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Capacitor directly connected to a DC battery.

  1. Sep 4, 2012 #1
    HeyoO everyone (^_^) and Thanks for your time.

    1. The problem statement, all variables and given/known data
    This question is about the "intial" point of time i.e. as soon as the key is closed.
    Before the current starts to flow which charges the capacitor till the voltage of capacitor is equal to voltage across the battery.

    A capacitor,C is directly connected to a DC battery of (say) 10V.
    If the plates of the capacitor are taken to be at 0V, then what would be potential of "negative" and "positive" terminal
    of battery relative to the plates?


    2. Relevant equations
    Electrons likes to flow from "Low" to "High" potential.

    Finally, the plate A will be "Positively" charged and plate B will be negatively charged.
    That means Electrons flowed from A -> X -> Positive terminal -> Negative terminal -> Y -> B which created those charges on Plate A and B.

    This implies that Initially (the time at which this question is to be answered) Plate A [ 0V ] was at a lower potential than the positive terminal of the battery....... And......... Plate B [ 0V] was at a higher potential than the negative terminal of the battery.

    A battery is of 10V implies that the potenital difference between the +ve and -ve terminal is of 10V.



    3. The attempt at a solution
    I think that the +ve terminal will be at +5V and -ve terminal will be at -5V.
    That is +"Half of the voltage" on positive terminal and -"half of voltage" on negative terminal.

    [Q.01] Am i right (o_O) ??


    [Q.02] Will other pairs of (+ve Terinal , -ve terminal) work out?? :
    Like (+9V , -1V) or (+4V, -6V).


    [Q.03] The pair (+13V , +3V) will not be correct right ??
     

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  3. Sep 4, 2012 #2

    CWatters

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    Ok I see the switch is in red.

    So the initial conditions are switch open and the capacitor discharged.

    It's necessary to state that the following voltages are measured with respect to the -ve terminal of the battery...

    The A terminal of the capacitor (and point X) is the same voltage as the +ve terminal as the battery eg 10V.

    The B terminal of the capacitor (and point Y) is the same voltage as the A terminal (eg 10V) because the capacitor is discharged.
     
    Last edited: Sep 4, 2012
  4. Sep 4, 2012 #3

    CWatters

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    The way to think of this is that at t=0 when the switch is closed/ON the B terminal of the capacitor will be "pulled down" from 10V to 0V.
     
  5. Sep 4, 2012 #4

    CWatters

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    Consider this circuit.

    The discharged condition is represented by SW1 closed and SW2 open.
    The charged condition is represented by SW1 open and SW2 closed.
     

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