1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Doubling voltage and power in a light bulb

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A light bulb is designed for use in a 120 V circuit. If by mistake it is plugged into a 240 V circuit, what will happen to the power of the light bulb? Explain.

    2. Relevant equations
    I = V/R
    P = VI

    3. The attempt at a solution
    Since I = V/R and the resistance of the lightbulb does not change, doubling V also doubles I. Pf = 2V2I = 4P0. Alternatively, P = V2/R, so doubling V quadruples P. Where I'm getting stuck is on explaining this. What are the practical implications of quadrupling the power? What is the physical representation of quadrupling it - I mean, aside from the equations, WHY is it quadrupled? I'm really stuck on this.
  2. jcsd
  3. Jun 22, 2010 #2
    Tungsten actually has a pronounced positive change in resistance with increasing heat, so that a cold lamp suffers an 'inrush' current as it comes to its working temperature and resistance.
    If the voltage is doubled from the rated voltage, less than 4 times the rated power is dissipated in the filament. However, experience shows that commercial tungsten filament lamps are not designed to withstand such treatment: we know that the bulb will fail due to melting of the filament and consequent lack of a conduction path.

    I hate to leave you feeling stuck on the matter of why. All I can say is that if you had to carry twice as much water, twice as high, to generate the potential to run some machine, you'd soon understand what 'quadrupling' the power means.
    Last edited: Jun 22, 2010
  4. Jun 22, 2010 #3
    Ok, so lets say that the light bulb is 100 ohms. At 120V it would consume 1.2A and 144 Watts. We connected by mistake to 240V, then it consumes 2.4A and 576 Watts. Most likely the light bulb is not designed to take that much current and it fries.
  5. Jun 22, 2010 #4
    Light bulbs are heated to a high temperature and their resistance is much greater than the resistance of a "cold" light bulb. Therefore, assuming R = const. is not valid.

    Let us assume that the resistance is proportional to the aboslute temperature and that the temperature determines the radiated power according to Stefan's Law, i.e. the radiated power is proportional to the 4th power of the absolute temperature. These two proportionalities would mean that the resistance is proporional to the fourth root of the radiated power.

    In equilibrium, the radiated power is equal to the power consumed. Using the two above equations to eliminate I, you should get:

    I = \frac{V}{R}

    P = V I = \frac{V^{2}}{R}


    P R = V^{2}

    But, [itex] R \propto P^{1/4}[/itex], so:

    P^{5/4} \propto V^{2}

    P \propto V^{8/5}

    So, if you double the voltage, this "improved" model would predict that the power should increase by a factor of:

    2^{8/5} = 2 \, 2^{3/5} = 3.03

    One should perform an experiment to see if the above scaling arguments holds.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook