For A.C. through a capacitor, current_voltage.

Explanation

Let us consider the electric circuit shown below. We have a capacitor and an AC voltage V, represented by the symbol ~, that produces a potential difference across its terminals that varies sinusoidally. Here, the potential difference or the AC voltage can be given as,v=vmsinωtHere, vm is the amplitude of the oscillating potential difference and the angular frequency is given by ω. The current through the resistor due to the present voltage source can be calculated using the Kirchhoff's loop rule, as under,∑V(t)=0For the given capacitor we can write, v=qCAccording to Kirchoffs law we can write from the above circuit,vmsinωt=qCThe current through the circuit can be calculated using the relation,i=dqdt⇒i=d(vmCsinωt)dt=ωCvmcosωt⇒i=imsin(ωt+π2)[Usingtherelation,cosωt=sin(ωt+π2)]Here the amplitude of the current can be written as,im=ωCvmor else we can write it as,Im=vm1ωCHere, we can see that the term 1/ωC can be said to be equivalent to the resistance of this device and is termed as the capacitive reactance. We denote the capacitive reactance of the device as XC.XC=1ωCAnd thus, we can say that the amplitude of the current in this circuit is given as,Im=vmXCIn the above equations, the dimension of the capacitive In the above equations, the dimension of the capacitive reactance can be seen to be the same as that of resistance, and also, the SI unit of capacitive reactance is given as ohm. The capacitive reactance restricts the passage of current in a purely capacitive circuit in the same way as resistance hinders the passage of current in a purely resistive circuit.Here we say, that the capacitive reactance is inversely proportional to the frequency and the capacitance. We also see from the above equations that the current in a capacitive circuit is π/2 ahead of the voltage across the capacitor.The instantaneous power supplied to the capacitor can be given in terms of the current passing through the capacitor as,Pc=iv=imcosωtvmsinωtPc=imvm2sin2ωtHere, the average power supplied over a complete cycle can be given as,P=imvm2sin2ωt=0Concluding , we can say that in the case of a capacitor the current leads the voltage by π/2.