Voltage and Current

for the inductive current lag-inductor are connected parallel to the source which make the inductor possess the same voltage as the source, hence the voltage leads and current lags.

for capacitive current lead- current flows in the capacitor from the source and charges it up, then making it to possess same voltage with the source. Hence capacitive current leads as the capacitor is been charged and discharged.

Mathematically e = L(di/dt), e = voltage applied, i is the current through the inductor, (di/dt) is rate of change of current through the inductor. In frequency domain d/dt translates to the operator j [which has a value of 90 deg ahead]

hence the voltage leads the current.

Similar explanations can be offered in the case of capacitor, voltage and current

If we catch the signals of the voltage and currents in an osciloscope, we find voltage leading current by a quarter cycle [90 deg].

Sir ,

your answer cleared my doubt mathematically . will u pls elaborate it physically i.e what exactly happens inside the conductor ?

When an AC voltage is applied to an inductor a flux is developed in the inductor which opposes the flow of current.

Assuming that we have a pure inductor without any resistance, in the absence of this induced emf, we will get an infinite current flowing through the inductor as there is no resistance in the inductor.

However, the current is limited by the induced voltage as given by the equation e(t)=d/dt(i(t))

this opposition is dependent on the value of inductance and frequency of supply voltage and in steady state equation terms we call this reactance XL = 2(pi)*f*L

where pi = 3.14.. f, applied voltage frequency, L inductance etc.

In steady state equation terms, the voltage drop across the inductor is given by (R+jXL)*I

You may refer to fundamental texts for more explanations.