Why current leads voltage by 90 degree in capacitor?

Thread Starter

movva

Joined May 16, 2007
1
Hello,

Can anybody tell me why current leads voltage by 90 degree in capacitor and vltage leads current by current in inductor?


Also please suggest me some good books regarding this.




Thanks in advance,
Movva.
 

recca02

Joined Apr 2, 2007
1,214
this quote explains it well:

The current through a capacitor due to an AC source reverses direction periodically. That is, the alternating current alternately charges the plates: first in one direction and then the other. With the exception of the instant that the current changes direction, the capacitor current is non-zero at all times during a cycle. For this reason, it is commonly said that capacitors "pass" AC. However, at no time do electrons actually cross between the plates, unless the dielectric breaks down. Such a situation would involve physical damage to the capacitor and likely to the circuit involved as well.

Since the voltage across a capacitor is proportional to the integral of the current, as shown above, with sine waves in AC or signal circuits this results in a phase difference of 90 degrees, the current leading the voltage phase angle. It can be shown that the AC voltage across the capacitor is in quadrature with the alternating current through the capacitor. That is, the voltage and current are 'out-of-phase' by a quarter cycle. The amplitude of the voltage depends on the amplitude of the current divided by the product of the frequency of the current with the capacitance, C. ----------wiki
also i think similar topic has been discussed here before (it had a cool java simulation) i hope u find that using forum search.

as far as i understand, the voltage across the capacitor increases till the current changes direction, hence when current reaches zero(before direction change) voltage is maximum across the capacitor .this also explains why voltage is proportional to the integral of current.
 

BladeSabre

Joined Aug 11, 2005
105
And for an inductor, changes in current cause there to be a voltage across it (for magnetic field reasons). It happens that the voltage across an inductor is proportional to the derivative of the current.
 

recca02

Joined Apr 2, 2007
1,214
oh yes forgot abt the inductor,sorry!
as bladesabre san said voltage across inductor is proportional to derivative of current.
the emf induced in a inductor is proprtional to rate of change of current(self induced) thru it.
for mutual inductance it is proportional to rate of change of flux linked with it.
 

alifriend7

Joined Nov 13, 2012
1
Hello,

Can anybody tell me why current leads voltage by 90 degree in capacitor and vltage leads current by current in inductor?


Also please suggest me some good books regarding this.




Thanks in advance,
Movva.
(Q = VC).
dQ=CdV

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°.
( d means delta )
 

JDT

Joined Feb 12, 2009
657
Non mathematically, it's because the current depends on the rate of change of voltage, not the actual voltage across the capacitor.

So with a sine wave, maximum current occurs when the voltage is passing through zero. Draw a graph and you will see the 90deg phase difference!

With a inductor, the voltage depends on the rate of change of current - due to self induction. Or to put it another way: the rate of change of current depends on the actual voltage across the inductor.

See?
 

crutschow

Joined Mar 14, 2008
25,398
Let's try a little though experiment to see if that helps. Think about what happens when a sinewave voltage is applied to a capacitor or inductor.

With a capacitor, as the voltage increases there is charge (current) going into the capacitor (think of it as a bucket for charge) until it reaches the positive peak voltage. At that point the charging (current) stops. As the voltage now starts to reduce, charge (current) now flows out of the capacitor and continues until the voltage reaches the negative peak, at which point the charging (current) again stops. If you trace this out on a graph you will see that the current is leading the voltage by 90°, reaching zero current at the peaks of the voltage and maximum current as the voltage is going through zero.

With an inductor, with a positive voltage applied, the inductance (think of it as sort of an inertia to current) causes the current to increase as the voltage goes through the peak and continues to increase until the voltage returns to zero, which is the maximum positive current point. Now voltage going negative causes the inductive current to start to slow down and reverse until the voltage returns to zero from the negative peak, at which point the current is now at the negative current maximum point. It you trace this out on a graph you will see that the inductor current is lagging the voltage by 90°, reaching the peak values 90° after the peak voltages.

Does that make sense?
 

Veni

Joined Jan 7, 2013
3
If current lags voltage then why is vout in phase with vin at low frequencies for a pass filter. The current through the resistor is always in phase with the voltage across it, which is in phase with both Vin and Vout. As it's a series circuit, then the current through the capacitor is the same as that of the resistor, which means current through the capacitor is also in phase with Vout. This seems contradictory to these previous posts.

http://upload.wikimedia.org/wikipedia/commons/thumb/e/e0/1st_Order_Lowpass_Filter_RC.svg/250px-1st_Order_Lowpass_Filter_RC.svg.png

Veni
 
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crutschow

Joined Mar 14, 2008
25,398
If current lags voltage then why is vout in phase with vin at low frequencies for a pass filter. The current through the resistor is always in phase with the voltage across it, which is in phase with both Vin and Vout. As it's a series circuit, then the current through the capacitor is the same as that of the resistor, which means current through the capacitor is also in phase with Vout. This seems contradictory to these previous posts.
If you measure the voltage and current phase between a filter's input and output, then yes, the phase changes with frequency, due to the effect of the resistance in the circuit. But the voltage measured directly across the capacitor is still 90° out of phase with the current through the capacitor, independent of frequency (of course the voltage isn't). It's the phasor addition of that phase shift and the resistor phase shift that determines the filter output phase shift. For LP RC connection you posted, at high frequencies the capacitor reactance is much lower than the resistor value so the phase shift approaches 90 degrees. At low frequencies the capacitor reactance becomes very high and so the phase shift approaches zero degrees.

