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Conceptual question: light bulbs in simple DC circuits

  1. Oct 9, 2008 #1
    I have two quick questions about how light bulbs (or any resistor) react in (DC) circuits, as my book's not being very helpful.

    A) If a light bulb is removed from its socket, ceteris paribus, does current stop flowing through the socket apparatus entirely, and if it precedes all other elements in a circuit, will it therefore cut current to the rest of the circuit? I'm pretty sure that the potentials will all still be there, correct? I mean, that's why you don't stick your finger in the socket when the switch is still on--you don't want to complete the circuit, right? :)

    B) Two identical bulbs are connected in series to a DC emf. If a single earth ground connection is placed between the two bulbs, why are they not affected? I think I'm misunderstanding how ground wires work. I understand from the facts I'm given that the ground wire can't suck up all of the electrons because some are still obviously reaching the wire, but why is this? Is there not a junction made here with one path leading toward zero potential, which is the preferred state? It's very confusing because it's not an electrostatic situation.

    Thanks for any answers to this stuff. :)
  2. jcsd
  3. Oct 9, 2008 #2


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    Homework Helper

    Welcome to PF.

    A filament is a near short, as the light power is given by I2R, so for low R you get loads of I. Removing it from the circuit results in an open circuit across the socket unless there are other paths through other elements.

    When you have two series bulbs across a battery and short the middle, then you have supplied an alternate path for current to flow to and from ground. Moreover you have established a reference for the voltage source. So long as there is potential across the bulb, as there will be by placing the ground reference midway between the + and - potential posts of the source, there will be current flowing through both as before. The effect will be however that the negative terminal will read as -V/2 and the Positive will read V/2. Whereas if you had attached ground to the negative terminal of the battery then you would have V - V/2 across the first bulb and V/2 across the second.
  4. Oct 9, 2008 #3
    Just a small note.
    I think you mean that "is a near short" when is cold. When the bulb is ON the resistance is not so small. It's 144 Ohm for a 60W bulb and more than 200 Ohm for a 60 W bulb.
    The current is less than 1 Amp. (0.5 A for 60W).
    Only when you turn the bulb ON you have a large current, for a fraction of second.

    Of course, it has no consequence for the following analysis.
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