March 31, 201016 yr So, I need help from anyone who is good with electricity. I need to make a simple circuit powered by a 9V battery that can power 4 very small lightbulbs (49s), and if I disconnect one of the lightbulbs, the rest stay lit. Problem is, it's a series circuit that must be made (all 4 bulbs must be connected by wire), and with series circuits, you can't disconnect one and have the others stay lit. Can anyone tell me how to wire it like a chistmas light thing? I know you need a 3rd wire to complete the circuit if the bulb is out, but how do I do that? ~ 3,072nd to 99 Mining on August 30th, 2009 ~~ 112,084th to 99 Magic on April 16th, 2011 ~~ 131,681st to 99 Crafting on March 29, 2019 ~~ 178,385th to 99 Prayer on April 2, 2019 ~~ 234,921st to 99 Defence on May 9, 2019 ~~ 173,480th to 99 Herblore on June 21, 2019 ~~ 155,160th to 99 Smithing on July 16, 2019 ~
April 2, 201016 yr You could try using a resistor to "short" the bulbs, except that you use one with a greater resistance than the actual bulbs. This way, they only kick in when the bulb dies. [hide=Stats][/hide]"One death is a tragedy. A million deaths is a statistic."-Joseph Stalin
April 8, 201016 yr Resistor is close, but if each of the 4 bulbs will run at 2.25V (1/4 of the 9v), then you could have a 2.7V Zener diode across each, to complete the circuit with a little more votage drop - NB. if the supply voltage exceeded 10.4V, the string of zeners would conduct and overload. Not sure if they are available in 2.4V versions (for less voltage loss on bulb fail, but a supply voltage maximum of only 9.6V).
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