Make sense?
 
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WBahn

Joined Mar 31, 2012
26,144
The current through the resistor is always in phase with the voltage across it, which is in phase with both Vin and Vout.
No. How do you get from the voltage across the resistor (i.e., Vin-Vout) being in phase with the current in it to a claim that it is now somehow always in phase with either Vin or Vout, let alone both?
 

Veni

Joined Jan 7, 2013
3
Hmm. Vin and Vout are in phase at low frequencies. There will come a time when Vin and Vout are both 0v at the same time. At this time, if current in capacitor is leading by 90 deg, then the current in capacitor is peaking. If it's a series circuit, current through resistor should be peaking too. But by definition both ends of the resistor are now at 0v (Vin = Vout = 0v). How can current through it be peaking, it would have a voltage drop across it?

Veni
 

WBahn

Joined Mar 31, 2012
26,144
Non mathematically, it's because the current depends on the rate of change of voltage, not the actual voltage across the capacitor.

So with a sine wave, maximum current occurs when the voltage is passing through zero. Draw a graph and you will see the 90deg phase difference!

With a inductor, the voltage depends on the rate of change of current - due to self induction. Or to put it another way: the rate of change of current depends on the actual voltage across the inductor.

See?
While this explains why voltage and current are 90 degrees out of phase in either a capacitor or an inductor, it falls short of explain why voltage leads current in one and the opposite in the other. In either one, when current is at a maximum (amplitude) the voltage is zero and, likewise, when current is zero voltage is at a maximum (amplitude).

To explain (without pulling out the math) the difference requires a bit more detail. So let's focus on the same point in one of the waveforms, say when the voltage is zero have been negative and now getting ready to go positive. In both a capacitor and an inductor the current is at a maximum amplitude, but in one it is positive and in the other it is negative (polarities assigned per the passive sign convention). So which is which?

In a capacitor, if the voltage was negative and now is going positive, that means that there was a negative charge on it and now the charge is going positive which, in turn, means the current is positive. This means that, one quarter of a cycle earlier, the current was zero and becoming positive, thus the current is leading the voltage.

In an inductor, if the voltage was negative and now is going positive, that means that the current was previously becoming more negative and is not going to start becoming less negative, meaning that the current was at a negative maximum. This means that, one quarter of a cycle later, the current will reach zero and start becoming positive, thus the current is lagging the voltage.
 

WBahn

Joined Mar 31, 2012
26,144
Hmm. Vin and Vout are in phase at low frequencies. There will come a time when Vin and Vout are both 0v at the same time. At this time, if current in capacitor is leading by 90 deg, then the current in capacitor is peaking. If it's a series circuit, current through resistor should be peaking too. But by definition both ends of the resistor are now at 0v (Vin = Vout = 0v). How can current through it be peaking, it would have a voltage drop across it?

Veni
The situation you describe only occurs at DC (frequency equals zero) in which case the capacitor looks like an open circuit, no current is flowing in the resistor, and Vout equals Vin.
 

GopherT

Joined Nov 23, 2012
8,012
The original poster dropped this question here in May 2007 and never made another post. He/She was probably notified by mail.yahoo.com that the question was answered again and they are saying, "WT@, why do people keep answering this question? I dropped out of the EE program 5 years ago!"
 
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MKCheruvu

Joined Nov 20, 2012
30
Let us take an example of a RC Circuit being driven by battery connected thru a switch.
The series circuit current at any given point of time is
Ic(t) = [Vbattery - Vcapacitor(t)]/R
Let us consider various operating conditions in the circuit :
1) @ t=0 (ie when the switch is closed) ,initial capacitor voltage Vc = 0 and the capacitor current Ic(0+) = Vb/R which is the Maximum
2)@ t > 0- as the charge(current) flows in the series circuit the capacitor voltage Vc will increase.
3) @ t=T -as the capacitor is fully charged to Vb , the Current Ic(T) = 0
4) ie Capacitor experiences a condition of - Vc=0, Ic=Max and Vc=vb(Max) ,Ic=0 and this happens in a period T is explained in time domain.
The period T converted to frequency domain for periodic Time varying signal(say - sinewave), is the Phase Angle (in degrees).
Hence there is shift(Phase) of Min & Max conditions of Vc/Ic to occur in the capacitor.
 
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