Yes - the bulb's brightness depends on how much current passes through it. Think of the bulb as a resistor. Then it obeys Ohm's law (it doens't strictly speaking because its resistance changes as it heats up), so the power used by the bulb is proportional to the current squared times the bulb's resistance. THis power comes off as light and heat so the more current the more light.
2007-07-17 05:56:01
·
answer #1
·
answered by nyphdinmd 7
·
3⤊
0⤋
If there are no changes to the circuit other than limiting the current, then the bulb will be dimmer. The effective brightness can be changed by 1) changing the voltage level, or 2) limiting the current to the bulb.
This of course means that you have designed the circuit to limit the current and still maintain the same voltage level at the bulb.
One other point to consider in your question is the vise/versa, trying to force extra current in to the bulb won't make it any brighter. You can not exceed the ratings of the bulb, a pesky thing called Ohms Law comes into play.
2007-07-17 06:14:34
·
answer #2
·
answered by MtBikr 7
·
1⤊
0⤋
Yes, current affects the brightness and current is affected by voltage or power factor if an AC circuit containing an inductance is used. For resistive circuits, I = V / R where I is the current, V the voltage and R, the resistance. Power consumed by the bulb is V x I = V^2 / R. So, for the light to become dimmer, we need to reduce the voltage or increase the resistance (night lamps have higher resistance). Or we use a dimmer circuit, which either introduces a resistance in the circuit (not energy efficient) or changes the angle between voltage and the current by using an inductance.
2007-07-17 05:57:00
·
answer #3
·
answered by Swamy 7
·
1⤊
0⤋
Yes. Consider a flashlight with a 2-1.5V batteries in series with the bulb. The voltage across the lightbulb is always 3V. However, the current passing through it will depend on the resistance of the lightbulb filament wire. The power dissipated by the lightbulb is P = I*V. The voltage won't change, but the more current that passes through the lightbulb, the more power it will use, i.e. it will get hotter and brighter. Lower the current, and bulb will use less power.
2007-07-17 06:32:12
·
answer #4
·
answered by Anonymous
·
1⤊
0⤋
Given a standard light bulb, the resistance is fixed. The two variables that may be changed are force (V) and flow (A). Flow is a dependent variable of force. Thus only by changing the input voltage can you change the flow. If you raise the voltage, it will be brighter, and burn out much sooner. If you lower the voltage, you lower the flow, and extend the life of the light at the sacrifice of brightness.
2007-07-17 06:02:11
·
answer #5
·
answered by muddypuppyuk 5
·
1⤊
0⤋
Yes, the electrons are moving slower through the element of the bulb or resistance. If the current increases, the resistance gets higher making the bulb brighter.
2007-07-17 06:08:37
·
answer #6
·
answered by rwd46jr 1
·
0⤊
1⤋
Yes that is definately correct. for more info check out bitesize revision site. it makes everything really simple rather than using technological language!!
2007-07-18 06:42:24
·
answer #7
·
answered by Anonymous
·
0⤊
0⤋