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Resistance - Wattage?

  1. Jul 20, 2006 #1
    Hi everyone !

    First time to post! I am taking a distance course of a physics class now.
    I am bit confused about the resistance & wattage.

    In the circuit with the bulb, when there is a greater resistance with the bulb the more energy dissipate, therefore it shines brighter... I think.

    But, when I read through my tutorial book talking about the wattage of the bulb for example, 60W has less resistance than 40W bulb.

    Umm.. I thought "more resistance, more enegery dissipate" so I thought 60W has more resistance than 40W bulb to dissipate more = to mae it shine brighter...?

    Could anyone advise?
    Cheers!!:smile:
     
  2. jcsd
  3. Jul 20, 2006 #2

    rcgldr

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    When the resistance is less, then more current flows through the light bulb when the voltage is constant (as in this case). The more current, the more power, which is converted into light and heat by the light bulb.
     
  4. Jul 20, 2006 #3

    Kurdt

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    Consider the equations of power.

    [tex] P=V I[/tex]

    [tex]P=I^2R[/tex]

    For the domestic bulb it usually operates at 120V (varies depending on which country you come from) so we will take this as the operating voltage.

    Now for a 40Watt bulb the Current is from the first equation 1/3 Amp and for a 60 Watt bulb it is 1/2 Amp.

    Taking these and putting them in the second equation and solving for R you get for the 40 Watt bulb R=360 ohms and for the 60 watt bulb R=240 ohms.

    Do these calculations and have a play about with the power equations to convince yourself.
     
  5. Jul 20, 2006 #4
    Hi !

    Thanks for the replies.

    I understand the above formula and theory. Thanks for the example. But I think I might have confused this with the following;


    P = IV:
    So, more current means less resistance (I=V/R) means more power.
    So, 60W has less resistance, that makes sense to me now. :)

    I was confused the wattage to the following sentence I read on my tutorial;

    For most conductors, increased tempreture means increased resistance

    I thought increased tempreture is voltage, is it so?
    Because R = V/I. More tempreture / voltage creates more resistance, then, when thinking of P=IV, more voltage means more power but also it means more resistance.

    But I think your explanation is when voltage (V) = constant.


    So... they use tungsten for a filament because of its high resistance, if it has no resistance it doesnt dessipate the heat / light.
    So its resistance has to be high to shine bright.

    I cant quite put these two theory together... high resistance, brighter it shines... but more power has less resistance....
     
  6. Jul 20, 2006 #5

    Kurdt

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    If you look at the second power formula I gave above perhaps that can clear up your misunderstanding. If you want more power what is easiest to do? Increase the current (I) or the resistance (R) with the constraint that voltage is essentially constant.

    But also consider that high resistance causes heat and light because the electrons in the current are supposedly bouncing off the atoms in the wire. If you increase the current there are more electrons and therefore more collisions. Consider this scenario fora relatively low resistance. If you put a massive current through a low resistance wire it will still heat up because there are so many electrons flowing through the wire that even with the relatively low chance of hitting atoms (think of resistance as probability of hitting an atom) it still occurs in large enough quantity to produce heat and light.

    Perhaps my wording or analogies are awful and you still have questions but I hope this explains it.
     
  7. Jul 20, 2006 #6

    Office_Shredder

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    Are you sure about this? From what I know, they use tungsten because it melts at a high temperature. This allows them to make a lightbulb that gives off more heat
     
  8. Jul 20, 2006 #7
    Hi Kurd & OS

    [tex]P = I^2R[/tex]

    For this formula, unfortunatly I couldn't find in my tutorial. No explanations.. but looking at this formula doesn't the both increase in I and R increase the power?

    But then, increase in the R means decrease in I.
    Or should I not mix these two theories together..?

    I checked with my all time favourite Wikipedia and it says,

    .....is remarkable for its robust physical properties, especially the fact that it has a higher melting point than any other non-alloy in existence.

    Yes, the Tungsten has high melting point which I think it means it stands up to the high tempreture without melting and that's why I think it's used in the filament.

    But here...the tutorial note says:


    Tungsten wire is capable of glowing white hot without melting and is used as the filaments of light bulbs. Light and heat are generated as the current heats the high resistance tungsten filament.

    The hotter the filament, the brighter the bulb. That means a bright bulb has a lower resistance than a dimmer bulb. Just as water flows with more difficulty through a thinner pipe, electrical resistance is greater for a thinner wire.


    So.. tungsten is used because it's high resistance and the following sentence says, The hotter the filament, the brighter the bulb

    I combined these two idea and thought high resistance material produce more power. But then the following sentence doesnt match to this theory. (That means a bright bulb has a lower resistance than a dimmer bulb. )

    Do you or do you not think this explanation is bit mismatch..?


