Homework problem HELP!

Discussion in 'Homework Help' started by viksanity, Oct 7, 2013.

  1. viksanity

    Thread Starter New Member

    Oct 7, 2013
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    The switch in the circuit shown (see attached pic) has been open for a long time, and is closed at T = 0.

    a) What is the voltage difference between node A and node B when T<0? What is the charge stored on the capacitor? What is the current provided by the battery?

    b) What is the voltage difference, stored charge, and current at T = 0?

    c) What is the voltage difference, stored charge, and current at T --> ∞ 0?

    THANK YOU!
     
    Last edited: Oct 7, 2013
  2. viksanity

    Thread Starter New Member

    Oct 7, 2013
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    Here is an isolated pic of the circuit.
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
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    What is, What is, What is ?

    This is homework help

    and what is your atttempt at these problems?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    How about a pic of YOUR attempt to work YOUR homework problem?
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I will take a swing at part a.

    Voltage difference between A and B is 5 volts.
    Charge in capacitor is 0.
    Current provided by battery is 0.
     
  6. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Please do not give the OP answers. We should only point them in the right direction, THEY need to do the work.

    The point of the homework forum is to help TEACH the students, not give them the answers.
     
  7. shteii01

    AAC Fanatic!

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    I know, that is why I only did part a. I find part b and c far more interesting.
     
  8. DerStrom8

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    You still gave him answers, which we don't do on this forum.
     
  9. shteii01

    AAC Fanatic!

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    Are you saying my answers are correct?
    :D
     
  10. DerStrom8

    Well-Known Member

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    Are you saying they are not? If that's the case, why'd you give them at all?

    I didn't check if they were right or wrong, I was simply speaking about the principle. We do not give the students answers, only guidance in finding them for themselves.
     
  11. WBahn

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    Mar 31, 2012
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    Strongly second DerStrom8's remarks and reinforce them with the fairly solid expectation that the OP make SOME effort to work the problem themselves before we provide much help -- if for no other reason than that we need that to provide a good stepping stone.

    If you want to post hints, then go for it. One way is to work the problem yourself and then ask the OP questions that mirror the thought process that you went through to solve it.

    Having said all that, I looked at the problem and don't like it. There is an assumption that has to be made that is not valid for an ideal capacitor in order to get the answer they are obviously looking for (based on their hint for part b) for part a.
     
  12. viksanity

    Thread Starter New Member

    Oct 7, 2013
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    Hey, sorry guys! New to the forum. My attempt was this:

    Voltage difference = 0, since no current is flowing in the open circuit, so point A and point B would have a voltage of 0
    Stored charge = 0 since there is no current flowing cross capacitor
    Current = 0 since open circuit
     
  13. viksanity

    Thread Starter New Member

    Oct 7, 2013
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    Hey, sorry guys! New to the forum. My attempt was this:

    Voltage difference = 0, since no current is flowing in the open circuit, so point A and point B would have a voltage of 0
    Stored charge = 0 since there is no current flowing cross capacitor
    Current = 0 since open circuit
     
  14. viksanity

    Thread Starter New Member

    Oct 7, 2013
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    b) What is the voltage difference, stored charge, and current at T = 0?

    Voltage diff = Abs [
    (I*R at point B) - (I*R at point A)] = | 0 - (0.1)*50 | = 5V

    Stored charge = 0, since no time as elapsed for charge to be stored

    Current =
    V/R = 5V/50 ohms = 0.1 Amps

    c) What is the voltage difference, stored charge, and current at T --> ∞ 0?

    Voltage diff = Abs [(I*R at point B) - (I*R at point A)] = | 0 - (0.1)*50 | = 5V

    Stored charge = 49.999nano-coulombs

    Current =
    V/R = 5V/50 ohms = 0.1 Amps
     
  15. WBahn

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    Mar 31, 2012
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    No problem -- live and learn. As long as you come back and "make amends" all is forgiven immediately -- and it's a shame how few do that.

    Your reasoning on the first two is very faulty. Let's examine that.

    If you are saying that the voltage from point A to point B would be zero because no voltage is flowing, then would you also say that the voltage between the terminals of a battery is 0V when it is not connected to anything or that the voltage at the wall socket is 0V when nothing is plugged in?

    You are falling into the fallacy that everything somehow obey's Ohm's law. But the only things that obey Ohm's law are the things that obey Ohm's law. Most things don't. Let's call the node at the bottom right of the circuit Node C.

    Q1) What is the voltage at Node A relative to Node C (i.e., across the battery)?

    Q2) What is the voltage at Node C relative to Node B (i.e., across the resistor)?

    Hint: For one of the questions above, the component in question is ohmic (i.e., obey's Ohm's law) and the other is not.

    Your answer to the second question suffers the same fallacy as the first -- a capacitor is not an ohmic component, so Ohm's law simply doesn't apply to it. If I charge up a capacitor and sit it on a shelf, there will be no current flowing in it and, as long as there isn't, it will hold the charge indefinitely if it is treated as an ideal capacitor (which the probably the case for the capacitor in your problem). Real capacitors have an effective parallel resistance which slowly bleeds the charge off over time. Most capacitors will lose a good fraction of their charge through self-discharge effects in a matter of tens of minutes, but others can hold it much longer. I don't know what the record is.

    So while you might argue that the capacitor in this problem has been disconnected for so long that it has surely bled down to zero charge, you cannot make that claim based on the fact that it has zero current in it.

    Your third answer is correct and the reasoning is valid.
     
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  16. WBahn

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    Mar 31, 2012
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    Please refrain from using formatting in your posts unless it is really important to get the point across, particularly in posts that are likely to be quoted in responses. It's a hassle for those responding.

    Where does this "equation" come from?

    What is the basis for claiming that the voltage difference is the absolute value of anything? The voltage difference between the positive terminal of your car battery and the negative terminal is NOT the same as the voltage difference between the negativer terminal and the positive terminal. Just fail to take that into account the next time you install a battery or jump start a car and you will learn that lesson all too well.

    What does it mean to use "I*R at point A". What I do you use? What R do you use? What voltage does this give you?

    Ohm's law is very, very, very specific. V=IR means "The voltage ACROSS a resistor is equal to the produce of the current THROUGH THAT resistor times the resistance of THAT resistor (with proper care take for polarity according to the passive sign convention)." You can't just grab any voltage -- it must be the voltage difference across THAT resistor. You can't just grab any current -- it must be the current through THAT resistor. You can't just grab any resistance -- it must the the resistor through which the current is passing and across which the voltage is taken.
     
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  17. ninjaman

    Member

    May 18, 2013
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    guidance? that's an interesting word!
    berate, chide, scold. now these are words your looking for. mer her her herrrrrr
    can I join your club now?

    no, seriously. I come on here for exactly what your talking about. a point in the right direction. an indirect point. a slightly confusing, "this isn't helping" point in a direction slightly off that which I require. but a point none the less
     
  18. WBahn

    Moderator

    Mar 31, 2012
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    And your "point"? :rolleyes:
     
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  19. viksanity

    Thread Starter New Member

    Oct 7, 2013
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    Thanks for the reply. But Im not sure how to answer part B to your question. I get your point across the battery, so A would be 5V correct? I need some more help with the B
     
  20. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    For Part B.
    You have just closed the switch. The capacitor is charging, which makes it a short circuit. So you have 5 V from battery and a voltage drop across resistor. To me, Voltage from A to B=5 volts + IR. If I stick a voltmeter from A to B, I will see voltage from the battery plus voltage across resistor.
     
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