< Return to Video

L7 6 1 Phasor Diagrams

  • 0:01 - 0:06
    >> In this video, we're going to
    demonstrate how phasor diagrams
  • 0:06 - 0:09
    or showing the phasors
    in the complex plane can
  • 0:09 - 0:13
    demonstrate the relative phases
  • 0:13 - 0:17
    and magnitudes of voltages and
    currents within a circuit.
  • 0:17 - 0:21
    To do this, we're going
    to remind ourselves that
  • 0:21 - 0:27
    phasor V is equal to the
    phasor I times the impedance.
  • 0:27 - 0:29
    Now for example, in a resistor,
  • 0:29 - 0:34
    the impedance is R and V then is simply
  • 0:34 - 0:38
    equal to I times R. R is a real number.
  • 0:38 - 0:42
    So there is no phase term associated
    with the impedance of a resistor,
  • 0:42 - 0:46
    and we say that the voltage is in
  • 0:46 - 0:56
    phase with the current in a resistor.
  • 0:56 - 1:05
    So if we have just arbitrarily choose
    the current to have a zero phase angle,
  • 1:07 - 1:11
    then the phasor voltage
  • 1:11 - 1:12
    associated with the resistor or the voltage
  • 1:12 - 1:15
    across the resistor call it V sub r,
  • 1:15 - 1:18
    will have the same angle as I,
  • 1:18 - 1:20
    be in the same direction,
  • 1:20 - 1:25
    but just have a different magnitude than I.
  • 1:25 - 1:31
    In fact the magnitude of this
    will be R times I. All right.
  • 1:31 - 1:34
    Unlike in a resistor, a capacitor
    introduces a phase difference,
  • 1:34 - 1:41
    we know that V across the
    capacitor is equal to I times Z.
  • 1:41 - 1:45
    Well, the impedance of
    a capacitor is one over
  • 1:45 - 1:49
    j Omega c. We bring that
    j up in the numerator
  • 1:49 - 1:52
    with a minus sign and we'll
    have then this is equal to
  • 1:52 - 1:58
    I times a negative j over Omega c. Well,
  • 1:58 - 2:06
    that negative j in rectangular
    coordinates is the same thing as
  • 2:06 - 2:10
    a minus 90 degree phase term
    in polar coordinates
  • 2:10 - 2:15
    or we have then that V is equal
  • 2:15 - 2:25
    to I times one over Omega c
    times e to the minus j90.
  • 2:25 - 2:29
    So whatever the phase of the current is,
  • 2:29 - 2:31
    when we multiply the current
    times the impedance there is
  • 2:31 - 2:35
    a negative 90 degree phase term
    that gets added to the current.
  • 2:35 - 2:42
    So once again if we let the current
    have a zero reference angle,
  • 2:42 - 2:47
    then the voltage in a
    capacitor is going to be
  • 2:47 - 2:52
    90 degrees less than
  • 2:52 - 2:54
    the phase of the current or we
  • 2:54 - 2:59
    say the voltage is 90 degrees
    behind the current,
  • 2:59 - 3:04
    or we can say the current is
    90 degrees ahead of the voltage.
  • 3:04 - 3:08
    Finally, we look at the
    voltage across the inductor,
  • 3:08 - 3:11
    and we have phasor V. So that one,
  • 3:11 - 3:13
    of course these are all
    phasor terms over here.
  • 3:13 - 3:19
    Phasor V sub l is equal to I times Z.
  • 3:19 - 3:24
    Well the Z the impedance of
    an inductor is j Omega L,
  • 3:24 - 3:32
    which is equal to I times Omega L times j.
  • 3:32 - 3:39
    Well, j is the same thing
    as e to the positive j90.
  • 3:39 - 3:41
    So in an inductor
  • 3:41 - 3:45
    the voltage will be 90 degrees
    ahead of the phase of the current.
  • 3:45 - 3:47
    Whatever the phase of the current
  • 3:47 - 3:50
    is and it gets multiplied by
    the impedance of the inductor,
  • 3:50 - 3:53
    which has this positive
    90 phase term with it,
  • 3:53 - 3:55
    that positive 90 gets added to
  • 3:55 - 3:57
    the phase of the current
    and the voltage will be
  • 3:57 - 4:02
    90 degrees ahead of the current.
  • 4:02 - 4:06
    So over here we had that being
    I once again we'll reference I,
  • 4:06 - 4:12
    just arbitrarily choosing
    it to have a zero phase.
  • 4:12 - 4:19
    I then V sub l will
  • 4:19 - 4:25
    be 90 degrees ahead of phasor I.
