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Simple Curcuit Question (Theory)

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

    a.) The circuit above shows four identical bulbs connected to an ideal battery. Rank the bulbs in order from brightest to dimmest

    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.
  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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