# confused about the bulb glow

Thread Starter

#### e-learner

Joined Apr 25, 2015
30
Is it the power,voltage or the current which is reponsible for the glow of the bulb?

#### WBahn

Joined Mar 31, 2012
26,398
All three are interrelated.

This sounds like a homework problem. Is it?

#### GopherT

Joined Nov 23, 2012
8,012
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?

Thread Starter

#### e-learner

Joined Apr 25, 2015
30
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

Thread Starter

#### e-learner

Joined Apr 25, 2015
30
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 ?

#### WBahn

Joined Mar 31, 2012
26,398
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?

#### KeepItSimpleStupid

Joined Mar 4, 2014
4,749
the TS said:
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.

#### WBahn

Joined Mar 31, 2012
26,398
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.

Thread Starter

#### e-learner

Joined Apr 25, 2015
30
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?

#### WBahn

Joined Mar 31, 2012
26,398
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.

Thread Starter

#### e-learner

Joined Apr 25, 2015
30
I think I got your point. I hope I 'll be able to apply this in problems also.