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Circuit with 1 capacitor

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Two identical bulbs (bulb A and bulb B), a capacitor, and a switch are connected to a battery as shown at right. The switch has been open for a long time. The battery is ideal and has a voltage (Vbat) of 4.0 volts.

    The switch is now closed.


    Just after the switch is closed, rank the magnitudes of the four voltages Vbat, VA, VB, and Vcap.

    A) Vbat=VA=VB= Vcap
    B) Vbat=VA=VB>Vcap
    C) Vbat=VA=Vcap>VB
    D) Vbat>VA=VB>Vcap
    E) VA=VB>Vcap>Vbat

    After the switch has been closed for a long time, rank the magnitudes of the currents ibat, iA, and iB. (Assume that the battery is ideal.)

    A) ibat = iA = iB
    B) ibat > iA = iB
    C) ibat = iA > iB
    D) iA > iB > ibat

    Just after the switch is opened, the voltage across bulb A (in V) is:

    2. Relevant equations
    V=IR; C=Q/V; Kirchhoff Loop Rules

    3. The attempt at a solution
    For the first, I know that in parallel, voltages will be equal. The capacitance is at zero at this moment, thus less than.
    Second, after a long time, the circuit will act as a loop w/o capacitor since the capacitor will now be fully charged.
    I bolded my answers. Wanted a check to see if my thought processes were indeed correct. If not, I would love to learn from my mistake. Thank you!
  2. jcsd
  3. Feb 18, 2013 #2


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    Science Advisor
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    Gold Member

    All your bold answers look ok to me.


    You mean the voltage on the capacitor is zero. The capacitance of the capacitor is unchanged.

    I wouldn't use the words "w/o capacitor" (because of question 3). I think it's better to say...

    After the switch has been closed for a long time the capacitor becomes fully charged and Ib falls to zero. Hence Ia>Ib.
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