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Alternating Current(AC)

  1. Aug 25, 2013 #1
    when an AC passes through a capacitor,if the expression for alternating voltage is V=E×Sinωτ,the expression for current is (I×sin(ωτ+∏/2)).This shows that current leads voltage by a phase ∏/2.Also at τ=0,we have voltage=0 but there is a finite value of current.In absence of voltage at T=0,where from this current comes? please give detailed explanation & please don't consider it to be a home-work question.
     
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  3. Aug 25, 2013 #2

    Andrew Mason

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    Current only requires a voltage if there is resistance. In a superconductor (0 resistance), for example, current flows with no applied voltage. So the question you should be asking is what would stop the current from flowing in this circuit?

    This is an ideal capacitor so there is effectively no resistance. The only thing that will stop the current is a voltage in the opposite direction. That does not occur until ∏/2 after the voltage reaches 0.

    AM
     
  4. Aug 25, 2013 #3

    Dale

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    You also have a 0 voltage and non-0 current at τ=2π/ω. Where does the current come from then?
     
  5. Aug 25, 2013 #4

    ehild

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    AC current flows if the capacitor is connected to some AC source. It is of the sinusoidal form I0sin(wt+pi/2) if some time passed after switching on. Before that, there is a transient current, but that transient period is very short.

    The circuit usually has some other elements, and at the beginning, when the voltage source is connected, the capacitor behaves as a short-circuit. The current is maximum when no charge on the capacitor is present yet. We say that the current leads the voltage. The voltage across it is zero and the current is maximum. The current charges the capacitor, and the voltage increases while the current carries more and more charge on it. There is current till the source voltage differs from the voltage across the capacitor. When the voltage of the capacitor reaches its maximum the current is zero. The source voltage varies periodically with time, increases, then decreases. As the source voltage becomes lower than the voltage of the capacitor, the current changes direction, the capacitor discharges and the voltage decreases across it. As the source voltage changes sign, the capacitor also gets oppositely charged after some time. That is a periodic process, the voltage across the capacitor changes while current flows, and reaches maximum positive or negative value when the current is zero. The voltage across the capacitor varies according to the function UC=U0sin(wt).

    The source voltage is not in phase with the capacitor voltage as there are also resistances in the circuit. If nothing else, it is the internal resistance of the source. The source provides the current and the voltage across the capacitor lags behind that current.

    ehild
     
  6. Aug 25, 2013 #5
    I know that there will be no current in the absence of voltage...but if we put τ=0 in the expression of
    current,we get a definite value but the voltage is zero.what does it physically imply (give physical interpretation)?
     
  7. Aug 25, 2013 #6

    Andrew Mason

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    You explained very well why the current leads the voltage when a potential difference is applied to a capacitor. I am not sure why you added this last paragraph. Addition of resistance would make the phase angle less than π/2 ie. rather than cause lag, it would reduce the phase lag of voltage to current ie the phase angle: [itex]\tan\phi = (X_L - X_C)/R[/itex]

    There is a discontinuity at t=0. You have to look at t>0.

    AM
     
  8. Aug 25, 2013 #7

    ehild

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    I wanted to emphasis the difference between the source voltage and voltage across the capacitor. The source is the cause of the current which sets up the voltage across the capacitor. The phase of current and the voltage across the capacitor differ by pi/2.

    ehild
     
  9. Aug 25, 2013 #8

    Dale

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    This is not true in general.

    Please answer my question to you in post 3. Then, think about what that means, keeping in mind that a sin wave is periodic, both into the future and also into the past.
     
  10. Aug 31, 2013 #9
    I think it is not correct to state that current leads the voltage by a phase of ∏/2.According to me,the answer to my question is that current lags voltage by a phase of 3∏/2.

    current=I Cosωt=I sin(wt+∏/2)=I Sin(wt-3∏/2)

    The last expression shows that current lags voltage(=E Sinωt) by a phase 3∏/2.

    Is the answer correct???
     
  11. Aug 31, 2013 #10
    The source voltage is not in phase with the capacitor voltage as there are also resistances in the circuit. If nothing else, it is the internal resistance of the source. The source provides the current and the voltage across the capacitor lags behind that current.

    ehild[/QUOTE]

    you said that voltage across the capacitor lags behind that current.But my question was regarding the source voltage that lags behind current (according to the equation derived,which is impossible.....)
     
  12. Aug 31, 2013 #11

    vanhees71

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    The current is given by the time derivative of the charge on the capacitor plates, i.e.,
    [tex]i(t)=\dot{Q}(t)=C \dot{U}(t).[/tex]
    Now if
    [tex]U(t)=U_0 \sin(\omega t)[/tex]
    you find
    [tex]i(t)=C U_0 \omega \cos(\omega t)=C U_0 \omega \sin(\omega t+\pi/2).[/tex]
    So the phase of the current is advanced by a phase shift of [itex]\pi/2[/itex] compared to that of the voltage.
     
  13. Aug 31, 2013 #12

    ehild

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    The phase difference between source voltage and current depends on the resistance in the circuit. There is always some resistance also, when you connect a capacitor to a voltage source. If nothing else, the source has some internal resistance.
    When you switch on the AC source to the capacitor, there is a transient current, and after a very short time the current follows the periodicity of the source voltage. We speak about phase difference in that state.

