Parallel and Series Circuit

Discussion in 'The Projects Forum' started by Lightfire, Nov 20, 2010.

  1. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Hello,

    I am currently working on my own project which the power source is only a battery. Anyway, here we go.

    I made a parallel circuit.

    Bulb-2 (which is both 12.0 volts)
    Battery (12.0 volts) (wet cell) I admit that its battery of vehicle...

    Then when I turn off (or remove) the other one bulb, the other one bulb is going brightly, and when I turn on (or put) the other one bulb, the other one bulb is going looks like not bright anymore...

    Is there any way to make it balance which both bulb is going brightly. I mean if I put a one other bulb, the other bulb will not be affected...

    Thanks! Sorry for my bad English and bad Explanation!!! Hope you understand!
     
  2. TeslaTux

    New Member

    Nov 20, 2010
    2
    1
    Hey Lightfire,

    If I understand this correctly you have two lightbulbs connected in parallel to a 12volt battery and when you pull one of them out the other lightbulb glows brighter than it did with the other one plugged in. There are a few things that could be going on and my best guess is that there is not enough current to make them both run brightly.

    However, In a lot of automotive lightbulbs there are two filaments a bright one and a dimmer one simply put. If they are both connected to say the dimmer one and they dim by a lot I would go back to I do not believe the battery can output enough current to completely light them.

    I hope I did not confuse you, If I did I'll try my best to clarify whatever you don't understand.

    Respectfully,
    TeslaTux.
     
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  3. retched

    AAC Fanatic!

    Dec 5, 2009
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    There is no regulation in your circuit.

    One bulb is "regulating" the other.

    If you had a current regulator for each bulb they would stay at an even brightness regardless of the other bulbs status.

    When you pull a bulb, the voltage in the circuit increases. This is because a light bulb is essentially a resistor. There is a voltage drop across each bulb, leaving less voltage for the other bulb.

    Pull one bulb and you increase the voltage. More voltage to a resistor, the more current. More current, more light.

    Remember Ohms law Wattage = Amperage x Voltage

    By removing a resistor(bulb) you remove a voltage drop.
     
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  4. TeslaTux

    New Member

    Nov 20, 2010
    2
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    Given this is a parallel circuit wouldn't the current be the only thing to drop while voltage stays constant?

    Respectfully,
    TeslaTux
     
  5. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    It sounds like the bulbs are high power/very bright?

    The bulbs themselves may not be perfectly matched, but when both are connected, the battery may be in a low state of charge so that when a high drain of current (to light the bulbs) is applied, the voltage sags.
     
  6. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
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    Hello,

    Thanks for help!!!

    My battery is 12.0. And my bulbs (2 pieces) is 12.0 volts too at the same time. I connected them as parallel circuit. Then, when I remove one bulb, the another one is going more brightly than earlier...

    Is there any way to make them balance. I mean if I will add more bulbs, the bright of current bulb will not be affected...

    Anyway, isn't it,,, 12.0 divided by 2 = 6 volts. So means, that, both bulbs are only consuming 6 volts instead of 12 volts as they are two pieces>>>>




    THANKS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

    If you may notice, the "CIRCUIT 1" is more brighter. However, the "CIRCUIT 2",,, isn't...
     
    Last edited: Nov 21, 2010
  7. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Do you have a DMM (Multimeter)?

    What is the terminal voltage of the battery? Is the battery fully charged?

    Then (without power), what is the resistance of bulb 1 and of bulb 2, individually?

    The way you show them hooked up in the drawing, they should both be getting 12V. If the brightness is going down when both are turned on, it would seem the battery is sagging. Do the wires or the battery itself get warm?
     
  8. windoze killa

    AAC Fanatic!

    Feb 23, 2006
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    No is the simple answer. This would be true if you put the bulbs in series. When they are in parallel then they will always have the same voltage across them. In this case it will be 12V.

    If for example you have a 50W bulb @ 12V it will draw ≈4.2A to give full brightness. If you now put another bulb in parallel with it as you diagram shows you now have 100W @ 12V which means you must be able to supply a total of 8.4A to obtain full brightness.

    Now to the possible reasons you may not be getting full brightness when you add the secind bulb.

    1. Battery is going flat.
    2. Battery is too small and can't supply enough current.
    3. The wires you are using to connect it all up are too thin and can't suuply enough current. The resistance of the wire means you aren't actually supplying the full 12V to the bulbs.
    4. The joints to the bulbs aren't very good. Make sure the connections are good and solid.

    Of course these are only guesses. Tell us the size of the battery and the type of bulbs you are using and we maybe able to assist a bit more.
     
  9. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Hello,

    We did project in our school earlier. I asked my teacher why does the one bulb is going brighter if I will pull out the another one. She said that that's why it's like that because the batteries are supplying volts to a two bulbs so means that she kinda like to say that the batteries are being fraction or divided depends on the number of the bulbs or even device. Is that true???

    P.S. I will take a picture of my project but I think, I can only uploaded it tomorrow or next next day???

    Thanks!
     
  10. retched

    AAC Fanatic!

