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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?
"My textbook says a bulb glows brighter due to its high resistance." That's a strange and (misleading) thing for an author to write.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?
But how to achieve the condition where both current and resistance are high?Both. P=IV, so you need both resistance and current.
that is a contradiction ... higher resistance = lower currentBut how to achieve the condition where both current and resistance are high?
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?"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 ..
Thanks:)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 = V^{2}/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 = I^{2}R 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.
hope above helpsSo the essential point is that it converts electrical energy to thermal energy?
Yes ! That's the essence of it.
Resistance is analogous to friction , rub your hands together and feel the heat.
The harder you press the more resistance to motion you'll feel and the more heat you'll make.
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?
Yes ! That's the essence of it. Power is the product of resistance and the square of current
power equals resistance X current^{2}
and that power leaves the filament as radiated heat and light ,
in SI units
power is watts
current is amps
resistance is ohms
So that's why both high resistance and large current are needed?
Yes, with this caveat..
In the grand scheme of things lightbulbs are fairly puny handlers of power
To give you some perspective.......
one amp through 100 ohms would be 100 X one ^{2} = 100 watts, once a very typical bulb for a household room ceiling fixture
an automobile headlamp is around 35 watts low beam, nominal 12 volts & 3 amps; 50 watts high beam
the element in your water heater is probably 2500 watts
and the motor in your vacuum sweeper is perhaps 700 watts.
Thanks a lot! Now I can understand it better:)hope above helps
old jim
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.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?
Why? Because of the formula: P=I^2R? So under a fixed voltage, I affects more due to the power of 2?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.
Right. And V=IR (I=V/R) so decreasing R increases I by the same amount and thus increases P.Why? Because of the formula: P=I^2R? So under a fixed voltage, I affects more due to the power of 2?
Thanks :)Right. And V=IR (I=V/R) so decreasing R increases I by the same amount and thus increases P.
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.But how to achieve the condition where both current and resistance are high?
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.Current Voltage and Resistance are related by simple ratios and that's Ohm's law.
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.
OK. Thanks very much:)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
We have covered Ohm's Law, also only ohmic conductors obey Ohm's Law.OK. Thanks very much:)
However I have a question- it is said that no matter ohmic or non-ohmic conductors, the resistance generally increases with increasing temperature, but I find that according to Ohm's Law, if V is fixed, the larger the current, the lower the resistance( R=V/I). As I believe that larger current means higher temperature, the whole thing becomes contradictory. How is it that possible?OK. Thanks very much:)
My only point is that a simple ratio does not make a LAW. It's just the Definition of Resistance.
You know i have a soapbox about the term "Ground"My only point is that a simple ratio does not make a LAW. It's just the Definition of Resistance.
no, it's a fine point that should be made just as Mr Davis did for us boys back in 1962.Am I really being too picky here? Go on, you can tell me.
Abbé de Condillac adds: "But, after all, the sciences have made progress, because philosophers have applied themselves with more attention to observe, and have communicated to their language that precision and accuracy which they have employed in their observations:
In correcting their language they reason better."