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>> We'll do that same example now only with

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the actual values for the circuit
elements and the source here.

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So again, the first thing we do is convert

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the time-domain source to
its phasor representation

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and then calculate the impedances
associated with the capacitor,

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resistance and inductor here.

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So this the phasor
representation V we'll call it

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cap V or phasor V is
equal to 10E to the J30.

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Now, let's be explicit here.

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Omega is 1,000 radians per second.

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Radians per second.

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Z_C is equal to one over J Omega C
which is equal to one over

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J times Omega which is

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1,000 times C which is 10 times
10 to the minus six to six micro,

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10 microfarads and that then
equals negative J 100 Ohms.

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That's a negative J 100 Ohms.

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Z_R is just a value of
the resistor which is

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50 Ohms and Z_L is equal to J
Omega L which is equal to J times

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Omega which is 1,000 times L which is

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75 millihenries and that
gives us a positive J 75.

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So this is a J 75 there.

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This is R which is equal to the impedance
that is equal to 50 ohms and this,

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the impedance of the capacitor
is a negative J 100.

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We'll go ahead and define the current

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flowing through here or the phasor
representation of the current going

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through there call it phasor I and let's

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write a Kirchhoff's voltage law equation
going around that loop.

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Once again, adding up the voltage drops.

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So, a voltage increase will be a negative,

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that would be a negative 10 E to the J
30 plus going across this capacitor,

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the voltage drop across that capacitor will

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be the current I times the impedance.

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So, that will be I times
the impedance which is

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a negative J 100 plus the voltage drop
across the resistor which is going

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to be I times 50 plus the voltage drop
across that inductor is going to be

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I times the impedance of
that inductor which is

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J 75 and the sum of those
terms then equals zero.

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We want to solve for I
the current in this circuit.

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So we'll factor out the I
from these common terms

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here and bring this term here on
over other side and we get then

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that I times a negative J 100 plus 50

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plus J 75 is equal to
a positive 10 E to the J 30.

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We have a negative J 100 plus J 75
that gives us a negative J 25 plus 50.

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Dividing both sides by that term
then gives us that I is equal

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to 10 E to the J 30 divided by,

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writing the real part first 50 minus J 25.

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You get out your calculator.

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First of all, let's convert
just this denominator here to polar form.

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It's a little bit easier to see
what's going on here if we do so.

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In polar form, that's equal to 55.9 E

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to negative J 26.57.

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Now, we can see there the division,

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it's 10 E to the J 30 divided
by 55.9 E to the minus J 26.57.

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So 10 divided by 55.9
gives us the magnitude.

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The magnitude of the numerator
divided by the magnitude of

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the denominator gives us 0.179.

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Now, the phase E,

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it will be E to the J 30 in the numerator,

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E to the minus J 26.57 in the denominator.

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We've already demonstrated that when
dividing numbers involving exponents,

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it's the exponent of the numerator
minus the exponent in the denominator.

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So it's J 30 minus and minus
it becomes plus a 26.57 that

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gives us a positive J 56.57 degrees.

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This entire thing here is
the phasor representation of the current.

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We can now take this back
into the time domain

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by recognizing that the
amplitude of the current

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is going to be the magnitude of
this phasor or let's be explicit now.

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I of T then is equal to amplitude of 0.179.

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It will be a cosine term.

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We came in as a cosine will come back
to the time domain as a cosine term.

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Cosine, the frequency
doesn't change that's still

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1,000 T. But now the phase of the
current is equal to 56.57 degrees.

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So plus 56.57 degrees.

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That then is the time domain
expression for the current.

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The voltage, the source driving
this circuit is oscillating at

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1,000 radians per second and the current
is oscillating 1,000 times per second.

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It's going back and forth changing at

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the same frequency that
the source is going into it.

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The source has an amplitude of 10 volts.

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The current flowing has
an amplitude of 0.179 Amps.

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So, the amplitude of the current
is smaller than the amplitude of

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the voltage and finally
let's look at the phase.

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The phase angle of the voltage source
is 30 degrees, positive 30 degrees.

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The angle of the current is 56.57 degrees.

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So, we'd say then that the current
has been shifted ahead

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of the source by the 26.57 degrees.

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Here are plots of those,

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the voltage in the red,

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V of T is equal to 10 cosine
of 1,000 T plus 30 degrees.

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I of T we have now found
to be equal to 0.179,

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cosine of 1,000 T plus 56.57 degrees.

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So here's the voltage.
It's got an amplitude of

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10 volts and it has been shifted
to the left by 30 degrees.

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The current has an amplitude of

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0.179 and it has been shifted
to the left 56.57 degrees.

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So the difference, I can
draw this very well.

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I don't know if I can or not.

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That distance right there is

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a phase shift of the current
ahead of the voltage.

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We say that the current
is leading the voltage,

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the current peaks out
before the voltage peaks.

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The current crosses the zero before
the voltage crosses the zero.

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The current is leading the voltage by that

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much which corresponds to
again that 26.57 degrees.

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Just one note about what appears
to be an inconsistency of units.

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Not only does it appear to be
an inconsistency of units,

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it isn't inconsistency of units.

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What are the units of omega,
the radial frequency?

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That's 1,000 radians per second.

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Yet we're specifying the
phase shift of 30 degrees.

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We're specifying it in degrees.

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To be consistent and frankly,

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we're going to leave it in this form
I think simply because it's

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a lot easier for us to visualize.

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We have a better intuitive feel for

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phase shifts in terms of degrees than
we do radians but if wherever we're

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doing calculations we'll need
to convert E to the this into

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degrees per second or
change this to radians.

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Each course the change of radius
is going to just be simply

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the 30 degrees times Pi divided by 180.

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So that phase shift there,

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the 30 degree phase shift represents

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Pi sixth phase there

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and somebody could do the same for
the phase shift and the current.

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But the point is let's
just drive this home.

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This circuit here, the circuit that we just

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analyzed did not change the frequency.

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The frequency of the current flowing in

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here is the same as
the frequency of the voltage.

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What did change was the
amplitude and the phase.

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Using phasor analysis we were
able to determine the phase

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on the amplitude and phase of the current.