    Dec 5, 2009
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    I read "series" in the title and started spouting off at the keyboardin my first reply ;)

    now:

    The battery acts as a supply and a resistor.

    Regulation at the bulb would help the problem, however, the parallel configuration allows the bulbs to pull directly from the battery, individually.


    The internal resistance of the battery changes as the draw increases. SO, when you removed the second bulb, the draw decreased and the resistance decreased, Increasing the supply available to the remaining bulb.
     
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  11. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Thanks retched, I understand now because of you. :p

    But is there any way to make it balance. I mean nothing will change on the bulb if I will add a new one or remove the existing one. Just the same bright.
     
  12. wayneh

    Expert

    Sep 9, 2010
    12,155
    3,061
    The only ways to hold them steady would be to 1) Use a more steady source, for example several batteries in parallel so that the small incremental power drawn by an extra bulb is not noticeable. 2) Use bulbs that draws less power, for instance an LED with a current limiting resistor. 3) Regulate the existing bulbs down to such a low draw that it doesn't matter whether one or both are connected.

    Option 3 would give very dim bulbs and might not even be possible. One way to play around with this approach would be to put bulbs in serial, for passive regulation. For instance, 3 of the bulbs in serial would draw much less current than one, and they'd all be dim. Adding another string of 3 bulbs MIGHT not dim the first very much, although I suspect you'd still notice a change.

    Active regulation could solve this; for instance if you used a voltage regulator to feed your bulbs 9v instead of whatever the battery is supplying. As long as the drain on the battery did not pull the voltage below ~11v, the regulator would hold the voltage across the bulbs at a steady level.

    The current required by your bulbs would make active regulation more than a simple project.
     
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  13. windoze killa

    AAC Fanatic!

    Feb 23, 2006
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    Or find a battery with a much lower internal resistance so the perceived change in brightness would be much less. Make sure the battery can still supply the current required to power both bulbs.
     
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  14. retched

    AAC Fanatic!

    Dec 5, 2009
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    The thing is, incandescent bulbs change their resistance as they heat.

    If you were to measure the resistance of one of those bulbs while removed from the circuit and cold, you would get a pretty low reading. This low startup resistance is what causes a big draw when the bulb is first turned on.

    As the bulb heats up, the resistance increases humongously (is that a word?). The increased resistance basically regulates itself.

    It will hit equilibrium and regulate itself. If there is not enough current available from the source, the bulb never reaches proper operating resistance. A dimmer ads the additional resistance when you wish to intentionally dim a bulb. In your case, you only have the battery internal resistance and the bulbs resistance.

    The dual bulbs cause an increased draw from the battery, increasing battery resistance, working like a dimmer switch.

    If you had a battery with less internal resistance, the lamp would not dim as much. At least not until the battery ran down.

    The more drained a battery, the higher the internal resistance.

    Self-regulation is why you can plug a bulb directly into the wall and it does not just blow.

    It heats and increases the resistance until it is happy. This is what a regulator would do.

    A stand alone bulb does what it wants. A bulb connected to a regulator does what the regulator wants.
     
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  15. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Now, I'm pretty sure that I understand now... But anyway, I tried it on the electricity and no change happen. (I use parallel circuit)... I observed the other one bulb while putting up a new one, and yes, nothing change happen. just still the same. How to do that with batteries???

    OFFTOPIC: (I did it because I think no need to create more thread just for this question, I guess. :p)

    1. What is the name of clip that are inserting in a battery (vehicle/motor battery). It's looks like clip but I want to clarify.....

    2. Is there any thing that can remove the rubber from the wire instead of using scissors. But I want no copper (I mean the bronze, I guess the conductor) will remove because if I use scissor, frequently there are some copper cut...
    Any help would really be appreciated.. :))))))))))))))))))))))

    Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
     
  16. retched

    AAC Fanatic!

    Dec 5, 2009
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    Wire strippers. Buy yourself a decent pair of wire strippers. They will remove the insulation from the wire without damaging the conductor.
    http://en.wikipedia.org/wiki/Wire_stripper

    Just be sure the wire gauge you are working with is accepted by the strippers you buy.

    Also, the clips are usually 'spade connectors' for smaller lead-acid batteries.

    The post type used on most cars and trucks, are typically SAE posts.

    Here is info on different connector types on those types of batteries:
    http://en.wikipedia.org/wiki/Battery_terminal
     
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  17. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
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    It can be cut correctly. I mean if I want to cut the rubber of wire just 1 inches only, it is okay? Also, may I ask if this item is use in manufacture/company that about system like cellphone?

    Thanks!
     
  18. retched

    AAC Fanatic!

    Dec 5, 2009
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    Wire strippers are used in EVERY industry that involves wire. Most cellphones use circuit boards only and there is VERY LITTLE wire. But If you see insulated wire, it has to be stripped. (In most cases)

    Here is a link to a set:
    http://www.grainger.com/Grainger/KLEIN-TOOLS-Wire-StripperCutter-4A854?Pid=search

    "Automatic" wire strippers are nice. I own a pair or two. They cut and remove the insulation in one squeeze. neat.
     
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