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Can't understand voltage equation for capacitor

Discussion in 'General Electronics Discussion' started by boogyman19946, Apr 4, 2012.

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  1. boogyman19946

    boogyman19946

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    May 2, 2011
    In a book I'm reading (and kind of trying to employ as I go), I'm finding a certain concept concerning capacitors hard for me to figure out.

    The book presents a circuit that comprises of two components: a capacitor and a resistor in a series loop (that is, they are connected to each other with both ends). I figure the capacitor starts out fully charged.

    The book states that using capacitor rules, the following can be derived:

    C(dV/dt) = I = -V/R

    where C is the capacitance of the cap, dV/dt is the change in voltage, I is the current, V is the voltage, and R is the resistance of the capacitor.

    So the I = -V/R equations looks a little like Ohm's Law, but all calculations afterwards rely on the fact the ratio V/R is negative, whereas Ohm's Law is not. I suspect that the book is trying to maintain the right "direction" of current so when a voltage is reduced from a charged cap, it's going to provide "negative voltage." But the circuit makes no mention of any voltage being taken away. It doesn't even say the cap is charged. It just presents the circuits, doesn't give any starting conditions, writes out some formula and kind of relies on the reader to either take a leap of faith or go on a forum to ask the pros ^.^

    That being said, my question is mainly this: How come I = -V/R and not I = V/R ?
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    11,524
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    Nov 17, 2011
    You're on the right track towards the answer.
    Normally you would label the voltages and currents in a circuit by arrows indicating positive voltage and positive current. As it happens, during operation of a circuit these can change direction (e.g. a cpacitor can be charged by a positive current, but discharged by a negative current through the same wire). Or you may not know beforehand (that is befor you analyzed the circuit) which direction is positive, so you place the arrow in an arbitrary direction.
    In both cases you wouldn't want to redraw the circuit diagram each time a current or voltage changes direction. So you do the necessary calculations taking the initial arrows as positive. Now, if a voltage or current changes direction, it just becomes negative.

    Reconsider the example of a capacitor charged and later discharged through a resistor. Assume the plus pole of a volatge source is connected to the resistor, ressitor to capacitor, the other end of the capacitor to the - pole of the voltage source.
    You start by pointing the current arrow from the resistor to the capacitor. Thus a current flowing from the voltage source through the resistor into the capacitor is positive.
    Now assume the voltage source is set to 0 V (effectively a short circuit). Now the charge flows from the capacitor to the voltage source. The current arrow still points from the resistor to the capacitor, but the direction of the current is reversed. Thus the current is now negative.

    Its like your bank account:
    Money you earn is accounted positive, money you spend is accounted negative, although there is no such thing as negaive money.
    Hey, i like that comparison :))

    There is a tutorial here http://www.electronics-tutorials.ws/dccircuits/dcp_1.html that may help you along.

    Harald
     
  3. Laplace

    Laplace

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    Apr 4, 2010
    The resistance R is discharging the capacitor C, so the voltage V across the capacitor is decreasing to zero. Therefore dV/dt must have a negative value.

    The equation is also related to the definition of capacitance:

    C=Q/V = dQ/dV = Idt/dV or C(dV/dt)=I
     
  4. boogyman19946

    boogyman19946

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    May 2, 2011
    I think I'm starting to get the picture. Later sections of the book that talk about charging a capacitor write the equation out normally so I think I'm getting on the right track. Although, I still wish there was a little footnote that mentioned that little detail ^.^

    @ Laplace, but if dV/dt is negative, then shouldn't the equation be

    C(dV/dt) = -I = -V/R

    I mean, I know what a derivative is and realize that a decreasing voltage means dV/dt < 0 but if I = V/R uses negative current (as per C * dV/dt where dV/dt is negative), the formula should stay the same, only the values you plug in should be negative; otherwise it would mean that negative current still presents positive voltage.
     
    Last edited: Apr 4, 2012
  5. boogyman19946

    boogyman19946

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    May 2, 2011
    Woot, I think I get what's going on here!

    Ok, for all this time I was assuming that V is the voltage drop across the resistor because R is involved in the equation, but V is actually the voltage across the capacitor!

    Now, in a circuit with only a resistor and a charged capacitor, using Kirchhoff's voltage laws we know that the Vr = -Vc because Vr + Vc = 0. That means that the voltage given across the capacitor is the negative voltage across the resistor! If there was a voltage, like Harald said, the equation would spawn

    (E - Vr) = Vc but there is no voltage E so that means E is 0 which gives

    0 - Vr = Vc
    -Vr = Vc
    Vr = -Vc

    If we use the resistor to define current we get

    V/R = I

    but if we substitute the capacitor voltage for V we get

    -V/R = I

    :D Well then... that makes matters a whole lot clearer I think! Can anyone confirm I'm still sane with this idea ? XD
     
  6. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Now consider how those same equations for discharging a capacitor apply to charging the capacitor. Also realize that this discussion only looks at instantaneous time. When looking over some duration of time, the value of V and I are changing so that one must resort to using calculus.
     
  7. boogyman19946

    boogyman19946

    38
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    May 2, 2011
    If I'm getting this right, then the equations themselves don't change. What changes is the results given an input, so when a capacitor is generating a positive current, it means that the voltage that's being applied is rising and cramming charge onto one plate while it is leaving from the other. When the voltage is lowered, some of the charge leaves from one plate and is being drawn back onto the other generating a negative current. That's why when we derive Q = CV in terms of time and q=-q on both plates it has the same effect as if charge was flowing through that point, giving charge/time which is current, and so
    dQ/dt = d(CV)/dt
    I = C(dV/dt)

    I also have another question regarding "storing charge" in a capacitor. I mean, if for any voltage V, the charges on both plates are q = -q then has the capacitor really stored any charge? I mean, whatever charge enters onto one plate, it leaves out the other, so does it make sense to say that a capacitor stores a charge?
     
    Last edited: Apr 4, 2012
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