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>> In this video, we're going to
demonstrate how phasor diagrams
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or showing the phasors
in the complex plane can
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demonstrate the relative phases
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and magnitudes of voltages and
currents within a circuit.
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To do this, we're going
to remind ourselves that
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phasor V is equal to the
phasor I times the impedance.
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Now for example, in a resistor,
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the impedance is R and V then is simply
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equal to I times R. R is a real number.
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So there is no phase term associated
with the impedance of a resistor,
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and we say that the voltage is in
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phase with the current in a resistor.
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So if we have just arbitrarily choose
the current to have a zero phase angle,
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then the phasor voltage
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associated with the resistor or the voltage
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across the resistor call it V sub r,
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will have the same angle as I,
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be in the same direction,
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but just have a different magnitude than I.
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In fact the magnitude of this
will be R times I. All right.
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Unlike in a resistor, a capacitor
introduces a phase difference,
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we know that V across the
capacitor is equal to I times Z.
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Well, the impedance of
a capacitor is one over
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j Omega c. We bring that
j up in the numerator
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with a minus sign and we'll
have then this is equal to
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I times a negative j over Omega c. Well,
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that negative j in rectangular
coordinates is the same thing as
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a minus 90 degree phase term
in polar coordinates
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or we have then that V is equal
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to I times one over Omega c
times e to the minus j90.
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So whatever the phase of the current is,
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when we multiply the current
times the impedance there is
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a negative 90 degree phase term
that gets added to the current.
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So once again if we let the current
have a zero reference angle,
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then the voltage in a
capacitor is going to be
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90 degrees less than
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the phase of the current or we
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say the voltage is 90 degrees
behind the current,
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or we can say the current is
90 degrees ahead of the voltage.
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Finally, we look at the
voltage across the inductor,
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and we have phasor V. So that one,
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of course these are all
phasor terms over here.
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Phasor V sub l is equal to I times Z.
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Well the Z the impedance of
an inductor is j Omega L,
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which is equal to I times Omega L times j.
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Well, j is the same thing
as e to the positive j90.
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So in an inductor
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the voltage will be 90 degrees
ahead of the phase of the current.
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Whatever the phase of the current
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is and it gets multiplied by
the impedance of the inductor,
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which has this positive
90 phase term with it,
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that positive 90 gets added to
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the phase of the current
and the voltage will be
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90 degrees ahead of the current.
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So over here we had that being
I once again we'll reference I,
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just arbitrarily choosing
it to have a zero phase.
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I then V sub l will
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be 90 degrees ahead of phasor I.
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So we say that the current lags the voltage
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in an inductor or the voltage
leads the current in an inductor.
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Notice the voltages of the capacitor and
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the inductor are 180 degrees
out of phase with each other.
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The voltage and the current in the voltage
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of the capacitor lags by 90 degrees,
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the voltage in the inductor
leads the current by 90 degrees.
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There's a mnemonic that
goes along with this,
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today's clear back to the late 1800s
when much of electrical work that was
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being done was associated with
refrigeration and creating ice,
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and the mnemonic is ELI, the ICE man.
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What this is suggesting is that E
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represents voltage as in
physics we talk about the EMF.
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So E is voltage,
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I is current, L is an inductor.
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So in an inductor,
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the voltage leads
the current by 90 degrees.
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In an inductor, the voltage
leads the current by 90 degrees.
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In a capacitor, the current
leads the voltage by 90 degrees.
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So in the capacitor c,
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the current is 90 degrees ahead.
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So as in the word ICE,
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I comes ahead of E. So I is
leading the voltage over here.
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In ELI, the voltage is leading the current.
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Now let's demonstrate how
these phasor diagrams can be
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used to demonstrate the various or
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the relative phasors of different voltages
and currents within a circuit.
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Here we have a single loop circuit
and writing one KVL,
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we know then that phasor
V sub s is equal to.
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Well, first of all V sub s sum of
the voltage is equal to V sub r,
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plus V sub c,
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plus V sub l and of course
these are all phasor voltages.
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Now let's go ahead and put
in that V sub r is equal
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to I times R. In this case,
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R is eight ohms
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plus the voltage across the capacitor
is just going to be I times Z sub c,
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which is I times a negative j8 plus
the voltage across the inductor.
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That's a J2 ohm inductor.
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So the voltage across
that will be I times J2.
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So we have three terms here.
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Three different phasors.
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These are phasor voltage is equal to 8I,
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minus J8I, and positive J2I.
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Showing this explicitly with the phases,
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this will be I times eight.
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There is no phase difference
between the voltage
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across the resistor and the current.
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On the other hand we have I times eight
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times e to the minus j90 in the capacitor,
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plus I times two times e to
the positive j90 for the inductor.
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So again, what this is saying is that
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the phase of the voltage
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across the inductor is 90 degrees
greater than the phase of the current.
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On the other hand, the phase
of the voltage across
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the capacitor is 90 degrees
behind the phase of the current.
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So let's just start out and
say that this is the current.
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I will say that it has
a magnitude I naught.
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The voltage across the resistor
is eight times that
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or V sub r is equal to eight
times I naught in length,
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and it has the same angle as I.
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The capacitor on the other hand
is 90 degrees behind the current.
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So the voltage V sub c is 90 degrees
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behind the current V sub c,
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and it's down here at
negative 8I naught, and the voltage.
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Let's see that's the
voltage of the capacitor,
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and the voltage of the inductor
is 90 degrees ahead of V sub l,
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the length of it is equal to two times
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I naught and it is at an angle of
90 degrees ahead of the current.
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So our three voltages.
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The voltage across the resistor
is in phase with the current,
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capacitor is 90 degrees
behind the current and
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the inductor voltage is 90 degrees ahead.
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Now we have here that V sub s,
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the source voltage is equal to
the sum of those three terms.
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So when adding phasors,
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you do it just like you do vectors
and that is tip to tell them.
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Here's V sub r, add in V sub c,
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that comes down here like that
and then add in V sub l. That
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brings us back two units back that way
to this point here at negative 6I0,
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and so we end up at that point
and that is V sub s then.
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Let see how good of a job I
can do drawing a vector down,
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they're not very good.
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That's a straight line vector.
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That's better I guess.
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So V sub s then is equal to V sub r,
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plus V sub c, plus
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V sub l. This results in phasor
is V sub s. What does it tell us?
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Well, it tells us that V sub s is something
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less than 90 degrees
behind the current or that
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the current in this circuit I is leading
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the voltage source by some angle
less than 90 degrees.
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It also tells us the relative length of
this voltage is longer than V sub r,
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it's shorter than V sub c because
V sub c and V sub l are opposites.
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I guess to be more accurate if
you to say that V sub c and
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V sub l are in opposite directions
and they tend to cancel each other.
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Let me say that one more
time just a little bit
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differently and see if I can
make it a little more clear.
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V sub s the source voltage is equal to
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the phasor sum of V sub l, plus V sub r,
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plus V sub c. By diagramming
or drawing in V sub l,
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V sub r and V sub c and then
adding them tip to tail,
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we can see the relative length and phase
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of the source to the voltage
across the resistor,
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which is in phase with the current.
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The relative length and
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the relative angle of the source
relative to the voltage across
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the inductor and the relative length and
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relative phase of the source relative
to the voltage across the capacitor.
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Again, this doesn't give
us a real accurate,
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but it does give us some feel
for the relative phases,
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the phases of one of each of
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these voltages relative to the current
and relative to the source.