Capacitor connected to AC voltage.

Discussion in 'Homework Help' started by cdummie, Apr 15, 2016.

  1. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    In the picture bellow there's capacitor connected to AC voltage. Phase of voltage is known and it's value is θ. Effective values of charge is Q and period is T (both are considered as known values). Find the expression for current i(t). Directions of current and voltage are shown in the picture.

    kondenzator.png

    Now, this is very easy problem, since I=Q/T it means that i have effective value of current. Now i only need it's phase, since passive element is capacitor it means that current is π/2 "ahead" of voltage so it's phase should be φ=θ+π/2 so

    i(t)=Qsqrt(2)/T cos(ωt+θ+π/2). But if i look at the picture the direction of current opposes direction of voltage, does it means that this is not the correct phase of current, because, as far as i know, it works this way if directions are the same?
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    You may want to look up "passive sign convention". That is a convention where if the current enters the top node of a component that is absorbing energy then the top node is taken to be the most positive voltage node, using conventional current flow.

    So for an element with terminals A and B, if current enters terminal A then terminal A has the most positive voltage so terminal B will be more negative than A, or at least taken to be more negative than A.

    Note that elements that supply energy to the circuit will be opposite to this. The current will flow out of the most positive node.
     
    Last edited: Apr 15, 2016
  3. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    What it means in my case, i mean, i know the definition of KVL is Σ(E,-ZI)=0, here, i have only capacitor and directions of current and voltage are opposite so IZc+U=0 which means that U=-IZc (these are all complex values).
    Also, since i only have capacitor, i know that phase of current is π/2 + phase of voltage. But, i ma still wondering, is this correct if directions are opposite, i believe something should be different, since absolute values are the same even if directions are opposite, it means that something should happen with phases, does this means that phase of current is π/2 ahead of negative value of phase of voltage?
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    Current leads voltage in a capacitor. The relatively simple analysis goes like this...

    We have a voltage E, capacitor C, frequency w, current is:
    E/(1/jwC)=E*jwC

    The phase of the current is:
    Ph=atan2(E*w*C,0)

    Since E is taken to be positive and w and C are positive, this means the phase is:
    Ph=atan2(positivenumber,0)

    which always equals pi/2, so the phase Ph of the current is pi/2.

    With a phase shift of pi/2 we would have in the time domain:
    I=A*sin(w*t+pi/2)
    which is equal to:
    I=A*cos(w*t)

    If you look at a cosine wave with a sine wave you immediately see that the cosine wave is ahead of the sine wave by pi/2. So the current leads the voltage.

    It has to be this way because the capacitor has to accumulate charge in order to show any voltage.
     
  5. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1

    I don't actually understand this completely, how come phase of current is always π/2, shouldn't it always be π/2 ahead of voltage if passive element is capacitor? Does this means that my calculations were correct?
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    I dont understand your question. You are asking why the phase is always pi/2 and then stating that it should be pi/2, so you are making a question out of the same answer you are providing which does not make sense.

    If your calculations came out to pi/2 then you must be correct. Since they appear to come out to pi/2, it appears that you are correct. Why you still ask the question then is a mystery.

    In time it is sin(w*t+pi/2) or for a voltage that is already out of phase by 'ph' then as you noted sin(w*t+ph+pi/2).
    The main reason is because the current is considered to be completely imaginary which puts it into another dimension which is 90 degrees away from the original.

    All this is for a perfectly ideal capacitor of course, because even the smallest resistance in series with it causes a slightly different phase angle:
    atan2((w*C*E)/(w^2*C^2*R^2+1),(w^2*C^2*E*R)/(w^2*C^2*R^2+1))

    which may simplify to:
    atan(1/(w*C*R))

    and here we see that the values can change the phase shift, unlike the previous solution which was always pi/2.
     
  7. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1
    Well, i think you haven't understood what i meant to say, or i expressed it badly, anyway, here is what i meant. You said that it's always π/2, but if phase of voltage is π/3 for example (in this case it's some θ), then phase of current would be π/3 + π/2 which isn't π/2, that's why i said that it's π/2 ahead of voltage phase.