  • 4:25 - 4:29
    So we say that the current lags the voltage
  • 4:29 - 4:34
    in an inductor or the voltage
    leads the current in an inductor.
  • 4:34 - 4:38
    Notice the voltages of the capacitor and
  • 4:38 - 4:42
    the inductor are 180 degrees
    out of phase with each other.
  • 4:42 - 4:45
    The voltage and the current in the voltage
  • 4:45 - 4:48
    of the capacitor lags by 90 degrees,
  • 4:48 - 4:53
    the voltage in the inductor
    leads the current by 90 degrees.
  • 4:53 - 4:55
    There's a mnemonic that
    goes along with this,
  • 4:55 - 5:00
    today's clear back to the late 1800s
    when much of electrical work that was
  • 5:00 - 5:05
    being done was associated with
    refrigeration and creating ice,
  • 5:05 - 5:13
    and the mnemonic is ELI, the ICE man.
  • 5:13 - 5:17
    What this is suggesting is that E
  • 5:17 - 5:21
    represents voltage as in
    physics we talk about the EMF.
  • 5:21 - 5:22
    So E is voltage,
  • 5:22 - 5:25
    I is current, L is an inductor.
  • 5:25 - 5:27
    So in an inductor,
  • 5:27 - 5:34
    the voltage leads
    the current by 90 degrees.
  • 5:34 - 5:40
    In an inductor, the voltage
    leads the current by 90 degrees.
  • 5:40 - 5:48
    In a capacitor, the current
    leads the voltage by 90 degrees.
  • 5:48 - 5:50
    So in the capacitor c,
  • 5:50 - 5:52
    the current is 90 degrees ahead.
  • 5:52 - 5:54
    So as in the word ICE,
  • 5:54 - 5:59
    I comes ahead of E. So I is
    leading the voltage over here.
  • 5:59 - 6:04
    In ELI, the voltage is leading the current.
  • 6:06 - 6:10
    Now let's demonstrate how
    these phasor diagrams can be
  • 6:10 - 6:13
    used to demonstrate the various or
  • 6:13 - 6:20
    the relative phasors of different voltages
    and currents within a circuit.
  • 6:20 - 6:25
    Here we have a single loop circuit
    and writing one KVL,
  • 6:25 - 6:32
    we know then that phasor
    V sub s is equal to.
  • 6:33 - 6:40
    Well, first of all V sub s sum of
    the voltage is equal to V sub r,
  • 6:40 - 6:41
    plus V sub c,
  • 6:41 - 6:49
    plus V sub l and of course
    these are all phasor voltages.
  • 6:49 - 6:53
    Now let's go ahead and put
    in that V sub r is equal
  • 6:53 - 6:58
    to I times R. In this case,
  • 6:58 - 7:01
    R is eight ohms
  • 7:02 - 7:07
    plus the voltage across the capacitor
    is just going to be I times Z sub c,
  • 7:07 - 7:14
    which is I times a negative j8 plus
    the voltage across the inductor.
  • 7:14 - 7:16
    That's a J2 ohm inductor.
  • 7:16 - 7:22
    So the voltage across
    that will be I times J2.
  • 7:22 - 7:25
    So we have three terms here.
  • 7:25 - 7:28
    Three different phasors.
  • 7:28 - 7:31
    These are phasor voltage is equal to 8I,
  • 7:31 - 7:36
    minus J8I, and positive J2I.
  • 7:37 - 7:40
    Showing this explicitly with the phases,
  • 7:40 - 7:43
    this will be I times eight.
  • 7:43 - 7:45
    There is no phase difference
    between the voltage
  • 7:45 - 7:48
    across the resistor and the current.
  • 7:48 - 7:54
    On the other hand we have I times eight
  • 7:54 - 7:59
    times e to the minus j90 in the capacitor,
  • 7:59 - 8:10
    plus I times two times e to
    the positive j90 for the inductor.
  • 8:10 - 8:13
    So again, what this is saying is that
  • 8:13 - 8:21
    the phase of the voltage
  • 8:21 - 8:26
    across the inductor is 90 degrees
    greater than the phase of the current.
  • 8:26 - 8:31
    On the other hand, the phase
    of the voltage across
  • 8:31 - 8:36
    the capacitor is 90 degrees
    behind the phase of the current.
  • 8:36 - 8:42
    So let's just start out and
    say that this is the current.
  • 8:42 - 8:46
    I will say that it has
    a magnitude I naught.
  • 8:46 - 8:52
    The voltage across the resistor
    is eight times that
  • 8:52 - 8:59
    or V sub r is equal to eight
    times I naught in length,
  • 8:59 - 9:03
