Register to reply 
Why does current lead the voltage in capacitor? 
Share this thread: 
#1
Aug2611, 07:43 AM

P: 951

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. 


#2
Aug2611, 07:57 AM

Sci Advisor
PF Gold
P: 1,776

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. 


#3
Aug2611, 07:59 AM

P: 951




#4
Aug2611, 08:18 AM

Sci Advisor
HW Helper
Thanks
P: 26,157

Why does current lead the voltage in capacitor?
Hi Bassalisk!
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), 


#5
Aug2611, 08:20 AM

P: 951




#6
Aug2611, 11:01 AM

P: 1,036

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 nonzero, i.e. P = dW/dt. But power P = I*V. So, I & V must both be nonzero during the transition so that power is nonzero & 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 


#7
Aug2611, 11:07 AM

P: 951

Thank you very much mr. Claude ! 


#8
Aug2711, 05:29 AM

Sci Advisor
PF Gold
P: 1,776

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. 


#9
Aug2711, 05:44 AM

P: 951




#10
Aug2711, 08:38 AM

P: 1,036

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 


#11
Aug2711, 08:44 AM

Sci Advisor
PF Gold
P: 11,942

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.



#12
Aug2711, 12:55 PM

P: 1,036

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 


#13
Aug2711, 01:15 PM

P: 3,883

Hi Cabraham, how are you? Haven't seen you around for a while. Got your PhD yet?



#14
Aug2711, 01:15 PM

Sci Advisor
PF Gold
P: 11,942

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) 


#15
Aug2711, 01:17 PM

P: 951




#16
Aug2711, 01:20 PM

P: 1,036

Claude 


#17
Aug2711, 01:26 PM

P: 1,036

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 


#18
Aug2711, 01:29 PM

P: 951

Thank you mr. Claude. And thank you sophiecentaur, You have been both, more than helpful. 


Register to reply 
Related Discussions  
Capacitor in AC Circuit: Why there is such relation of current and voltage?  General Physics  8  
Current/Capacitor/voltage problem. Please Help.  Calculus & Beyond Homework  0  
What is the correct voltage current eq. for a variable capacitance capacitor?  Classical Physics  1  
Integrating CurrentVoltage Relationship for a Capacitor  Engineering, Comp Sci, & Technology Homework  1  
VoltageCurrent lead/lag  General Physics  2 