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BIT1749

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BIT1749

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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

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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?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?

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ehild

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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 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

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BIT1749

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current,we get a definite value but the voltage is zero.what does it physically imply (give physical interpretation)?

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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]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.

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

current,we get a definite value but the voltage is zero.what does it physically imply (give physical interpretation)?

AM

- #7

ehild

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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]

AM

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

ehild

- #8

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This is not true in general.I know that there will be no current in the absence of voltage..

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.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)?

- #9

BIT1749

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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?

- #10

BIT1749

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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...)

- #11

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[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.

- #12

ehild

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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...)

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)

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

- #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).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...)

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

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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.

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cabraham

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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

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[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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