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Parallel resistance and heat problem

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A heater joined in parallel with a 60W bulb is connected to the mains. If 60W bulb is replaced by a 100W bulb, will the rate of heat produced be more/less/remain the same ?


    Potential Difference = 220V

    2. Relevant equations
    H= VIt = I^2Rt = V^2*t/R
    P= VI

    3. The attempt at a solution
    R= (V^2/P)
    = 48400/60
    =806.6 ohms

    R = V^2/P
    = 48400/100
    = 484 ohms

    I=V/R
    = 220/806.6
    = 0.27 A
    I = V/R
    = 220/484
    = 0.45 A

    I am stuck here. I noticed that it is given 'rate of heat' which is rate of energy which is again nothing but power.
    I don't know what I'm finding in the problem.:confused:
    How do I calculate the rate of heat of the heater in parallel with the bulb ?
    Has parallel connection of the bulb and heater anything to do with the problem, or is it just to confuse us?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 11, 2010 #2

    lewando

    User Avatar
    Gold Member

    You must first decide if the question is asking about the heat generated by the heater or by the total system (heater plus bulb).
     
  4. Sep 11, 2010 #3

    rl.bhat

    User Avatar
    Homework Helper

    H = v^2*t/R.

    In parallel combination, V is the same for all components. So by connecting 60 W bulb or 100W bulb will not change the resistance of the heater or the voltage across. So H remains the same.
     
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