L7 3 2 Example 1 No numbers

>> 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.