    So assume the emf of the AC source is E (the time dependence is Esin(ωt)). The circuit contains the source, the capacitor C and the resistor R, connected in series. The net impedance is Z=R-j/(ωC), with magnitude √((RωC)2+1)/(ωC) and phase θ=arctan(-1/(RωC)). The current is I=E/Z, which has opposite phase as the impedance: ψ=arctan(1/(RωC)). That is positive, so you can consider that the current leads the source voltage.
    This is valid only when some time has elapsed after connecting the source to the capacitor. Initially, the capacitor is not charged and it behaves like a short - a wire, so the initial current is E/R, in phase with the source voltage. After a while the current follows the periodicity of the source voltage, with the phase difference ψ=arctan(1/(RωC)) - leads the voltage, and the phase difference is close to pi/2. It means that the time dependence of current is I=Iosin(ωt+ψ). But you get the same function if you subtract 2pi from the phase I=Iosin(ωt+ψ-2pi), and then the current lags behind the voltage. :tongue: That the current leads the voltage does not mean that causality is hurt.

    ehild
     
  14. Aug 31, 2013 #13

    Andrew Mason

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    To add to what others have said, current leading voltage does not imply anything that is impossible (eg. effect occurring before cause).

    If you apply a sinusoidally varying voltage to a capacitor (with negligible resistance) and plot the applied voltage as a function of time on the same graph as the current as a function of time (after initial transient effects have disappeared), the voltage applied will have peaks that occur 1/4 of a complete cycle or at a time Δt = π/2ω AFTER the current peaks. So we say that current leads voltage in a capacitor by π/2 radians or 90 degrees.

    AM
     
  15. Sep 1, 2013 #14

    vanhees71

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    Of course there is no violation of the causality principle. What I've calculated is the stationary state which is valid only after the transient state is damped out due to some resistance, as has been emphasized in other postings.

    To solve for the problem with resistance you have to solve a differential equation
    [tex]R i + Q/C=U \; \Rightarrow \; R \dot{Q} + Q/C=U.[/tex]
    with some initial condition, e.g., [itex]Q(t=0)=0[/itex].

    The general solution is given by the complete solution of the homogeneous eqaution, i.e., for [itex]U=0[/itex] and one special solution of the inhomogeneous equation. The homogeneous equation is an exponential decay. After a long time, only the special solution of the inhomogeneous equation survives, and that's the stationary state one usually looks at in circuit theory.
     
  16. Sep 3, 2013 #15
    Causality is not violated. But why would one think otherwise? Current and voltage are inclusive and mutual, neither is generally the cause of the other. Working on the lab bench with a signal generator and scope can easily affirm that the current in the cap reaches its full value immediately and voltage takes time to catch up with the input source. If you observe transients and steady state I & V on the cap, you will see that I leads V by almost 90 deg, and that I does not lag V by 270 deg or whatever.

    This issue was thoroughly examined in the 19th century, and all experiments since have affirmed the published findings. I leads V by 90 deg in an ideal cap, slightly less than 90 deg in a real world cap due to inclusions such as esr, and esl, dielectric absorption, etc. Any good undergrad text on circuit theory solves the question in detail. I suggest referring to such a text, then we can clarify if needed. Best regards.

    Claude
     
  17. Sep 3, 2013 #16

    vanhees71

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    For causality reasons you must have
    [tex]G(t,t')=0 \quad \text{for} \quad t'>t.[/tex]
    Thus we make the ansatz
    [tex]G(t,t')=\Theta(t-t') g(t-t').[/tex]
    That [itex]G[/itex] should be a function of [itex]t-t'[/itex] is clear from the translation invariance of the equation of the Green's function.

    Now we have
    [tex]\partial_t G(t,t')=g(t-t') \partial_t \Theta(t-t')+\Theta(t-t') \partial_t g(t-t')=g(0) \delta(t-t')+\Theta(t-t') \partial_t g(t-t').[/tex]
    Plugging this into the equation gives
    [tex]R g(0) \delta(t-t')+\Theta(t-t') [R \partial_t g(t-t') + g(t-t')/C]=\delta(t-t').[/tex]
    Thus we must have
    [tex]g(0)=1/R[/tex]
    and
    [tex]R \partial_t g(t-t')+g(t-t')/C=0.[/tex]
    This ODE is easily solved by
    [tex]g(t-t')=\frac{1}{R} \exp \left (-\frac{t-t'}{RC} \right ).[/tex]
    The general solution of the original equation thus is
    [tex]Q(t)=Q_0 + \frac{1}{R} \exp \left (-\frac{t}{RC} \right ) \int_0^t \mathrm{d} t' \exp \left (\frac{t'}{RC} \right ) U(t').[/tex]
    Now let's see what happens for the example of a harmonic voltage, switched on at [itex]t=0[/itex] with the capacitor having charge [itex]Q_0[/itex] at [itex]t=0[/itex]. Then we have to set
    [tex]U(t)=U_0 \cos(\omega t)=U_0 \mathrm{Re} \exp(\mathrm{i} \omega t),[/tex]
    because then we can easily evaluate the integral to get
    [tex]Q(t)=Q_0 + \frac{1}{R} U_0 \exp \left (-\frac{t}{RC} \right ) \text{Re} \int_0^t \mathrm{d} t' \exp \left (\frac{t'}{RC}+\mathrm{i} \omega t' \right ).[/tex]
    Evaluating the integral gives finally (vor [itex]t>0[/itex])
    [tex]Q[t]=Q_0+\frac{C U_0}{1+(\omega R C)^2} \left [-\exp \left(-\frac{t}{RC} \right )+\cos(\omega t)+R \omega C \sin(\omega t) \right ].[/tex]
    This shows that the transient state has decayed within a time scale of about [itex]t_{\text{relax}}=R C[/itex]. For [itex]t \gg t_{\text{relax}}[/itex] you come to a stationary state, i.e., a current that is harmonic with the same frequency as the external harmonic voltage.

    The current is found by taking the time derivative of [itex]Q[/itex]:
    [tex]i(t)=\dot{Q}(t).[/tex]
    You can easily figure out the phase shift in the stationary state using the complex form by calculating the argument of the complex number in front of [itex]\exp(\mathrm{i} \omega t)[/itex].
     
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