To put this relationship between voltage and current in a capacitor in calculus terms, the current through a capacitor is the derivative of the voltage across the capacitor with respect to time. Or, stated in simpler terms, a capacitor’s current is directly proportional to how quickly the voltage across it is changing.
It is the inverse of the sin (or cos) wave form between the peak voltages. Because AC voltage varies in a sinusoidal waveform, the derivative at any point is the cosine of the value. Well, when I think of derivative with respect to time I think of something changing and when voltage is involved I think of capacitors.
‘C’ is the value of capacitance and ‘R’ is the resistance value. The ‘V’ is the Voltage of the DC source and ‘v‘ is the instantaneous voltage across the capacitor. When the switch ‘S’ is closed, the current flows through the capacitor and it charges towards the voltage V from value 0.
If you derive with respect to time you get the current through the capacitor for a varying voltage: d dt Q(t) = C d dt V (t) This equation tells you that when the voltage doesn’t change across the capacitor, current doesn’t flow; to have current flow, the voltage must change. (I hope it helped) This only applies to Alternating Current.
So the formula for charging a capacitor is: vc(t) = Vs(1 − exp(−t/τ)) v c (t) = V s (1 − e x p (− t / τ)) Where Vs V s is the charge voltage and vc(t) v c (t) the voltage over the capacitor. If I want to derive this formula from 'scratch', as in when I use Q = CV to find the current, how would I go about doing that?
Rearranging the equation gives us the capacitor voltage: Initially V c (t) is 0, however as current decreases, the voltage dropped across the resistor R decreases and V c (t) increases. After 4 time constants, it has reached 98% of its final value.