    For the second part, i don't know how the fact that voltage and current are in opposite directions can affect their phase difference, more precisely, my question is:

    If phase difference between current and voltage is π/2 when the are in the same direction does that means that their phase difference is π/2 even when they have opposite directions, like in this example?

    Sorry if i am bothering you too much, i am just trying to understand this.
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hello again,

    Using conventional current flow if we draw the capacitor as you have it in your diagram in the first post, and the current enters the left side and the voltage u is drawn with positive on the left side, then the voltage u is positive which would be zero degrees.
    If the current enters the left node and the voltage u is drawn as positive on the right side, then the voltage u is negative which would mean 180 degrees out of phase with a voltage that was positive, and that has to figure in with the phase shift of the capacitor obviously.

    Phases are measured with respect to some reference phase. Here we use the passive sign convention to establish the reference phase which would be u with positive on the left because that is how the voltage would establish the current as entering the left side of the capacitor. If we flip the voltage REFERENCE but keep the current the same, then the actual voltage must be 180 degrees out of phase to being with, or else the current would not flow like that. Thus,the 180 degree phase shift has to be included, but only if we draw u like that.

    All this changes if we use electron current flow because the direction of the current is taken to be opposite to the conventional current flow assumption. In most of circuit analysis and control theory however conventional current flow is used.

    Does this make more sense now? If not, we can do a couple more examples.
    When we talk about things like this everything has to be well drawn out. It would be good to draw two or more drawings to show each case and label each voltage and current clearly.
     
    cdummie likes this.
  9. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1

    Ok, let's see if i understood this:

    kondenzator2.png
    In this case, current and voltage are in the same direction, which means that in this case φ=π/2+θ (where φ is the phase of current and θ is the phase of voltage).

    Now, the second case, the one in the original problem

    kondenzator.png

    Since voltage is not in the same direction as the current is and since it's 180 degrees out of phase with the voltage in the previous picture it means that φ=-π/2+θ (where φ is the phase of current and θ is the phase of voltage).

    Is this correct?
     
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,959
    1,097
    There is only one a very special "excitation voltage" which "creates" a 90 degree phase shift between voltage and the current in capacitor.
    And this special "voltage" is an sine wave.
    Because for "normal" situation the capacitor current is proportional the rate of voltage change across it (proportional to how quickly the voltage across capacitor is changing). I = C * dV/dt
    This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand if voltage is kept constant no current will flow no matter how large the voltage.
    Likewise if the current through a capacitor is found to be zero, this means that the voltage across it must be constant, not necessarily zero.
    http://forum.allaboutcircuits.com/t...-capacitors-in-ac-circuits.48225/#post-318956
    http://www.allaboutcircuits.com/textbook/direct-current/chpt-13/electric-fields-capacitance/
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    Yes that looks right now, and we could graph the voltages and currents to make this even more clear.
    If you care to do that, you might see it more clearly too.

    I do want to point one thing out here, and that is that when we talk about the phase shift in the current that is relative to the voltage which is normally taken to have a phase of zero, so we dont need to show the other phase you draw as Theta. That would only really be necessary if there was a second voltage source we used as the reference phase. For example if v was at 0 degrees then we might say that u is at a phase of Theta, and then Theta+pi/2 would be the phase of the current in the first case. Without that second v phase however it is sufficient to talk about the voltage phase as being zero degrees, which means the phase of the current is simply pi/2 in the first case.
    If you want to include that anyway i guess it is OK but it's just unnecessary.

    The reason why the second case is different than the first is because we consider the reference phase to be positive to the right, so that the 'real' voltage must be positive to the left in order to get the current to flow in the direction drawn (to the right for conventional current). This means the voltage phase is flipped with respect to what we are calling the reference phase now (positive to the right).

    I also might ask what brought this question up? Maybe there is more we can look at to help in this area.

    Of course i assume all along we are dealing with sinusoidal waves only; both current and voltage.
     
Loading...