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series parallel RC circuit help

Discussion in 'Electronics Homework Help' started by bottomsup, Oct 28, 2015.

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  1. bottomsup

    bottomsup

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    Oct 28, 2015
    need a bit of assistance on this problem if you guys wouldn't mind. See attachment
     

    Attached Files:

  2. Colin Mitchell

    Colin Mitchell

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    Remove all the capacitors and you have three resistors in series.
    Add the values together.
    You have 120v and xxx ohms.
    Suppose you have 240 ohms.
    Across each ohm you have 0.5v.
    This means you divide xxx ohms by 120v to get yyy volts across each ohm.
    Now multiply yyy by 46 to get a result
    Multiply yyy by 30 to get a result
    Multiply yyy by 164 to get a result
    This will give you the voltage across the 80u and 20u.
    You can then perform simple calculation to find the value across the 10u
     
    hevans1944 likes this.
  3. ramussons

    ramussons

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    Jun 10, 2014
    A capacitor that's fully charged has Infinite resistance.

    Presume that there are no capacitors, calculate the voltages at the various Nodes. Those Voltages will be the same after "All the Capacitors Have Fully Charged".
     
  4. Ratch

    Ratch

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    The following points can be made.

    1 Capacitors are not ever "charged" (with what?). They are "energized" to a certain energy level (joules or voltage). Saying capacitors are charged is using technical slang that does not literally describe what is happening.

    2), A capacitor does not block current by presenting an infinite resistance to a voltage. It blocks current by presenting a counter voltage. This method of current blockage is not the same as an open circuit or infinite resistance. A capacitor supports charge flow (current) if the voltage changes, whereas no current exists in an open circuit.

    Ratch
     
    Arouse1973 likes this.
  5. dorke

    dorke

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    Jun 20, 2015
    The wording of the question is not correct.
    Instead of "fully charged" it should be in "steady state(DC)",
    or better yet "t(time after switch is closed) --> infinity"
     
  6. Martaine2005

    Martaine2005

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    May 12, 2015
    Ah, but the wording was "fully charged".
    I agree with my limited knowledge that all responses here are correct!".
    The poor OP had a question. The only way to answer that is by using the questions format!.
    If not he/she gets more confused. There is a reason for explaining physics, however, there is also a good reason to keep within the format of the question too.
    After all, how do we know that the teacher hasn't given the correct information but used analogy to keep it simple? Or hundreds of examples so they all understood in their own way!.
    I (wrongly) use fully charged and discharged every day!. But it's easier for people to understand immediately. So I can only imagine a school wants the pupils to get it quickly! Then the student/pupil who initiates further questions has obviously seen the light and gets that there is more to it!
    Which is why schools don't teach physics first. I bet two out of ten get the physics explanation. But an analogy or 6, all students get it in their own way!.
    Now, this is where I get shot down by the OP. He says, "I have studied physics for five years. There is nothing to charge a capacitor! They are energized. And then continues to explain the physics of an inductor!!.
    Wow, did I really type all that?...

    Martin
     
  7. Ratch

    Ratch

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    If you said "fully charged" to me, I would ask you what the caps are charged with. The slang and idiomatic expressions are more confusing than the correct description. It is not a matter of knowing the physics. Describing something correctly avoids confusion.

    Ratch
     
    Martaine2005 likes this.
  8. Martaine2005

    Martaine2005

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    I don't, or never would disagree witth you!.
    But some people simply cannot understand the physics! I am one of them!
    To feed children in a class with physics of a capicitor is obviously good! But the majority wont be head strong to understand the physics! Hence an analogy works! Then the explanation as to why can commence.

    I wrote a whole lot more here. In fact a full A4 page lol. I think the moderators saw my weakness and did their job!!.

    Martin
     
    Last edited: Oct 30, 2015
  9. dorke

    dorke

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    Jun 20, 2015
    Martin,
    It is not the physics you need to understand.
    It is the mathematical behavior of the model of an electronic(electrical)device.
    In this case the intuitive understanding of beginners should be this:

    1. We are talking here of a DC Circuit.
    2. Hence at steady-state(t-->infinity after the switch is closed) we have f=0Hz.
    3. Looking at the capacitor reactance formula for steady-state we have:
    Xc=1/[2*pi*f*c] ,with f=0 that means Xc=infinity (that is what we call DC block).
    see, no need to use "fully charged".
    4.The DC (steady state) current doesn't flow in the caps (DC block,Xc=infinity) ,
    hence you can omit the caps from the circuit,and are left with a resistor divider only.

    BTW,
    this approach is great for understanding intuitively and quickly what type of filter(LP,BP,BS,HP) a circuit with a bunch of Capacitors and Inductors is.
    (XL=2*pi*f*L --> HF block,DC short)
    (Xc=1/[2*pi*f*c] --> DC block,HF short )

    You should try it and see how "enlightening ,quick and simple" it is :cool:
     
    Martaine2005 likes this.
  10. Ratch

    Ratch

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    Mar 10, 2013
    Everyone can understand physics if it is kept simple enough. Analogs are good for illustrating a point, but not for describing how something works. When that happens, the explanation usually has to be unlearned and a correct explanation assimilated.

