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Power and Circuit

  1. Feb 23, 2007 #1
    1. The problem statement, all variables and given/known data

    How many 90-W, 120-V light bulbs can be connected to a 20-A, 120-V circuit without tripping the circuit breaker? (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.)

    2. Relevant equations

    See below.

    3. The attempt at a solution


    P_total = I*V = 25 W*120 V = 2400 W

    On 120 V line, a 90 W, 120 V bulb has a dissipation of 90 W?

    Number_bulbs = 2400 W/90W = 26.7 bulbs

    So maximum is 26 bulbs?

    As a check, compare current values?

    I = P_total/V = 26 bulbs(90 W)/(120 V) = 19.5 A, where 19.5 A < 20.0 A

    Thanks.
     
  2. jcsd
  3. Feb 23, 2007 #2

    ranger

    User Avatar
    Gold Member

    Well, they already said its a 90W bulb.

    Yup, your calculations are correct.

    Although I wouldnt have found the number of bulbs in that way. I would simply do the following:
    Isingle bulb = P/V = 90W/120V = .75A
    20A/.75A = 26.6; so we can have 26 bulbs with a total current of 26*0.75A = 19.5A
     
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