>> In this video, we're going to
demonstrate how phasor diagrams

or showing the phasors
in the complex plane can

demonstrate the relative phases

and magnitudes of voltages and
currents within a circuit.

To do this, we're going
to remind ourselves that

phasor V is equal to the
phasor I times the impedance.

Now for example, in a resistor,

the impedance is R and V then is simply

equal to I times R. R is a real number.

So there is no phase term associated
with the impedance of a resistor,

and we say that the voltage is in

phase with the current in a resistor.

So if we have just arbitrarily choose
the current to have a zero phase angle,

then the phasor voltage

associated with the resistor or the voltage

across the resistor call it V sub r,

will have the same angle as I,

be in the same direction,

but just have a different magnitude than I.

In fact the magnitude of this
will be R times I. All right.

Unlike in a resistor, a capacitor
introduces a phase difference,

we know that V across the
capacitor is equal to I times Z.

Well, the impedance of
a capacitor is one over

j Omega c. We bring that
j up in the numerator

with a minus sign and we'll
have then this is equal to

I times a negative j over Omega c. Well,

that negative j in rectangular
coordinates is the same thing as

a minus 90 degree phase term
in polar coordinates

or we have then that V is equal

to I times one over Omega c
times e to the minus j90.

So whatever the phase of the current is,

when we multiply the current
times the impedance there is

a negative 90 degree phase term
that gets added to the current.

So once again if we let the current
have a zero reference angle,

then the voltage in a
capacitor is going to be

90 degrees less than

the phase of the current or we

say the voltage is 90 degrees
behind the current,

or we can say the current is
90 degrees ahead of the voltage.

Finally, we look at the
voltage across the inductor,

and we have phasor V. So that one,

of course these are all
phasor terms over here.

Phasor V sub l is equal to I times Z.

Well the Z the impedance of
an inductor is j Omega L,

which is equal to I times Omega L times j.

Well, j is the same thing
as e to the positive j90.

So in an inductor

the voltage will be 90 degrees
ahead of the phase of the current.

Whatever the phase of the current

is and it gets multiplied by
the impedance of the inductor,

which has this positive
90 phase term with it,

that positive 90 gets added to

the phase of the current
and the voltage will be

90 degrees ahead of the current.

So over here we had that being
I once again we'll reference I,

just arbitrarily choosing
it to have a zero phase.

I then V sub l will

be 90 degrees ahead of phasor I.

So we say that the current lags the voltage

in an inductor or the voltage
leads the current in an inductor.

Notice the voltages of the capacitor and

the inductor are 180 degrees
out of phase with each other.

The voltage and the current in the voltage

of the capacitor lags by 90 degrees,

the voltage in the inductor
leads the current by 90 degrees.

There's a mnemonic that
goes along with this,

today's clear back to the late 1800s
when much of electrical work that was

being done was associated with
refrigeration and creating ice,

and the mnemonic is ELI, the ICE man.

What this is suggesting is that E

represents voltage as in
physics we talk about the EMF.

So E is voltage,

I is current, L is an inductor.

So in an inductor,

the voltage leads
the current by 90 degrees.

In an inductor, the voltage
leads the current by 90 degrees.

In a capacitor, the current
leads the voltage by 90 degrees.

So in the capacitor c,

the current is 90 degrees ahead.

So as in the word ICE,

I comes ahead of E. So I is
leading the voltage over here.

In ELI, the voltage is leading the current.

Now let's demonstrate how
these phasor diagrams can be

used to demonstrate the various or

the relative phasors of different voltages
and currents within a circuit.

Here we have a single loop circuit
and writing one KVL,

we know then that phasor
V sub s is equal to.

Well, first of all V sub s sum of
the voltage is equal to V sub r,

plus V sub c,

plus V sub l and of course
these are all phasor voltages.

Now let's go ahead and put
in that V sub r is equal

to I times R. In this case,

R is eight ohms

plus the voltage across the capacitor
is just going to be I times Z sub c,

which is I times a negative j8 plus
the voltage across the inductor.

That's a J2 ohm inductor.

So the voltage across
that will be I times J2.

So we have three terms here.

Three different phasors.

These are phasor voltage is equal to 8I,

minus J8I, and positive J2I.

Showing this explicitly with the phases,

this will be I times eight.

There is no phase difference
between the voltage

across the resistor and the current.

On the other hand we have I times eight

times e to the minus j90 in the capacitor,

plus I times two times e to
the positive j90 for the inductor.

So again, what this is saying is that

the phase of the voltage

across the inductor is 90 degrees
greater than the phase of the current.

On the other hand, the phase
of the voltage across

the capacitor is 90 degrees
behind the phase of the current.

So let's just start out and
say that this is the current.

I will say that it has
a magnitude I naught.

The voltage across the resistor
is eight times that

or V sub r is equal to eight
times I naught in length,

and it has the same angle as I.

The capacitor on the other hand
is 90 degrees behind the current.

So the voltage V sub c is 90 degrees

behind the current V sub c,

and it's down here at
negative 8I naught, and the voltage.

Let's see that's the
voltage of the capacitor,

and the voltage of the inductor
is 90 degrees ahead of V sub l,

the length of it is equal to two times

I naught and it is at an angle of
90 degrees ahead of the current.

So our three voltages.

The voltage across the resistor
is in phase with the current,

capacitor is 90 degrees
behind the current and

the inductor voltage is 90 degrees ahead.

Now we have here that V sub s,

the source voltage is equal to
the sum of those three terms.

So when adding phasors,

you do it just like you do vectors
and that is tip to tell them.

Here's V sub r, add in V sub c,

that comes down here like that
and then add in V sub l. That

brings us back two units back that way
to this point here at negative 6I0,

and so we end up at that point
and that is V sub s then.

Let see how good of a job I
can do drawing a vector down,

they're not very good.

That's a straight line vector.

That's better I guess.

So V sub s then is equal to V sub r,

plus V sub c, plus

V sub l. This results in phasor
is V sub s. What does it tell us?

Well, it tells us that V sub s is something

less than 90 degrees
behind the current or that

the current in this circuit I is leading

the voltage source by some angle
less than 90 degrees.

It also tells us the relative length of
this voltage is longer than V sub r,

it's shorter than V sub c because
V sub c and V sub l are opposites.

I guess to be more accurate if
you to say that V sub c and

V sub l are in opposite directions
and they tend to cancel each other.

Let me say that one more
time just a little bit

differently and see if I can
make it a little more clear.

V sub s the source voltage is equal to

the phasor sum of V sub l, plus V sub r,

plus V sub c. By diagramming
or drawing in V sub l,

V sub r and V sub c and then
adding them tip to tail,

we can see the relative length and phase

of the source to the voltage
across the resistor,

which is in phase with the current.

The relative length and

the relative angle of the source
relative to the voltage across

the inductor and the relative length and

relative phase of the source relative
to the voltage across the capacitor.

Again, this doesn't give
us a real accurate,

but it does give us some feel
for the relative phases,

the phases of one of each of

these voltages relative to the current
and relative to the source.