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RC circuit graph analysis

  1. Oct 1, 2009 #1
    Three common light bulbs and two identical capacitors are connected to a battery as shown in the figure below. The switch has been open for a long time. At t = 0 the switch is closed. The light bulbs are all identical and the battery has a value V = 6 V. (The brightness of a light bulb depends on the power dissipated in the bulb - the more power dissipated, the brighter the bulb. Identical light bulbs have the same resistance.)

    http://i662.photobucket.com/albums/uu347/TwinGemini14/1CIRCUIT.gif

    -----
    1) Which graph best represents the brightness of light bulb 1 as a function of time (on the graph, a value of zero represents the bulb not being lit up):

    http://i662.photobucket.com/albums/uu347/TwinGemini14/1a.gif

    ::
    I thought the answer should be A. There is an immediate gain in V because is it directly in series with the battery, but will gradually lose brightness as the other capacitors charge.
    -----

    2) Which graph best represents the brightness of light bulb 2 as a function of time (on the graph, a value of zero represents the bulb not being lit up):

    http://i662.photobucket.com/albums/uu347/TwinGemini14/2a.gif

    ::
    I thought the answer should be A. Since initially, the bulb lights, but once C2 gets charged, no current will flow. So it will eventually fade out.
    -----

    3) Which graph best represents the brightness of light bulb 3 as a function of time (on the graph, a value of zero represents the bulb not being lit up):

    http://i662.photobucket.com/albums/uu347/TwinGemini14/3a.gif

    ::
    I thought the answer should be A again. Immediately, the current must split paths in the circuit. Once the capacitors charge, all the current will now flow through bulb 3. So it gets brighter with time.
    -----

    4) After a long time, compare the charge on C1 and C2:

    A) Q1 > Q2
    B) Q1 = Q2
    C) Q1 < Q2
    ::
    I would say that the answer is B. Since they have the same V and C, and Q=CV then they should be the same charge.
    -----

    5) Which of the six statements are correct statements about this circuit?

    Statement 1: Since voltage is proportional to resistance, when the current through C1 is a
    maximum the voltage across C1 is a maximum.
    Statement 2: Immediately after the switch is closed, neither capacitor has any charge
    (since charge on a capacitor does not change instantaneously) therefore the
    voltage drops across both of the capacitors is zero.
    Statement 3: The values of the capacitors (assuming they are non-zero and finite) do not
    affect the final brightness of the light bulbs.
    Statement 4: Since capacitors have two plates separated by air (or another non-conducting
    material), no current can ever flow through light bulb 2.
    Statement 5: Since this circuit also has capacitors (not just resistors), Kirchoff's Voltage
    and Current Laws are no longer applicable.
    Statement 6: After a long period of time, the capacitors both have a maximum charge on
    them and maximum voltage drop across them.

    A) 1, 3, 5
    B) 2, 4, 5
    C) 1, 4, 6
    D) 3, 5, 6
    E) 2, 3, 6
    ::
    I would think that the answer is E. I immediately realized that statement 5 is incorrect. Kirchhoff's voltage and current laws still apply here. This limits it down to C or E. Then I noticed that statement 4 is incorrect. Bulb 2 will still have a current through it. (Besides, then all options to question 2 above would all be invalid). Therefore, E is left.
    ------
    6) After a long time, the switch is opened. Immediately after the switch is opened: (Note immediately after means there has been time for the currents to change, but the charges on the capacitors are the same as just before the switch was opened)

    A)none of the light bulbs are lit
    B) light bulbs 2 and 3 are lit, but light bulb 1 is not lit
    C) only light bulb 3 is lit
    ::
    The answer must be B. Light bulb 1 is the only one not connected to a path containing a capacitor. The other two will still be lit, but will gradually fade out as the capacitors discharge.
    -------------------------------------
    I KNOW IT'S A LOT TO ASK, BUT CAN SOMEBODY PLEASE LOOK OVER THESE ANSWERS? THANKS GUYS. I REALLY APPRECIATE IT.
     
  2. jcsd
  3. Oct 2, 2009 #2

    Redbelly98

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    1 and 4 look good.
    I disagree with 2 and 3.
    I haven't really looked at 5 and 6.

    Hint for 2 and 3: Immediately after the switch is closed, what is the voltage across C1 and C2?

    That will help you figure out the voltages and currents of B2 and B3 immediately after the switch is closed.
     
  4. Oct 3, 2009 #3
    I see.

    2) So the answer will be C, correct? Since there is no potential difference across that capacitor right when the switch is closed, it will begin a 0. Then it will grow and once the capacitor is filled, no current will flow anymore so it dims. Is that right?

    3) Would the answer then be C as well? I'm not too sure about this one. Possibly because when the capacitors are empty, the current wants to flow in the direction of least resistence and with time, more current would want to flow to the 3rd bulb as the capacitors gradually fill. I'm not too sure about this one. The logic for choice A seems to make more sense to me here. Can somebody better explain this to me please?

    Thanks, I really appreciate the support.
     
  5. Oct 3, 2009 #4

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, it's C for both. For 3, the capacitor C1 is at 0 volts initially, therefore the parallel bulb B3 is at the same 0V.

    p.s. your answers to 5 and 6 look good as well.

    You're welcome!
     
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