brightness of bulb

Voltage and current.
With larger voltage, more and more etectrical energy is pushed into the bulb, the wire in the bulb will be heat up, when the heat increases, the wire lights up and becomes brighter and brighter. And there is the same theory to current. Current is the amount of electricity and voltage is the electricity push.
So in conclusion, when the voltage and current increases, the brightneess of light bulbs will be greater.

And also resistances will affect the brightness of bulbs. ( More resistance the less brightness of the bulbs)

I have just learnt to this topic, cheers!
圖片參考：http://l.yimg.com/f/i/tw/ugc/rte/smiley_1.gif

The Major Reason To Affect The Brightness f A Bulb Is Current~

But Why Some People Say Is Voltage?

It Is Because If The Resistance Remain Unchanges,,

The Current And Voltage Is Proportional~

Voltage Is The Energy,,

But Current Is The Speed Of Energy,,

If The Speed Of Energy Is Increase(Amperage High),,

Than The Bulb Will Bright,,

But If The Speed Of Energy Is Decrease(Amperage Low),,

That Mean The Bulb Not Bright Any More,,

So..

If R Remaain Unchange,,

V=A (Proportional),,

The Power Also Affect The Brightness Of A Bulb As Well

It depends on the condition that you are comparing the brightness.
For a given bulb, the larger the current that passes through the bulb, the brighter is the bulb.[Note: since voltage and current are proportional for a given bulb, larger current also indicates a larger voltage across the bulb]

For the same voltage applied to bulbs of different resistances, the one with lower resistane gives more brightness.
For the same current flowing through bulbs of different resistances, the one with larger resistance gives more brightness.
You try to think, from physics principles, why these are so.

As all we know light can produce heat and heat is a forms of energy therefore more energy/power the brighter the light!!
through eqn P=RI^2 and V= IR where P= power R=resistance I= current , V=volt
from the eqn V=IR shows that if there is a increases in (I )will gives a decrease in R (V/I=R), the same if there is a increase in R will gives a decrease in I (v/I=R). and from eqn P=RI^2 shows inorder to increase the power a increases of I and R is needed and the only way to increases the both with affacting each other is to increase the voltage!