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DC circuit question

  1. Aug 3, 2004 #1
    First one light bulb is connected to a battery. Then two are connected in series to
    the same battery.
    Here's the picture:

    A) When two are connected, the battery puts out
    a) less current
    b) more current
    c) less voltage
    d) the same current

    B) Which arrangement puts out the most light?
    a) The single bulb
    b) The two bulbs
    c) Both arrangements put out the same total amount of light
    I put (a) because [itex]V=IR_eq[/itex],
    so[itex]I_1 = \frac {V}{R} [/itex]
    and [itex]I_2 = \frac {V}{2R} [/itex]
    Therefore, the second circuit has less current.
    However, I heard because it's a serial circuit, so current should not change.
    But I don't know how to prove that..

    I got (a)
    From [itex]P = I^2 R [/itex]
    I got [itex]P_1 = \frac {V^2}{R} [/itex]
    and [itex]P_2 = \frac {V^2}{2R} [/itex]
    so the first circuit it brighter.
    But if the current is the same for both circuit, wouldn't the total light for the second circuit = the first circuit?
    I'm kind of confused now.... :confused:
    Could anyone help me out?
    Thank you very much!
  2. jcsd
  3. Aug 3, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You are correct in both cases.

    When you put two bulbs in series, you are doubling the total resistance as compared to a single bulb. This cuts the current in half (as Ohm's law indicates), and thus cuts the power in half also.

    - Warren
  4. Aug 3, 2004 #3
    I see..
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