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Power and Resistors

  1. Oct 23, 2006 #1
    Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 4.7 ohms and 17.7 W, 23.1ohms and 13.9 W, and 12.8ohms and 12.9 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?

    A - to find the greatest voltage, i needed the max current so i used P= IR^2 for each one to and find the current, the lowest current will provide the max current for the circuit without the resistor burning up. Then use the current multiple by the sum of resister to find the voltage.

    My question is , I though when resistors are placed in series (if voltage is given), the current is the same through out, how come in this problem, i had to find current separately. as oppose to finding the sub of the resistance and multiplying it by the voltage (if given)
     
    Last edited: Oct 23, 2006
  2. jcsd
  3. Oct 23, 2006 #2
    I believe that from P=VI and V=IR,
    you get P=(IR)I
    and finally P=I^2 * R
     
  4. Oct 23, 2006 #3

    berkeman

    User Avatar

    Staff: Mentor

    You approached it correctly. The series current is indeed the same in each resistor, so you can calculate the max current allowed based on the smallest power-handling resistor. That defines the current for the series resistors, which gives you the max total voltage of the battery.
     
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