2006-12-09 11:18:59 · 11 answers · asked by annabella 1 in Science & Mathematics ➔ Engineering
well....depends on what kind of light :)
incandescent: no - the circuit is broken - no electrical path.
fluorescent, compact fluorescent, metal halide, sodium vapor: the ballast will draw power even if the lamp is non-functional. *typically* there is a safety that detect this condition and stops consuming power after several seconds.
2006-12-09 11:28:29 · answer #1 · answered by mikesheppard 4 · 2⤊ 0⤋
nope. but if you microwave it for 5 seconds, you may get something interesting.
as an off note, for ohm's law, E = I * R, if R was zero, that would mean you have a short. Since it takes infinite effort (it's impossible) for electricity to go through a broken light bulb, R = infinity. The current flowing through is I = 0 obviously.
Put another way, conductance, which is 1/R is equal to zero. So you have,
E * (1/R) = I
120V * 0 = 0 current, 0 electricity.
2006-12-09 11:23:01 · answer #2 · answered by Nick C 4 · 0⤊ 0⤋
Ohms law, E = I x R I is the number of interest. It shows the load, expressed in amps. Switch on, E = 120 volts available. R = zero, since the bulb is burned out. Remember algebra? Isolate and solve for I.
I = 120/0 No I. No current draw. Work done, zero.
If you prefer power in watts. P = I x E
P = 0 x 120
P = 0
Zero watts, no electricity used.
Although I knew a woman that insisted it did. Anytime she found an empty light socket she would put a bulb in it, and if it lit, she would immediately turn off the switch, knowing that electricity leaked out of an empty socket and she was saving money.
2006-12-09 11:35:57 · answer #3 · answered by oklatom 7 · 0⤊ 2⤋
no
the circuit is broken by the burned (or broken) filament and a broken circuit will garauntee that it will naot be on the electric bill
if that bulb goes out then the circuit breaks because the filament burned but i will suggest turning it off still (you don't want to find out what happens when you complete a circuit with you fingers)
all that light socket is is just a space that needs conductivity
(until you flip the switch to it)
2006-12-10 03:00:05 · answer #4 · answered by macgyver 1 · 0⤊ 0⤋
No, a burned out bulb is an open in the circuit.
2006-12-09 11:22:07 · answer #5 · answered by Clycs 4 · 1⤊ 0⤋
Take the light bulb out and stick your finger in and test it yourself.
2006-12-09 11:45:56 · answer #6 · answered by thkiabdks 2 · 0⤊ 0⤋
No.
Doug
2006-12-09 11:38:18 · answer #7 · answered by doug_donaghue 7 · 0⤊ 0⤋
no it doesnt,though the elecricty still goes though it,afterall the lightbulb doesnt use elictricity if its burned out,it might pass through it but thats all
2006-12-09 11:22:54 · answer #8 · answered by Jaden B 3 · 0⤊ 4⤋
nah
2006-12-09 11:26:08 · answer #9 · answered by Anonymous · 0⤊ 0⤋
Nope, cicrcuit is dead.
2006-12-09 11:49:38 · answer #10 · answered by burton235 3 · 0⤊ 0⤋