone through the whole circuit, theyve dropped 12 V. So The voltage drop through Z plus the voltage drop through the parallel equivalent resistor is 12V

    Z + P_eq = 12

    Can you find P_eq? What do you know about the two?
  7. May 13, 2005 #6
    It should be even easier when you look at the fact that you have 12 volts, you want to have 6 volts across the bulbs, and you know you therefore need to drop the other 6 volts across the resistor.
  8. May 13, 2005 #7
    I was trying to have him realize that, but thanks for your input anywa.y
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