Why current leads voltage by 90 degree in capacitor?

Discussion in 'General Electronics Chat' started by movva, May 17, 2007.

  1. movva

    Thread Starter New Member

    May 16, 2007
    1
    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.
     
    Wasi-Ur-Rehman Khan likes this.
  2. recca02

    Senior Member

    Apr 2, 2007
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    this quote explains it well:

    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.
     
  3. BladeSabre

    Senior Member

    Aug 11, 2005
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    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.
     
  4. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    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.
     
  5. alifriend7

    New Member

    Nov 13, 2012
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    (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 )
     
  6. Blofeld

    Active Member

    Feb 21, 2010
    82
    18
    Nice answer alifriend7, and welcome to the forum.

    You seem to possess some magical powers that enable you to resurrect such an old thread from the dead :)
     
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  7. JDT

    Well-Known Member

    Feb 12, 2009
    658
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    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?
     
  8. Dodgydave

    AAC Fanatic!

    Jun 22, 2012
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    I remember the old rule " CIVIL " capacitor current before Voltage, current after voltage Inductor
     
  9. crutschow

    Expert

    Mar 14, 2008
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    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?
     
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  10. Veni

    New Member

    Jan 7, 2013
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    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/wikiped...svg/250px-1st_Order_Lowpass_Filter_RC.svg.png

    Veni
     
    Last edited: Jan 7, 2013
  11. crutschow

    Expert

    Mar 14, 2008
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    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?
     
    Last edited: Jan 7, 2013
  12. WBahn

    Moderator

    Mar 31, 2012
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    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?
     
  13. Veni

    New Member

    Jan 7, 2013
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    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
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    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.
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    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.
     
  16. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    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!"
     
    Last edited: Jan 8, 2013
  17. Veni

    New Member

    Jan 7, 2013
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    Thanks.

    Veni
     
    Last edited: Jan 8, 2013
  18. BillB3857

    AAC Fanatic!

    Feb 28, 2009
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  19. MKCheruvu

    Member

    Nov 20, 2012
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    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.
     
  20. Wasi-Ur-Rehman Khan

    New Member

    Jan 31, 2013
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    Please Tell me why Current Leads & Voltage Lags In Capacitor?
     
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