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Homework Help: Equation for the power a light bulb would radiate

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    An ordinary 150-Watt light bulb has a resistance of about 96 ohms.
    Suppose you were to connect a 2.0 V battery to such a light bulb. How much current would flow? Determine how much power the bulb would radiate? (probably too little for a person to detect). Use you own calculations for this question. The lab can be used to test your results by comparing a similar model.

    2. Relevant equations

    I = E (base eq) / R (base eq);
    P (base w) = I^2*R.

    I dont know the equation to find the power a light bulb would radiate
    in a given electrical circuit.

    3. The attempt at a solution
    If I hooked a 150 W bulb (about 96 Ohms) to a 2V Battery then:

    E (base eq) = 2V
    R (base eq) = 96 Ohms.

    I = E (base eq) / R (base eq) = 2V / 96 Ohms =~ 0.02080A

    0.02080 A is the current that a 150 Watt Light Bulb would draw from a 2V Battery.

    To find the power that would dissipate:

    P (base w) = I^2*R = (0.02080A)^2 * 150 Ohms =~ 0.06514;

    is the amount of power dissipated
  2. jcsd
  3. Feb 2, 2010 #2


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    Homework Helper

    Your calculation is correct. Curious - I haven't seen E used for potential since the 1960's - everyone seems to be using V now. If you put I = E/R into P=IE you get P = E²/R, which is the most efficient formula for this question.

    You are quite right to be puzzled by this question, which is much oversimplified. The bulb will not light up so all the power will be converted to heating and only a part of that will be radiated in the infrared - the rest will conduct away into the wires and fixture. Also, in the lab it will be important to know that the resistance of a light bulb is much lower when it is cold than when it is hot.
  4. Feb 2, 2010 #3
    Thanks Delphi,

    The E is used for Potential in, "College Physics" by Giambattista. I'm learning out of his text therefore using his assigned varaibles. And yes, I agree, he should use the more recent varaible names.

    Anyway, I do appreciate your reply, but I am still wondering how to determine the power the bulb would radiate, and what equation I should use..

    Again, thanks for your reply! :)
  5. Feb 2, 2010 #4


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    Homework Helper

    I'm sure your answer is the one expected.
    The true answer is impossible to find.
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