    Ratch
     
    hevans1944 and Martaine2005 like this.
  11. Laplace

    Laplace

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    Apr 4, 2010
    My physics textbook, "College Physics" by Sears & Zemansky, has this to say: The net charge on the capacitor as a whole is, of course, zero, and "the charge on a capacitor" is understood to mean the charge on either conductor without regard to sign.

    It does not seem too difficult to understand what is meant by physicists charging a capacitor. What are capacitors charged with? Obviously with protons and electrons, the ultimate natural unit of charge. One conductor will have an excess of protons while the other conductor has an excess of electrons. In electronics the term "fully charged" means that the capacitor has been charging in the circuit for a time in excess of six RC time constants. Really, that's what it means.
    Once a capacitor has been fully charged, what remains except leakage current through the dielectric? The dielectric insulator may not have infinite resistance, but it is pretty close. If the voltage changes in a circuit with a fully charged capacitor, then one must wait an additional six time constants to regain the infinite resistance.
     
  12. dorke

    dorke

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    Jun 20, 2015
    Laplace,
    There is a reason this web site is called "Electronicspoint" and not "Physicspoint".;)

    My 2 cents:
    Leave those "Sears & Zemansky"and other Physics text books on the shelf ,
    and move on to the Electronics ones...:rolleyes:

    The classic used to be "Basic Circuit Theory" by Desoer and Kuh.
    It is an engineering grade book and thus University grade math is needed.
    But,
    it will take your knowledge and understanding several order of magnitudes forward :confused:
     
  13. Ratch

    Ratch

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    That is a slang definition. A capacitor is a energy storage device, so its status should be defined with respect to the amount of energy it contains. That's "fully logical". You could say a capacitor is charged with energy, but then you may as well say it is "energized".

    Protons and electrons carry the smallest amount of discrete charge. MKS units of charge are coulombs.

    Who says? Why not 12 time constants. Or infinity? By the way, only electrons move in a conductor, not protons. Protons stay with their atoms, which form the ionic core of the material. Protons are not charge carriers in a conductor like electrons are. Therefore, the number of protons on each plate of a capacitor always remain the same, even if the distribution of electrons varies.

    An electric field, which causes a voltage across the plates.

    A fully energized cap is not the same as a open circuit, even if its dielectric if perfect. The circuit equations differ between the two, and the circuit performance will be different. Treating caps as an open circuit is a mathematical artifice used for calculating the final steady voltage or current. This is applicable to only a single particular circuit status.

    Ratch
     
  14. Ratch

    Ratch

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    Mar 10, 2013
    Not good. Physics is the foundation science of electronics, and cannot be ignored.

    So will lots of other books. No one book covers everything.

    Ratch
     
  15. Laplace

    Laplace

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    Apr 4, 2010
    Webster's New World College Dictionary gives the relevant definition of slang: the specialized vocabulary and idioms of those in the same work: now usually called shoptalk, jargon. I don't seem to have any problem with electronics professionals using specialized technical jargon.
    So we should change our terminology to state how many kilowatt-hours of energy are stored in a charged capacitor? How would that be at all useful? The capacitor is a charge storage device; equal and opposite amounts of charge are stored on each plate. It would make more sense to state the amount of charge stored, but even that would not be very useful because we have no way to measure the amount of charge stored. However, the amount of charge is proportional to the voltage of the charged capacitor, and we can measure the voltage. This is the reason for using the convention that a capacitor is charged to a particular voltage -- it is useful. So there is the question - Why would your proposal result in something that is even more useful? Show us how it would play out in solving circuit problems to specify capacitors energized in kilowatt-hours.
    I seem to have missed the point of this comment. Sorry???
    My statement is true based on the highest possible authority. "Electronics All-in-One For Dummies" by Doug Lowe states: "In theory, the capacitor will never be fully charged because with the passing of each RC time constant the capacitor picks up only a percentage of the remaining available charge. … For all practical purposes, you can consider the capacitor fully charged after five time constants have elapsed."

    Also, "Introductory Circuit Analysis, 3rd Ed" by Robert L Boylestad states (pg. 218) "For all practical purposes, therefore, a capacitor will charge to its final voltage in five time constants."

    I know these references state 5τ while I said 6τ, but I believe in being really sure.
    Even so, it is the protons that provide the positive charge on the positive plate of the capacitor.
    That seems strange. The capacitor is in a circuit with a voltage source charging the capacitor. What role does the voltage source have in causing the voltage across the plates? Does the voltage source cause the electric field? If there were no capacitor connected, but there are two wires connected to the voltage source (open circuit), is there an electric field between the wires?
    But this is a failure to understand. My comment only applied to the fully charged condition of the capacitor, when current flow has ceased and the circuit equations are no longer relevant. If there are two nodes in a circuit with a voltage difference between them yet no current flows at either node, how is that not the definition of an open circuit?
     
