# Bulb, light vs. burnt out

**URL:** <https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575>\
**Category:** Factual Questions\
**Created:** [April 28, 2003, 2:11am UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575 "2003-04-28T02:11:45Z")\
**Posts on this page:** 7\
**Page:** 1

<div class="post-metadata">

**Author:** ![bbs2k](https://avatars.discourse-cdn.com/v4/letter/b/e495f1/32.png) [@bbs2k](https://boards.straightdope.com/u/bbs2k)\
**Post date:** [April 28, 2003, 2:11am UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575/1 "2003-04-28T02:11:45Z")

</div>

This is pure curiousity.

Turning on a light (and keeping it on for that matter) takes up a certain amount of electricity right?

All right, the question: Would a burnt out bulb use up the same amount of power when turned on even though it doesn’t work? Or does the electricity just stop because it has nowhere to go, thereby using no electricity.

Thanks

---

<div class="post-metadata">

**Author:** ![Mr.Blue\_Sky](https://avatars.discourse-cdn.com/v4/letter/m/d26b3c/32.png) [@Mr.Blue\_Sky](https://boards.straightdope.com/u/Mr.Blue_Sky)\
**Post date:** [April 28, 2003, 2:20am UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575/2 "2003-04-28T02:20:22Z")

</div>

The use of electricity stops because the circuit cannot be completed.

---

<div class="post-metadata">

**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\
**Post date:** [April 28, 2003, 2:28am UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575/3 "2003-04-28T02:28:11Z")

</div>

Correct. Technically, there is a _very_ small current draw due to leakage through the gas in the bulb, but it’s measurable in nanoamperes and may as well be ignored.

---

<div class="post-metadata">

**Author:** ![Casey1505](https://avatars.discourse-cdn.com/v4/letter/c/dc4da7/32.png) [@Casey1505](https://boards.straightdope.com/u/Casey1505)\
**Post date:** [April 28, 2003, 2:53am UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575/4 "2003-04-28T02:53:31Z")

</div>

[Great minds post alike!](http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=180221) sort of… 😉

---

<div class="post-metadata">

**Author:** ![Una\_Persson](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/una_persson/32/346_2.png) [@Una\_Persson](https://boards.straightdope.com/u/Una_Persson)\
**Post date:** [April 28, 2003, 12:46pm UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575/5 "2003-04-28T12:46:00Z")

</div>

Also, in _very rare_ circumstances, the bulb can go “out” due to a high-resistance short via the metal and insulator. But for all practical purposes when the (incandescent) light is not shining, there isn’t any real current flowing.

---

<div class="post-metadata">

**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\
**Post date:** [April 28, 2003, 1:16pm UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575/6 "2003-04-28T13:16:01Z")

</div>

I think you need to explain that a bit better, **Anthracite**.

Your statement implies that the bulb goes out _because_ of the high resistance short.

Do you really mean that a high resistance short can remain _after_ the filament fails (presumably because of tungsten deposition on the central glass insulator)?

And **Q.E.D.** , leakage through the bulb should be negligible not only in absolute terms, but also relative to the dielectric losses in the cabling.

---

<div class="post-metadata">

**Author:** ![Una\_Persson](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/una_persson/32/346_2.png) [@Una\_Persson](https://boards.straightdope.com/u/Una_Persson)\
**Post date:** [April 28, 2003, 1:49pm UTC](https://boards.straightdope.com/t/bulb-light-vs-burnt-out/171575/7 "2003-04-28T13:49:16Z")

</div>

> [@](#):
>
> \*Originally posted by Desmostylus \*  
> \*\*I think you need to explain that a bit better, **Anthracite**.
> 
> Your statement implies that the bulb goes out _because_ of the high resistance short.
> 
> Do you really mean that a high resistance short can remain _after_ the filament fails (presumably because of tungsten deposition on the central glass insulator)?  
> \*\*

I meant that I have seen 10,000+ ohm or so shorts in the base of the bulb, before it gets to the filament, in a really crappy, heavily corroded bulb.
