>> In this video, we're going to demonstrate the actual process of
analyzing circuits using phasors and impedances,
analyzing circuits in the phasor domain.
We've already pointed out that in
the phasor domain or rather that we can analyze a circuit,
in the phasor domain by representing the voltages in terms of phasors,
the currents in terms of phasors,
and the impedances or representing the effects of these capacitors,
resistors, inductors in terms of their impedances.
So first of all,
let's just represent phasor V,
the source as having an amplitude V sub m e to the j Theta.
So, representing this time domain signal as
its phasor with an amplitude of V sub m and a phase shift of Theta.
Now the impedance of the capacitor is equal to one over
j Omega C. The impedance of a resistor is just equal to R,
and the impedance of the inductor is equal to j times
Omega times L. So we'll represent that or show that
here is this impedance is one over j Omega C. This is just R,
and this impedance here is j Omega
L. Now in the phasor domain
there will be a current flowing that we will represent as phasor I,
and with these representations let's go ahead now and write
a Kirchhoff's voltage law equation in terms of these phasers.
Starting at this point and going in this direction,
I'm going to add up the voltage
drops which means the voltage increase will be a negative.
So we'll have negative V sub m e
to the j Theta plus now going in the direction of current flow.
So direct and current flow represents a voltage drop and it will be a plus.
The voltage drop across the capacitor is
just the impedance of the capacitor times the current
or one over j Omega C times I,
plus the voltage drop across the resistor again going in the direction of
current flow will be just R times I.
Plus the voltage drop across the inductor will be,
the impedance of the inductor which is j Omega L times I.
The sum of those terms equals zero.
Now, this is not a function of I.
In fact, this is the constant associated with the source.
So let's bring it to the other side
and factor out the I that is common in each of those terms,
we have in that I not equal to I times one over j Omega C
plus R plus j omega L is equal to V sub m e to the j Theta.
We now solve for I by dividing both sides of
the equation by this term in parentheses, and we have then I,
the phasor representation of I is equal to V sub m e to
the j Theta divided by one over j Omega C,
plus R plus j Omega L.
Phasor I the phasor representation is
calculated by doing that complex division.