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Resistance, power and brightness of bulb?

  1. Jun 5, 2012 #1
    Hey,

    I have a question regarding practical electricity. Since we know that power (rate of energy released) determines brightness of bulb, then:

    Since P = VI and V = RI, it implies
    1. P =(V^2) / R
    2. P = (I^2) * R

    How can we tell if power increases or decreases when resistance (R) of the bulb increases? Using equation 1 it seems P decreases as R increases, but using equation 2 it seems P increases as R increases.

    Thanks!!
     
  2. jcsd
  3. Jun 5, 2012 #2
    Do you think V remains constant if you put various bulbs with different R's in an outlet?

    Do you think VIremains constant if you put various bulbs with different R's in an outlet?
     
  4. Jun 5, 2012 #3
    Surely you need resistance (R) for there to be power in the form of thermal energy release, but also as it increases the current flow must decrease given a fixed potential difference (V) being applied. You can see this clearly with EQ 1. In EQ 2, it may not be as obvious. (I) is not constant like (V) is in EQ 1, but rather dependent on (R) -- I = V/R.

    Considering that the (I) term is squared and the (R) term is not in EQ 2 and (I) is inversely proportional to (R); as (R) decreases, (I) squared "takes over". This is obvious in EQ 1. That is P -> ∞ as R -> 0.
     
  5. Jun 5, 2012 #4
    Hey guys,

    Thanks so much for the replies! Helped me alot :)

    Cheers :)
     
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