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Simple Curcuit Question (Theory)

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img153.imageshack.us/img153/8941/physicscircuits1ms3.th.png [Broken]
    a.) The circuit above shows four identical bulbs connected to an ideal battery. Rank the bulbs in order from brightest to dimmest

    http://img509.imageshack.us/img509/4655/physics3ya0.th.png [Broken]
    b.) Suppose that switch has been added to the circuit as shown. The switch is initially closed. When the switch is opened, will the current through bulb A increase, decrease, or remain the same?

    2. Relevant equations

    Inet = 0 @ a junction (Kirchhoff's Junction rule)

    3. The attempt at a solution
    a.) A = D > B = C
    For a, Bulbs A and D have the same current going through both of them since they are in series. Therefore they are equally bright. For the parallel network between bulbs A and D, the current going through the current equals that of the current going through bulbs A and D. The current is split between both bulbs B and C then (identical, so same resistance) and they are half as bright as bulbs A and D

    b.) Here is where I'm having trouble. I'm confused about the placement of the switch and the fact does current flow through bulb C still? If the switch were placed before bulb C , then the resistance in the circuit would increase since there is one less bulb in parallel right, meaning less current and a decrease in current through bulb A. But with the placement of the switch after bulb C, I thought, by Kirchhoff's junction rule that the current entering a junction must equal the current leaving or flowing out. Wouldn't bulb B now have more current flowing through it while Bulb A remains unchanged? Any help is appreciated.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 9, 2008 #2
    a) is OK
    b) do you mean it matters if the switch is above or below bulb C? In both cases, there can be no current going through bulb C if the switch is open.
    Kirchhof's junction rule tells you that I_A_closed = I_B_closed + I_C_closed
    and I_A_open = I_B_open
    I don't think you can draw any conclusions from that.
    (I_A_open is the current through lamp a if the switch is open)
  4. Mar 9, 2008 #3
    If no current is going through bulb C then does that mean that overall resistance in the circuit is increased since their resistances now add (bulbs A,B, and D are in series now)?
    Last edited: Mar 10, 2008
  5. Mar 10, 2008 #4
    I think you mean increased
  6. Mar 10, 2008 #5
    Woops... yeah meant increased! So if the equivalent resistance increases, then the total current in the circuit decreases and thus current through each bulb decreases?
    Last edited: Mar 10, 2008
  7. Mar 10, 2008 #6
    The current through B doesn't decrease because it now gets all the current that flows through the circuit.
  8. Mar 11, 2008 #7
    Ok gotcha. So the current through Bulb B increases since it now gets the current that flows through the circuit but this current is decreased for A and D as the equivalent resistance is now increased because all 3 bulbs are in series. So the total current in the circuit decreases, yet this is more than the current that bulb B had going through it before the switch was opened?
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