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1.When we mathematically derive the expression for the current from a sinusoidal voltage source (v=V sin(wt)), we take the derivative of q=cv where c is the capacitance. The final expression we get is i=I cos(wt) which we express as i=I sin (wt+90). We can very well express it as i=I sin (wt-90). So we can say that the current lags the voltage instead of saying what the standard is.

  1. What makes the current lead the voltage? If I have a manual voltage source which I can use to change the voltage across the capacitor as and when I desire (assuming no resistance), will the current still lead? How and why?
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3 Answers 3

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Both your questions relate to the special property of the sine wave, whose derivative is another sine wave shifted ahead by a quarter period, or as you write, 90 degrees, in advance.

You are right we can express the current as "i=I sin (wt+90)". But this is a different function than "i=I sin (wt-90)", which would have a wrong sign. Therefore it is only correct to say the current leads the voltage.

Everything above holds only for sine waves. If you have an arbitrary voltage source, the current has to be computed using derivatives of the waveform.

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  • $\begingroup$ And if we ask WHY the sign is as such, and not the other way around, we must trace it back into our established convention of the direction in which we decided to measure voltages and currents. $\endgroup$
    – dominecf
    Apr 9, 2016 at 19:43
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Your question is very reasonable. Indeed both solutions are allowed from a mathematical point of view. But at any time you check current and voltage with an oscilloscope you will get the same shift.

This has to do with the asymmetry of the Lorentz force for negative and positive charged particles. In a current carrying coil the magnetic field will be oriented always in the same direction for a current of electrons. But if it would be possible for you to have a current of positrons (in a world of antimatter) or of protons (ionised liquid in a spiral tube) you will get a magnetic field which will be in the opposite direction to the usual direction.

What that has to do with the shift of current and voltage in an alternating current? The point is that an alternating current is made by induction from a varying magnetic field (the opposite process of the Lorentz force). The magnet moves the electrons. If a magnet with the same direction of his magnetic field and the same rotational direction would induce a current of positrons in a coil (same direction of winding), you would see on your oscilloscope a shift of the waves you described as alternative to the usual result.

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  • $\begingroup$ The direction of the Lorentz force has to do it with the correlation between the directions of the particles magnetic dipole moment and their intrinsic spin. $\endgroup$ Apr 9, 2016 at 11:32
  • $\begingroup$ I am not sure whether this answer makes it 100% clear that any mechanism of current in a positive direction (i.e. positive particles forward and/or negative particles backward) creates always the same orientation of the magnetic field. In other words, exchanging a metallic coil with a thin tube of ionic liquid, or hypothetically an antimatter metal with free positrons, never reverses the magnetic field around it. $\endgroup$
    – dominecf
    Apr 9, 2016 at 19:33
  • $\begingroup$ Moreover I do not understand the third paragraph about generating AC fields with magnets. In my experimental approach, I would generate the AC current with an electronic sinewave generator and I would surely get the same result although no rotating coil nor macroscopic magnetic field are involved. $\endgroup$
    – dominecf
    Apr 9, 2016 at 19:39
  • $\begingroup$ @dominecf Please think about how the magnetic dipole moment is related to it's spin for a particle and for it's anti-particle. Due to the Einstein-de-Haas experiment the intrinsic spin indeed acts as a momentum. By this a magnetic field deflects a particle and it's anti-particle in opposite directions. $\endgroup$ Apr 9, 2016 at 19:52
  • $\begingroup$ This is true, but I do not see the relation to the original question. Maybe I miss somethting, but I still believe that the answer to the original question is basically this: d[sin(x)]/dx = cos(x) $\endgroup$
    – dominecf
    Apr 9, 2016 at 20:02
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Since the voltage on the capacitor comes from a buildup of current, do you find it surprising that the volatage lags the current? But that's the same as saying the current leads the voltage.

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