Which makes a bulb glows brighter?

[Drakkith]

Thx. I think my question was incorrect. Since V is fixed and we hv put a bulb in the circuit, the current should be fixed. There is no way to increase the current without changing the voltage, right?

You can change the resistance of the circuit. Increasing or decreasing the temperature of part of the circuit is one way to do this.

 

[sophiecentaur]

and Ohm's Law is the very basis of circuit analysis.

Hmmm. It isn't much use assuming Ohm's law applies to a diode is it - despite the fact that it will have a Resistance because it will have a V and an I? That will really mess up your analysis.

Let's go elsewhere then. Hooke's Law applies well to a metal spring but we know it doesn't work for a rubber band or a plastic bag. Would you say that the ratio for a rubber band (= force/extension), under a particular load "is Hooke's Law"? No; stiffness is the word that's used for it for it. I think that years of using a term inappropriately has actually done a disservice to a many students of EE. I'd hate to suggest that Mechanical Engineers are better disciplined than Electrical Engineers . . . . . . .Especially as I am / was a chartered Electrical Engineer.

 

[jim hardy]

Hmmm. It isn't much use assuming Ohm's law applies to a diode is it - despite the fact that it will have a Resistance because it will have a V and an I? That will really mess up your analysis.

Nope. You'll have to remember it's only that many ohms at that many amps. That's Ohm's Ratio (to use my term) .

But you have to remember that anyway.

 

[sophiecentaur]

You are incorrigible dear boy.
We're just going to have to disagree on this one.

 

[jim hardy]

We're just going to have to disagree on this one.

I thought we were agreeing. But I'm mildly dyslexic.

Some difference in our respective concept of "law" ?

 

[Angela Liang]

You can change the resistance of the circuit. Increasing or decreasing the temperature of part of the circuit is one way to do this.

How to increase or decrease the temperature?

 

[sophiecentaur]

How to increase or decrease the temperature?

You could increase the temperature of a light bulb filament by shining a high power light source on it. You could decrease the temperature by placing the bulb in a blackened container that's held a a very low temperature - say in liquid nitrogen and that would reduce the surface temperature by about 200°C.

 

[Angela Liang]

You could increase the temperature of a light bulb filament by shining a high power light source on it. You could decrease the temperature by placing the bulb in a blackened container that's held a a very low temperature - say in liquid nitrogen and that would reduce the surface temperature by about 200°C.

Wow. Thanks :)

 

[houlahound]

So power in a light is proportional to diameter of filament assuming constant length tungsten filament?

 

[Merlin3189]

So power in a light is proportional to diameter of filament assuming constant length tungsten filament?

I'm not sure which element of the thread this has come from - Ohm's ratio, temperature dependence, power in a resistor? But whichever it is, perhaps we need to be clear about the details of the context, else statements like this are hard to evaluate.
Assuming for the moment, it is talking about the light FROM an incandescent bulb, plugged into mains - approximating a constant voltage source, then I would think that:
increasing the diameter would lower the resistance in inverse proportion to the cross sectional area, so inverse proportion to the square of the diameter;
reducing the resistance increases the power dissipation according to V²/R, so inversely proportional to the resistance or proportional to the are or the square of the diameter;
greater power dissipation would increase the temperature until the radiated power equalled the new electrical power;
since the radiated power is also proportional to the surface area, which is proportional to the square of the diameter, so there is no need for any temperature rise to increase the radiated power.

Net result, increased light power is proportional to square of diameter for a constant length filament, so long as the the resistance of the filament remains much larger than the resistance of the power supply. (If you carry on increasing the diameter of your filament, eventually nearly all the power is dissipated in the supply and very little in the filament - or massive metal bar, as it might then be called.) The temperature does not change (if the filament is in empty space) so the colour does not change.

 

[sophiecentaur]

So power in a light is proportional to diameter of filament assuming constant length tungsten filament?

That's what the standard formula relating length CSA and Resistivity to Resistance says. Resistivity is temperature dependent. Edit: The Power will not have a simple relationship with Resistance because of this because the temperature changes with Power and the R will change accordingly etc...

 

[Merlin3189]

So power in a light is proportional to diameter of filament assuming constant length tungsten filament?

That's what the standard formula relating length CSA and Resistivity to Resistance says. ...

I hate to be pernickety, but CSA is proportional to D² not D, so I don't think it is what the standard formula says.

Possibly more interestingly your rider, that resistivity varies with temperature, provoked me to think that length and CSA would also vary with temperature. Then the thermal increase in length would be more than compensated by the thermal increase in CSA , leading to a lower resistance with increasing temperature. Unfortunately the effect of thermal expansion is about 1000x less than the increase in resistivity with temperature (for Tungsten), so I can see why no one ever mentions it.

 

[sophiecentaur]

I hate to be pernickety, but CSA is proportional to D² not D, so I don't think it is what the standard formula says.

Possibly more interestingly your rider, that resistivity varies with temperature, provoked me to think that length and CSA would also vary with temperature. Then the thermal increase in length would be more than compensated by the thermal increase in CSA , leading to a lower resistance with increasing temperature. Unfortunately the effect of thermal expansion is about 1000x less than the increase in resistivity with temperature (for Tungsten), so I can see why no one ever mentions it.

The expansion affects the radius which affects CSA and the length. But it wouldn't affect the number of atoms / electrons involved. Gets harder and harder , dunnit?
And then there's the increase in surface area of the whole filament. That will reduce the equilibrium temperature for a given rate of dissipation of energy.

 

[houlahound]

may have missed it, CSA is ?

 

[Drakkith]

may have missed it, CSA is ?

I believe it stands for Cross Sectional Area.

 

[David Lewis]

Ohm derived his law by parallel thought to contemporary experiments with conduction of heat through metal. He was using galvanic cells and wire.

Ohm used a thermocouple as his voltage source. (He realized source impedance of galvanic cells varied as a function of current and state of charge, which obscured the underlying principle he was trying to discover.)

 

[sophiecentaur]

Ohm used a thermocouple as his voltage source. (He realized source impedance of galvanic cells varied as a function of current and state of charge, which obscured the underlying principle he was trying to discover.)

That's a fascinating thing.
Life was tough for experimenters in those days. No radio Shack just down the road. Imagine having to wind your own insulation round wire that the local blacksmith would have probably drawn out for you.

 

[jim hardy]

Ohm used a thermocouple as his voltage source.

I believe you are right and thanks for the correction .

 

[sophiecentaur]

I now realize that Ohm can have had no particular Model for this - no idea of what was actually going on in the substance except the parallel with thermal conduction. The range of conditions that he could use must also have been fairly limited - how many Amps can you get out of a thermocouple (his particular type), I wonder? A bit of luck, you could say, that it later turned out to cover a huge range of current values.

 

[David Lewis]

Ohm inserted a separate term in his formula to represent source impedance. All test resistances were made up from the same spool of resistance wire, so that the only independent variable (during each run) was the length of the wire:

Current = voltage / (source impedance + wire length)