    and it has the same angle as I.
  • 9:03 - 9:10
    The capacitor on the other hand
    is 90 degrees behind the current.
  • 9:10 - 9:14
    So the voltage V sub c is 90 degrees
  • 9:14 - 9:21
    behind the current V sub c,
  • 9:21 - 9:30
    and it's down here at
    negative 8I naught, and the voltage.
  • 9:30 - 9:32
    Let's see that's the
    voltage of the capacitor,
  • 9:32 - 9:41
    and the voltage of the inductor
    is 90 degrees ahead of V sub l,
  • 9:42 - 9:48
    the length of it is equal to two times
  • 9:48 - 9:54
    I naught and it is at an angle of
    90 degrees ahead of the current.
  • 9:54 - 9:56
    So our three voltages.
  • 9:56 - 10:00
    The voltage across the resistor
    is in phase with the current,
  • 10:00 - 10:03
    capacitor is 90 degrees
    behind the current and
  • 10:03 - 10:06
    the inductor voltage is 90 degrees ahead.
  • 10:06 - 10:09
    Now we have here that V sub s,
  • 10:09 - 10:13
    the source voltage is equal to
    the sum of those three terms.
  • 10:13 - 10:15
    So when adding phasors,
  • 10:15 - 10:18
    you do it just like you do vectors
    and that is tip to tell them.
  • 10:18 - 10:22
    Here's V sub r, add in V sub c,
  • 10:22 - 10:30
    that comes down here like that
    and then add in V sub l. That
  • 10:30 - 10:38
    brings us back two units back that way
    to this point here at negative 6I0,
  • 10:38 - 10:44
    and so we end up at that point
    and that is V sub s then.
  • 10:44 - 10:46
    Let see how good of a job I
    can do drawing a vector down,
  • 10:46 - 10:48
    they're not very good.
  • 10:48 - 10:51
    That's a straight line vector.
  • 10:52 - 10:55
    That's better I guess.
  • 10:55 - 11:02
    So V sub s then is equal to V sub r,
  • 11:02 - 11:03
    plus V sub c, plus
  • 11:03 - 11:10
    V sub l. This results in phasor
    is V sub s. What does it tell us?
  • 11:10 - 11:14
    Well, it tells us that V sub s is something
  • 11:14 - 11:19
    less than 90 degrees
    behind the current or that
  • 11:19 - 11:22
    the current in this circuit I is leading
  • 11:22 - 11:29
    the voltage source by some angle
    less than 90 degrees.
  • 11:29 - 11:37
    It also tells us the relative length of
    this voltage is longer than V sub r,
  • 11:37 - 11:47
    it's shorter than V sub c because
    V sub c and V sub l are opposites.
  • 11:50 - 11:55
    I guess to be more accurate if
    you to say that V sub c and
  • 11:55 - 11:59
    V sub l are in opposite directions
    and they tend to cancel each other.
  • 11:59 - 12:02
    Let me say that one more
    time just a little bit
  • 12:02 - 12:04
    differently and see if I can
    make it a little more clear.
  • 12:04 - 12:08
    V sub s the source voltage is equal to
  • 12:08 - 12:11
    the phasor sum of V sub l, plus V sub r,
  • 12:11 - 12:16
    plus V sub c. By diagramming
    or drawing in V sub l,
  • 12:16 - 12:20
    V sub r and V sub c and then
    adding them tip to tail,
  • 12:20 - 12:24
    we can see the relative length and phase
  • 12:24 - 12:28
    of the source to the voltage
    across the resistor,
  • 12:28 - 12:30
    which is in phase with the current.
  • 12:30 - 12:33
    The relative length and
  • 12:33 - 12:37
    the relative angle of the source
    relative to the voltage across
  • 12:37 - 12:40
    the inductor and the relative length and
  • 12:40 - 12:47
    relative phase of the source relative
    to the voltage across the capacitor.
  • 12:47 - 12:49
    Again, this doesn't give
    us a real accurate,
  • 12:49 - 12:53
    but it does give us some feel
    for the relative phases,
  • 12:53 - 12:56
    the phases of one of each of
  • 12:56 - 13:00
    these voltages relative to the current
    and relative to the source.
Title:
L7 6 1 Phasor Diagrams
Description:

The use of phasor diagrams to analyze circuits operating in the Sinusoidal Steady State

more » « less
Video Language:
English
Duration:
13:01

English subtitles

Revisions