We discussed this manner in terms of inductor, not so long ago. I fully understood from many posts provided why does it lag. I mean, not everything can be fully understood but I got a good intuition about it. Question arose not so long ago, and I couldn't find anything good on the internet. Why the current lead voltage in capacitor? Does this has something to do with dielectric field? Can anybody provide me with some good links addressing this problem, or explain? Thanks.
It can help to think in terms of mechanical analogues. Compare an LC circuit to a mass spring system. velocity is the analogue of current, force the analogue of voltage. The mass corresponds to inductance, you must apply a force to get the massive object in motion just as you must apply a voltage to get current flowing in an inductor. A capacitor's analogue is a spring. You must displace the spring to increase the force and you must displace charge in a capacitor to increase the voltage.
Yes, thank you. I found that. But I want somebody to explain it in terms of physical nature. Like when we discussed that, in inductor, current makes a magnetic field which then induces current in inductor rendering it to lag etc etc.
Hi Bassalisk! (what's the dielectric field? ) Charge = voltage times capacitance (Q = VC). Current (I) = dQ/dt. So I = CdV/dt. If V is a sine curve, then I is the slope of the sine curve, which leads it by 90°. (In an inductor, L is proportional to dI/dt, instead of I being proportional to dV/dt in a capacitor, so L leads I, ie I lags L),
Well in capacitor, you have a dielectric. When u charge the capacitor, dielectric field forms around the dielectric. I am thinking that rearranging atoms(dipoles) in dielectric over and over in AC, has to something to do with it.
Best way is to consider an uncharged cap. A switch is closed & current enters the cap. The current is full value, a constant current source or a constant voltage source plus a resistor. At time t = 0+, the current i is maximum value, if the voltage source value is V, & resistance is R, then i(t=0+) = V/R. But the cap starts uncharged, Q=0, & V=0. As the voltage increases, the current decreases, eventually the cap reaches the source voltage & current tends to zero. In this case, current is maximum while voltage starts at minimum & increases. The current in the cap is said to lead the voltage. Another thought is that current in a cap can change quickly/abruptly but voltage in a cap changes gradually/slowly. Changing current involves little work, but changing voltage requires more work. Hence a current change is easy to do in a cap since work W = C*(V^2)/2, so that little work is done changing current. But changing cap voltage is changing its energy, needing work to be done. To change energy over a time interval, power must be non-zero, i.e. P = dW/dt. But power P = I*V. So, I & V must both be non-zero during the transition so that power is non-zero & work is done. In the ac domain. i = C*dv/dt. Cap voltage changes after charges have been transported to & from the plates. This transport of charge is the current, i = dq/dt, q = charge. So if v is a sine function of time, i is a cosine function of time. In the steady state with ac excitation, i leads v by 90 degrees. Did I help? Claude
YES! I think I understand this. I got the intuition for it. That is very important to me. Now I need a little time and work to understand the equation part. Thank you very much mr. Claude !
Another point. In just considering a capacitor in an unspecified circuit, the current need not lead voltage. For example if you short a charged cap. through a resistor the current and voltage will be proportionate (because that's the property of the resistor). The capacitor's property is that its voltage is proportionate to charge. How that relates to the rate of change of charge i.e. current is a function of other components in the circuit.
So, in a RC circuit(without any source), Charged capacitor discharges, and keeps its phase with resistor?
And if you connect a cap charged to voltage of value V to a voltage source of the same value through a resistor, the voltage stays at V & there is no current at all. Here is a special circumstance where V stays the same & I is zero. Examples like this can be found. Regarding the cap discharge through a resistor, it just happens that the time function involved is exponential in this case. Since i(t) = C*dv(t)/dt, the derivatiove of an exponential is an exponential with the same argument. The "lead" aspect is not apparent here. But as stated above, if we are dealing with any other functions, like trig, pulses, triangular, etc., the leading property is easily seen. With exponentials it isn't visible. Claude
You are again wanting to consider the device on its own. There is only a phase difference when you have a series resistance. From a voltage source the current will be infinite and instant when there's any change of voltage.
Actually. your argument supports my position. In the case of a voltage source & a cap w/ no R, the current goes to infinity instantly while the voltage changes gradually & finite. My point is just that. Current in your case does indeed lead the voltage. I made the point that changing a cap current involves only a small energy since the cap itself & the leads possess a small inductance. Changing current in this small inductance requires a small energy. But the energy stored in the cap is related to V^{2}. It takes a lot of work to change cap voltage. In your case of a voltage source driving a cap w/o R, ideally if L were 0 then I would be infinite. The work done is related to the change in V. I can change by a huge amount w/ no work done. That was my point. As far as phase difference goes, a cap always has another element in the picture. A voltage source connected to a cap via superconducting connections still possesses inductance in the overall circuit. This results in a phase difference. A constant current source, CCS, driving a cap w/ no other elements displays the I leading V property as well. If the CCS is dc (0 freq), the current is a step function, w/ the voltage being a ramp where It = Cv, or v = It/C. I leads V. If the CCS is a cosine, then V is a sine. Again I leads V. Any questions are welcome. Claude
I think the problem here is that you seem to want to discuss the effect of a step change and a sinusoid at the same time. 'Phase' only relates to AC. If you put an AC voltage across a capacitor then the charge in the Capacitor will be VC. This is all sort of degenerating into a mixture of ideal and practical circuit ideas and, Claude, I agree with what you have written butttttt I have a feeling that it can lead to confusion without actually drawing out the circuit and doing the sums, explicitly. Getting this all straight in ones own head involves a lot of going round the houses until the penny drops, I think. (to mix a few metaphors)
I'm fine thanks. I still have the dissertation to do, & the student teaching, which I am beginning this semester. All the course work is complete, & I passed the qualifier exam & took the needed make-up course & passed. Hopefully by the end of 2012 I'll graduate. Claude
It is usually understood that the terms "lead & lag" refer to trig functions like sines & cosines. But we can extend these terms to other functions like step, ramp, etc. Since a step acquires its final value immediately, & a ramp changes gradually, i.e. at a much slower rate, it is perfectly fair to say that a step "leads" a ramp. Again, we use the terms lead & lag loosely. With exponential waveforms, then "lead & lag" become ambiguous. But cap current always changes in proportion to the time derivative of the cap voltage. In the case of exponential waveforms, the derivative of the voltage & the voltage itself have the same math form, i.e. a constant times e^{-t/RC}. Anyway, I was only trying to elaborate on the "Eli the ice man" principle extending it beyond mere sine/cosine functions. Hopefully I succeeded & made matters easier to understand. If anything I said needs clarifying I will do so. Claude
Everything you say makes perfect sense. I got all needed intuition. I am only 1st year of EE, and I am yet to discover other forms of input voltages. Thank you mr. Claude. And thank you sophiecentaur, You have been both, more than helpful.