Which of the bulbs glow brightest and least bright?

[Jahnavi]

Homework Statement

Homework Equations

Power of a bulb = I²R = V²/R

The Attempt at a Solution

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From the ratings given on the bulbs using P=V²/R , we can conclude that resistances of the bulbs in decreasing order are R , P , Q i.e R has highest resistance and Q has lowest .

Now brightness is related to power dissipated .

Using P= I²R and the fact that both the current flowing in , as well as resistor of R is highest , so power dissipated in R should be highest .

Now potential difference across P and Q is same and resistance of P is more than that of Q , so using V²/R , power dissipated in P should be lower .

Hence least bright should be P . But R,P is not an option .

What is the mistake ?

 

[cnh1995]

I think your analysis is correct. It should indeed be R, P.

Current through R is the sum of the currents through P and Q, and its resistance is more than the resistances of P and Q. Hence, it should be the brightest. The dimmer one between P and Q is the dimmest bulb.

 

[magoo]

Your first conclusion would be correct if the bulbs all had the same voltage across them, but they don't.

 

[Jahnavi]

Your first conclusion would be correct if the bulbs all had the same voltage across them, but they don't.

Are you suggesting that R is not the brightest ?

 

[CWatters]

If you work out all the resistances, voltages and powers you get

P=2.4W
Q=4W
R=25.6W

So R, P is correct. The book answer is wrong.

 

[Asymptotic]

1. Bulb resistance calculated from rated wattage and voltage yields the hot resistance obtained when the bulbs are operated at rated voltage.
Are any of them operating at rated voltage?
2. What happens to resistance as filament temperature decreases? How does this affect the actual wattage dissipated by each bulb?

 

[Jahnavi]

I think your analysis is correct. It should indeed be R, P.

Current through R is the sum of the currents through P and Q, and its resistance is more than the resistances of P and Q. Hence, it should be the brightest. The dimmer one between P and Q is the dimmest bulb.

Thanks for confirming .

If you work out all the resistances, voltages and powers you get

P=2.4W
Q=4W
R=25.6W

So R, P is correct. The book answer is wrong.

Yes . I get the same result .

Thanks .

 

[cnh1995]

1. Bulb resistance calculated from rated wattage and voltage yields the hot resistance obtained when the bulbs are operated at rated voltage.
Are any of them operating at rated voltage?
2. What happens to resistance as filament temperature decreases? How does this affect the actual wattage dissipated by each bulb?

It is mentioned in the problem that the bulbs have a low thermal resistivity coefficient.
Doesn't it mean that we can assume the resistances to be temperature independent as the variation of resistance with temperature is small?

 

[Asymptotic]

It is mentioned in the problem that the bulbs have a low thermal resistivity coefficient.
Doesn't it mean that we can assume the resistances to be temperature independent as the variation of resistance with temperature is small?

Constantan, for example, has a negligible thermal resistivity coefficient, but isn't a practical material for incandescent filament manufacture. I cannot but concede the point as you are right, and a low thermal resistivity is specified, but I wonder whether this simplification does more harm than good when it comes to understanding the nature of how a filament lamp works.