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B Which makes a bulb glows brighter?

  1. Jul 21, 2016 #1
    My textbook says a bulb glows brighter due to its high resistance. But it also says a bulb glows brighter when a larger current flows through it. Which is correct?
     
  2. jcsd
  3. Jul 21, 2016 #2
    Both. P=IV, so you need both resistance and current.
     
  4. Jul 21, 2016 #3
    100W bulb has less resistance than 40W bulb.
    But P=IV = I^2R = V^2/R.
    If you connect them in parallel (as you would normally do), more resistance means less current at the same voltage and less power.
    However if you connect them in series then more resistance means more voltage at the same current and consequently more power.
     
  5. Jul 21, 2016 #4

    jim hardy

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    "My textbook says a bulb glows brighter due to its high resistance." That's a strange and (misleading) thing for an author to write.
    Can you provide the actual text ? Perhaps he said 'high temperature' ?
    Larger current will heat an incandescent filament to higher temperature increasing its light output, ie making it brighter ..
     
  6. Jul 22, 2016 #5
    But how to achieve the condition where both current and resistance are high?
     
  7. Jul 22, 2016 #6

    davenn

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    that is a contradiction ... higher resistance = lower current

    eg for a given voltage, as the resistance increases, the current will decrease

    but before we continue, answer Jim Hardy's question :smile:
     
  8. Jul 22, 2016 #7

    Merlin3189

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    I agree, the comment is nonsensical as written. I have many perfectly good bulbs which don't glow at all - because they are in their boxes in the cupboard. Whatever their resistance bulbs don't glow until you connect them to the electric power source. Therein lies the key to this issue.
    Mains bulbs are connected to what is essentially a constant voltage source. Therefore their power is determined by P = V2/R and (for the same physical size, which we haven't yet mentioned) brightness is inversely proportional to resistance.

    If you want power proportional to resistance, with P = I2R then you need a constant current source, which you come across in some LED drivers, but not for incandescent bulbs.

    A battery or dynamo with low internal series resistance is a good approximation to a constant voltage source. Constant current sources are more difficult to make and usually involve wasting power.

    Power inversely proportional to resistance is inherently more stable (for metallic filaments) because, if the filament temperature rises, the resistance increases, thence the power decreases and the temperature falls. This is better than rising temp -> rise in resistance -> increase in power -> rise in temp that would happen with a constant current source and metallic filament.

    Carbon filament bulbs have a falling resistance with temperature, so are happier with P ∝ R as with a current source. Carbon filament bulbs powered from mains are often run in series with a metallic resistor ballast (eg. an ordinary bulb!) This is usually thought of as a combined resistance having a neutral or slightly positive temperature coefficient of resistance, but could be thought of as moving a low resistance constant voltage mains supply towards a constant current supply by adding series resistance.

    Since real power supplies are neither constant voltage nor constant current, there is an optimum value of resistance where the power dissipated in the bulb is maximum. If the bulb has a resistance less than that, then it will get brighter if you increase its resistance. If the bulb has higher resistance than optimum, then you get more power by reducing the resistance. But you don't normally want maximum possible power: when connected to a mains supply that would be 10's of kW at least; with a battery not so much, but an equal amount of power is dissipated in the battery, so not a good idea.

    You may have noticed that I have vacillated between talking about power and brightness. This is because I am making the assumption that brightness is proportional to the power dissipated by the bulb, which may not be exactly the case.
    If you want to change the resistance of your filament, you may do this by changing the length or thickness of the filament. Much easier than making it out of a different alloy with different resistivity (I'd have thought.) Once you change the shape or size of your filament, the rate at which it radiates energy can change, thence the temperature at which it is in equilibrium for a given power dissipation. Different temperature gives different spectrum and different visual brightness. All a bit messy and probably not essential to your question.
     
  9. Jul 22, 2016 #8
    Ok. "The actual text is that Tungsten is used in light bulbs. It converts electrical energy to light and thermal energy due to its high resistance". So I interpreted it wrongly as it is suitable because it glows lighter. So the essential point is that it converts electrical energy to thermal energy? But you said 'larger current' will heat...' Do you mean that given its high resistance, more thermal energy is produced when a large current passes through? So that's why both high resistance and large current are needed?
     
