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.
- 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?
Best Answer
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.