    PS: Btw, this is not a home work question...:tongue: but it's move to this forum. Maybe anything to do with class or course should be poested in here? Then I will in the future.
     
  9. Jul 21, 2006 #8
    another derived equation using [tex] U = I \cdot R[/tex] is:

    [tex] P = \frac {U^{2}} {R} [/tex]

    The same as [tex] P = I^{2} \cdot R[/tex], they are derived from [tex] P = U \cdot I[/tex]:

    Replace U by [tex] I \cdot R[/tex] and I by [tex] \frac {U} {R}[/tex] and you'll find these formulas.
     
    Last edited: Jul 21, 2006
  10. Jul 21, 2006 #9

    Kurdt

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    The sentence is not contradictory. It says that tungsten has a high resistance and that the hotter it gets the brighter the bulb is. Where is the contradiction here? let me try and explain again. A greater current produces more heat because there are more electrons flowing through and colliding with atoms in the wire. You get more electrons flowing (i.e. more current) by lowering the resistance.
     
  11. Jul 22, 2006 #10
    No I dont see the contradiction when you explain like that... humm, where did I thinl there was a contradiction..? :uhh:

    If the Tunsgten had low resistance it would let more curent to flow but it would bright as much as it does with the high resistance, or does it? If there is no resistance, I think the current will "just flow without doing any works".

    I just wonder this "high resistancy" does work on emitting more energy to escape out as a heat / light and I wonder is that why,

    P = I^2R. To gain more power, more R is needed. So basically, more I and more R produces more P so it's not just I energy that produces more P.

    Bear me if I am being too dumb....
    I think I am more conecptual person than logical. But being logical you need the throughly understanding the concept so logical explanation is beyond concept and maybe it just take time for me to reach that extend? means a legally dumb? :bugeye:
     
  12. Feb 15, 2009 #11
  13. Feb 15, 2009 #12
    I think it's helpful to remember something about these two variables, voltage V and current I. The equipment will usually require a certain value for one of them, and then the other variable will be whatever it comes out to.

    If you have a voltage source, which is what a dry cell battery is, and also what an AC outlet on the wall is, it forces the device to have a particular value of V, so I would write [tex]P = \frac {V^{2}} {R}[/tex].

    However, if we had a current source, which is rare in practice, but common in textbook problems, it would force the device to have a particular value of I, so I would write [tex]P = I^{2} R[/tex].

    But here is an interesting result, which the position of R in either the numerator or denominator shows:

    With a voltage source: the higher the resistance, the lower the power.

    With a current source: the higher the resistance, the greater the power.
     
  14. Feb 15, 2009 #13

    Kurdt

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    The thread was started in 2006 guys. I guess it will be useful for any student that find it. Check the dates in future though.
     
  15. May 19, 2009 #14
    Simply I will put the question here. Becausein the question of bulbs of 40 and 60 Watts at very first we have got for a const V (i.e. 120 or whatever) resistance is less for more power. Then there is tungsten where more resistance means more power(light, heat). What is the matter can some one explain it accurately and helpfully without going in deep towards electrons etc of partial chem interest i.e i expect to get answer in physical terms. My friend Kurdt has written this but i have great confusion. Firstly let's read that :-
    "If you look at the second power formula I gave above perhaps that can clear up your misunderstanding.(i.e P= I^2R
    If you want more power what is easiest to do? Increase the current (I) or the resistance (R) with the constraint that voltage is essentially constant.
    But also consider that ' ' high resistance' ' causes ' ' heat and light' ' because the electrons in the current are supposedly bouncing off the atoms in the wire. If you increase the current there are more electrons and therefore more collisions. Consider this scenario for a relatively low resistance. If you put a massive current through a low resistance wire it will still heat up because there are so many electrons flowing through the wire that even with the relatively low chance of hitting atoms (think of resistance as probability of hitting an atom) it still occurs in large enough quantity to produce heat and light."

    But if you have high resistance then constant V means low current and bcoz power is more dependent upon I(square) therefore Power(heat,light) will be less. This simply implies that low resistance more power(Light,heat).Whatever i have read here i am upto only this conclusion that more R means less P.
    Ok, then my friend mikelepore writes concludarily that:-
    "With a voltage source: the higher the resistance, the lower the power.
    With a current source: the higher the resistance, the greater the power."

    But in both cases we have the same ""bulb"". where is the question of voltage source or current source.
    So please explain this, think that the question was posted by me.
    I will be really thankfull very very much to anyone who clears my querry.
    Please if you can copy and email your reply at shahrukhkhan1992@yahoo.com then do so. I will be glad
     
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