    Last edited: Jul 22, 2016
  10. Jul 22, 2016 #9
    Thanks:)
     
  11. Jul 22, 2016 #10

    jim hardy

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    hope above helps

    old jim
     
    Last edited: Jul 22, 2016
  12. Jul 22, 2016 #11
    Thanks a lot! Now I can understand it better:)
     
  13. Jul 22, 2016 #12

    russ_watters

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    I think their use of the word "high" was unnecessary and led you to the wrong interpretation. *Any* resistance results in heat dissipation and as it happens, lower resistance, not higher resistance, results in more heat dissipation(and light) at a fixed voltage.
     
  14. Jul 22, 2016 #13
    A lower resistance means a higher current will flow, but if you consider the whole circuit for a light bulb, the bulb filament should have significantly higher resistance than the wires in the lamp and wall. The light bulb is in series with the wires in the lamp and wall. Since the current through all elements in a series is the same (neglecting capacitance), the elements with higher resistance will dissipate more heat. You want the light bulb to heat up, not your wires in your walls (which can cause fires).

    It all goes back to the basics: ##P = IV##. For Ohmic resistors, you have ##V=IR##, so you can write it as ##P=I^2 R## or ##P=V^2 / R##. Now these equations work for either the single element or for the whole circuit. If you look at just the filament as a single element, you have ##P=V^2/R##, but the ##V## is the voltage across the filament, which is a little less than the voltage from the source, since some voltage is being dropped in the wires in the walls and lamp. Since ##I## is constant across the circuit, you see that the voltage drop is proportional to resistance, and the filament needs to have high resistance compared to the wires.
     
  15. Jul 22, 2016 #14
    Why? Because of the formula: P=I^2R? So under a fixed voltage, I affects more due to the power of 2?
     
  16. Jul 22, 2016 #15

    russ_watters

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    Right. And V=IR (I=V/R) so decreasing R increases I by the same amount and thus increases P.
     
  17. Jul 22, 2016 #16
    Thanks :)
     
  18. Jul 24, 2016 #17

    sophiecentaur

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    You feed it from a high voltage supply. Because it's made of Tungsten, its resistance will increase by ten times from cold to operating temperature (white hot). Despite that, given enough Volts you can achieve a high resistance and a lot of current - compared with the situation at room temperature.
    This is a strange way of looking at the situation, though.
     
  19. Jul 24, 2016 #18

    jim hardy

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    Angela,
    in your course
    have they covered Ohm's Law yet?
    have they introduced what are the basic units yet? Charge, Current, Voltage ?

    We who are decades into the subject tend to forget how awkward it was when we were just starting out.

    Current Voltage and Resistance are related by simple ratios and that's Ohm's law.
    You need them to be intuitive so you can work simple circuits in your head.

    If you need a hand with the basic concepts don't hesitate to ask.
    It's a helpful bunch here but sometimes we're not quite sure what to offer up.

    old jim
     
  20. Jul 25, 2016 #19

    sophiecentaur

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    There is always a caveat to be born in mind. Ohm's Law is not really a 'LAW'. It's more of a 'law', which just describes behaviour under certain circumstances. A light bulb filament does not follow Ohm's Law because the conditions of Ohm's Law are not followed in a glass envelope. Hence the reason for this ever lengthening thread. R is always the ratio of V/I but R is not always constant so it can't always be used on its own to jiggery pokery with V to get an I that you wanted.
    "R=V/I" is not Ohm's law; it just gives the Resistance of a component when it was measured.
    Otoh, Newton's gravitational LAW (under non-relativistic circumstances, at least): g=Gm1m2/r2 is a law because it applies to all combinations of m and r for all (as far as we know) values.
    I think this is important to point out but, of course, I can never hope to change the almost universal inaccurate usage of the term Ohm's Law. :frown:
     
  21. Jul 25, 2016 #20

    jim hardy

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    I'd say it's "inaccurate" only in that we don't finish the explanation by pointing out when R becomes a function of current f(I) , as in a lightbulb or semiconductor, it's no longer just a simple linear ratio . That's algebra not physics. well, maybe some physics and differential equations if we calculate resistance of the element as f(temperature), temperature in turn by heat balance with heat input f(I) .

    Yet an ohm is a volt per amp , simple ratio at the time of measurement
    as you said
    "...it just gives the Resistance of a component when it was measured..". which takes time out of the formula , but R can be a function of I .

    I was under impression OP might be struggling with basics so didn't go there yet. I might be wrong.

    old jim
     
    Last edited: Jul 25, 2016
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