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A 6v battery charges up a 470 microfarad capacitor in 1 second

  1. Mar 18, 2012 #1
    Hello to all, this may seem a very simple question but i cant work it out and have no clue where to start.
    My question is: A 6v battery charges up a 470 microfarad capacitor in 1 second, what is the final charge of the capacitor? and what is the current while the capacitor is charging?

    i know capacitance is charge/potential or c = Q/V
    i know charge is Q = CV

    Is the final charge in coulombs, 470 microfarad x voltage? or should i convert microfarad to farad?
    current i know means charge flowing per second but i do not know how to apply this to the question.
    The final part of the question asks for the value of the resistor, is this just ohms law? R = V/I?
    Any help would be much appreciated.
     
  2. jcsd
  3. Mar 18, 2012 #2

    gneill

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    Re: Capacitance

    Convert to farads if you want the result in coulombs. Otherwise you'll get your result in microcoulombs :smile:

    The problem you've described appears to be an RC series circuit where a capacitor is being charged through some resistance. You will have to review your notes on RC circuits and in particular the equations that describe charging and discharging a capacitor (time constants, etc.).
     
  4. Mar 19, 2012 #3
    Re: Capacitance

    I have reviewed my notes and there is little there to help me. I think time constant = resistance x capacitance? but i dont know the resistance.
    looking at the question is 1 second the time constant? this doesnt sound right but I literally have no idea.
     
  5. Mar 19, 2012 #4
    Re: Capacitance

    You have to remember that [itex]i=\frac{\delta q}{\delta t}[/itex] - so the current as you've put already is the rate of change of charge with respect to time... does this help?

    Is this all the information that is given in the question?
     
  6. Mar 19, 2012 #5

    gneill

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    Re: Capacitance

    Yes, the time constant would be ##\tau = RC##. 1 second was given as the time for the capacitor to charge. A basic "rule of thumb" you can use for these exponential decays is to assume that all the action is essentially over after five time constants have elapsed :wink:

    What's the equation that describes the current with respect to time for the circuit?
     
  7. Mar 19, 2012 #6
    Re: Capacitance

    Im sorry but neither reply is making much sense to me.
    gneill - do you mean after 5 constants the capacitor would be fully charged?
    As for the equation i have tried to find this but alot of what i see i dont understand and have never seen before.
     
  8. Mar 19, 2012 #7
    Re: Capacitance

    After 5 constants the capacitor would be fully charged? - Yes, practically

    I think the equation gneill means is [itex]i=i_{o}\exp^{(\frac{-t}{\tau})}[/itex] where t is the time, ... have you seen this? [itex]\tau[/itex] =RC and [itex]i_{o}[/itex] is the time at t=0 - I take it you've seen this before?
     
  9. Mar 19, 2012 #8

    gneill

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    Re: Capacitance

    After 5 time constants the charge on the capacitor will reach about 99.3% of its final value. Close enough to "done" for engineering purposes :smile:

    Your notes should contain formulas involving ##e^{-\frac{t}{\tau}}##. This is the basic "decaying exponential" that governs the way voltages and currents will evolve in a circuit containing resistance and capacitance (or resistance and inductance).

    If you can determine the time constant and hence resistor value, then you should be able to determine the initial current (based upon the characteristics of a capacitor at t=0) and to what value the current should eventually head as time goes on, and then select the appropriate equation to represent the current over time.
     
  10. Mar 19, 2012 #9
    Re: Capacitance

    Thanks for the replies, the reason im struggling because this is not even something i have been taught in class but it has come up in my coursework. I have no notes to call upon.

    This equation does e stand for eulars constant? minus t which is 0 divided by T which is 1 second?
     
  11. Mar 19, 2012 #10

    gneill

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    Re: Capacitance

    $$e^{\left(-\frac{t}{\tau}\right)}$$
    e is Euler's constant, e = 2.718...
    ##\tau## is the time constant
    t is time

    You can also write this as

    $$exp(-\frac{t}{\tau})$$
     
  12. Mar 19, 2012 #11
    Re: Capacitance

    thanks for the help so far but im still none the wiser. Is there anyway that you can help me further without providing the answer?
     
  13. Mar 19, 2012 #12

    gneill

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    Re: Capacitance

    How about you first describe, in your own words, how you think the current will behave in the circuit from the moment that the battery is first connected?
     
  14. Mar 19, 2012 #13
    Re: Capacitance

    There will be a current flow until the capacitor has reached its full charge, at this point it will act as an open circuit - there will be no electron flow.
    current flow should decrease over time?
     
  15. Mar 19, 2012 #14

    gneill

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    Re: Capacitance

    Yes, that is the general picture :smile:

    So, the current must start out with some initial value and then taper off as the capacitor charges and the potential difference across it reaches its maximum value-- the same as that of the battery which is driving that current. The current will have the form:
    $$ i(t) = i_o e^{-t/\tau}$$
    The trick, then, is to determine values for io and ##\tau##.

    If you happened to know the values of all the components in the circuit, how would you determine the initial current?
     
  16. Mar 19, 2012 #15
    Re: Capacitance

    if i knew all the values i would use I = V/R
     
  17. Mar 19, 2012 #16

    gneill

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    Re: Capacitance

    Right. So it would appear that the only required information you are missing is a value for R (since you are given the battery voltage V).

    Go back to the post I made about determining ##\tau## from the time to completion of charging. You should be able to find suitable values for ##\tau## and R.
     
  18. Mar 19, 2012 #17
    Re: Capacitance

    The way I interpret the question is that the capacitor reaches the battery voltage in one second and that it is not a basic RC circuit but it is a circuit which charges at constant current.
     
  19. Mar 19, 2012 #18

    gneill

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    Re: Capacitance

    What sort of circuit components might that comprise for an introductory electronics course? How would a student answer the question, "what is the current while the capacitor is charging?"
     
  20. Mar 19, 2012 #19
    Re: Capacitance

    Firstly my interpretation of the question is due to the way the question is worded for example with the wording "what is the current" the implication is(at least to me)that the current is steady.In the UK it is an A level physics experiment to charge a capacitor at constant current where the student adjusts a variable resistor.The current cannot be held very steady,it is a crude experiment but it demonstrates a principle.Students also investigate exponential charging/discharging.The current is,of course,the rate of flow of charge(Q/t)
    I may well be wrong in how I interpret the question but I suspect that the OP has summarised the question and may have left out some important details.
     
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