Is it the power,voltage or the current which is reponsible for the glow of the bulb?

 

All three are interrelated.

This sounds like a homework problem. Is it?

 

Is it the power,voltage or the current which is reponsible for the glow of the bulb?

You need heat to make the bulb glow. Heat comes from power. Power = Volts x amps. Amps = volts/resistance.

Rearranged,

power (watts) = voltage x voltage / resistance.

Resistance of a cold bulb filament is low at the start and gets higher as the bulb heats up (almost instantly - in a blink of an eye). Does that answer your question?

 

All three are interrelated.i

This sounds like a homework problem. Is it?

I came across many problems for eg., bulbs having same power rating but different voltages,or differnt power ratings but same voltage and I am really confused which formula to use i.e P=VI or P=(V^2)/R or P=(I^2)R

 

You need heat to make the bulb glow. Heat comes from power. Power = Volts x amps. Amps = volts/resistance.

Rearranged,

power (watts) = voltage x voltage / resistance.

Resistance of a cold bulb filament is low at the start and gets higher as the bulb heats up (almost instantly - in a blink of an eye). Does that answer your question?

So should I look for more resistance to glow more ?

 

I came across many problems for eg., bulbs having same power rating but different voltages,or differnt power ratings but same voltage and I am really confused which formula to use i.e P=VI or P=(V^2)/R or P=(I^2)R

All of the formulas apply (keeping in mind that the resistance of a bulb filament is a strong function of temperature).

So if you have the power rating and the voltage rating, you can find the current and the resistance of that bulb (at that power level).

If I give you two 1 W resistors and tell you that one is a 1 Ω resistor and the other is a 10 Ω resistor, what does that tell you about the maximum voltage and current ratings for those two resistors?

 

I am really confused which formula to use i.e P=VI or P=(V^2)/R or P=(I^2)R

Whichever values you have and the easiest to compute.

 

So should I look for more resistance to glow more ?

If you are forcing the same current through it, yes. But if you are applying the same voltage across it, no.

 

All of for the formulas apply (keeping in mind that the resistance of a bulb filament is a strong function of temperature).

So if you have the power rating and the voltage rating, you can find the current and the resistance of that bulb (at that power level).

If I give you two 1 W resistors and tell you that one is a 1 Ω resistor and the other is a 10 Ω resistor, what does that tell you about the maximum voltage and current ratings for those two resistors?

for one resistor : I=1 A and V=1 V and for the other V= √10 =3.16 V and I=√0.1 =0.316 A.
Does this mean that both will glow equally ( if assume a bulb) since both has same power rating? and if they are connected in a circuit :
In series with a votlage source of 1 v : for first bulb power =1 w and for other 0.1 W .Second bulb has less power but more resistance .will it glow less or more compared to other?

 

for one resistor : I=1 A and V=1 V and for the other V= √10 =3.16 V and I=√0.1 =0.316 A.
Does this mean that both will glow equally ( if assume a bulb) since both has same power rating? and if they are connected in a circuit :
In series with a votlage source of 1 v : for first bulb power =1 w and for other 0.1 W .Second bulb has less power but more resistance .will it glow less or more compared to other?

You can't really relate wattage to "bulb glow" in the first place. The power dumped in a resistor goes into heat and, in the case of a light bulb, some of it goes into light. But most of it goes into heat. So two bulbs with the same rating and being driven at that rating may not put out the same amount of light.

But, for discussion sake if nothing else, it is reasonable to assume that the more power actually being dumped into a light bulb the more light it will give off.

 

I think I got your point. I hope I 'll be able to apply this in problems also.