  16. Ratch

    Ratch

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    Slang is like a foreign language, or communicating with grunts and gestures. It usually is not descriptive to the layperson who is not knowledgeable about a subject. Those in the know can blow smoke rings at each other to communicate, but slang can be confusing or impede learning for beginners. Example: How many beginners will think that a capacitor has more electrons within it than it had before when they hear a capacitor is "charged"? The correct descriptive language is always understood the right way.

    I never said you have to give the amount of energy contained in a capacitor in energy units. There is no reason you cannot say that a capacitor is energized to a X number of volts or energized to a Y number coulombs of imbalance. I said that energizing is a more descriptive and less confusing word than charging.

    You said that protons and electrons were the ultimate natural unit of charge. I was pointing out that although they are the smallest discrete charge available, we do not use their charge as a standard unit. Their charge is defined in coulombs, a MKS unit.

    The number of time constants needed depends on the application, not the authors of electronics books. Suppose you were de-energizing a one million volt capacitor from a lightning machine. Wouldn't you want to go 12 time constants to get the voltage down to 6 volts, instead of 6 time constants and still have over 2400 volts across the capacitor?

    Yes, that is true, but the positive charge is usually referred as coming from a positive ion, not a proton. Protons are held tightly by the atomic nucleus, and are not free to roam like electrons are.

    The voltage imbalances the charge distribution between the plates. That causes an electric field to form. Anytime you have a electric field, which is a vector field, a scalar voltage field is also defined from a reference point. When a voltage is removed from the capacitor, the different charge distribution between the plates is still in effect, so the voltage remains on the capacitor along with the capacitor energy stored in the electric field..

    Yes it does, by causing the charges between the cap plates to be different. Anytime there is a different charge between two points, there be an electric field.

    Yes, a charged particle like a free electron outside the wire will be attracted to the positive wire and repelled from the negative wire. Ever notice how dust is attracted to voltage sources? That proves an electric field exists. Unlike a capacitor, the field will disappear when the voltage is shut off because the wires have no energy storage capability..

    You misstate the conditions. There are two nodes, but the capacitor is applying an opposite voltage equal to the applied voltage. The net voltage is zero, so no current is present even though a conducting path is in place. This is not the same situation as having an applied voltage with no counter voltage and no conducting path. In both cases, no current exists for that static state, but the circuits are different, and will have different responses to perturbations.

    Ratch
     
    Last edited: Nov 1, 2015
  17. Martaine2005

    Martaine2005

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    Hi Bottomsup, (culos arriba).
    That schematic is worded incorrectly. It says " fully charged" when it should read " energized". Your teacher should know better than to use 'slang' for teaching.
    I took the liberty to edit your schematic.
    Posts #2 and #3 are still appropriate. Although we all knew what it meant.

    caps.png
     
  18. hevans1944

    hevans1944 Hop - AC8NS

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    Capacitors that are "fully charged" (whatever that means) in a steady-state DC circuit are not conducting current. They can therefore be removed from the steady-state analysis as @Colin Mitchell stated in Post #2. Follow his instructions to find the potentials across all the resistors and hence the potentials of the capacitors connected in parallel with those resistances.

    As for energizing a capacitor, this is always done by moving charge from one plate to the other plate. Since the only charged objects that are free to move are electrons, an electron current is required. Current is measured in amperes, defined as the number of coulombs of charge passing a given point in one second. So as you watch the charged electrons moving from one plate to the other plate of the capacitor, leaving behind a positive charge on one plate and accumulating a negative charge on the other plate, I will forgive you if you want to think of this operation as "charging" the capacitor. A similar effect occurs as a result of ion chemistry when a battery is "charged"... er, energized. And a completely different effect occurs when an inductor is energized by passing a steady-state current through it.

    In all these cases (and many others) energy is being accumulated and stored, so it is important to understand the physics of how this occurs if you want to know what is really going on. However, to solve problems in Electricity 101, just follow the advice in Post #2 and Post #3. Make sure you fully understand Ohm's Law and Kirchoff's Laws first.
     
    Martaine2005 likes this.
  19. Martaine2005

    Martaine2005

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    That was my point Hop, he didn't 'think' anything. Just following a homework paper. It is not the fault of the pupil to show the 'actual' paper involved.
    My point was, they used 'fully charged' for that paper!.
    The responses should have been appropriate! Not a lecture on capacitors.
    We all use different terms, depending who we talk to. This is no different.
    And I certainly don't want this to be another transistor war!
    Fully charged, fully energized, reached it's full capacity.. They all mean the same thing..For home work.

    Martin
     
  20. hevans1944

    hevans1944 Hop - AC8NS

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    Yeah, lazy editing strikes again. This type of phrasing is common in physics problem texts. It is used as a clue here to mean the capacitors are not part of the steady-state analysis. Once the student realizes that, the "problem" devolves to finding the voltages developed across three resistors connected in series. I have no idea what the student is expected to learn from all this... maybe that some "problems" are "tricky" to solve?
     
    dorke likes this.
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