>> 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.