why current is leading in capacitor?

Question Posted / guest

1 Answers
6598 Views
PSU, I also Faced
E-Mail Answers

why current is leading in capacitor?..

In a close circuit [in a loop] the current flowing in the
conductor produces a magnetic flux [F]. F=L*i [L=circuit
inductance, i=instantaneous current intensity].
In a.c. the flux will change from maximum positive to
maximum negative and again to maximum positive-ideal sine
wave form-in a full wave time-and in turn will produce a FEM
[equal to supply voltage-if we neglect the conductor
resistance].
v=L*di/dt -current derivate in time.
If i=Imax*sin(w*t)[ considering i=0 when t=0] w=2*pi()
*f f=frequency[50 or 60 Hz]
Simplifying: d(sin(wt))/dt=w*cos(wt) and d(cos(wt))/dt=-
w*sin(wt).
v=L*w*Imax*cos(w*t) if L*w*Imax=Vm then:
v=Vm*cos(w*t)
Since at time 0 voltage v=Vm but i=0 and i=Im will be
only after 1/4 of a wave , voltage leads the current.
In a condenser -capacitor-an electric field is installed
[if in the circuit is a voltage supplied ] and the plates
will be charged in short time accumulating the current. In
d.c. after the charge is finited no current will flow
further.
In a.c. a steady state current will flow following the
voltage variation.
From the relation :Q=C*v Q=electric charge C=Capacity
V=voltage
By definition i=dQ/dt then i=C*dv/dt
since v=Vm*cos(w*t)
dv/dt=-w*Vm*sin(w*t) i=-C*w*Vm*sin(wt) .
At t=0 i=0 but before this at wt=-pi/2 rad[-90 dgr] i=Im
that means current leads[the voltage].
90drg. lead=360-90=270 dgr. lag., of course. Conventional
we take only up to 180 dgr. for the angle between voltage
and current of the same phase.