>> We'll do that same example now only with
the actual values for the circuit elements and the source here.
So again, the first thing we do is convert
the time-domain source to its phasor representation
and then calculate the impedances associated with the capacitor,
resistance and inductor here.
So this the phasor representation V we'll call it
cap V or phasor V is equal to 10E to the J30.
Now, let's be explicit here.
Omega is 1,000 radians per second.
Radians per second.
Z_C is equal to one over J Omega C which is equal to one over
J times Omega which is
1,000 times C which is 10 times 10 to the minus six to six micro,
10 microfarads and that then equals negative J 100 Ohms.
That's a negative J 100 Ohms.
Z_R is just a value of the resistor which is
50 Ohms and Z_L is equal to J Omega L which is equal to J times
Omega which is 1,000 times L which is
75 millihenries and that gives us a positive J 75.
So this is a J 75 there.
This is R which is equal to the impedance that is equal to 50 ohms and this,
the impedance of the capacitor is a negative J 100.
We'll go ahead and define the current
flowing through here or the phasor representation of the current going
through there call it phasor I and let's
write a Kirchhoff's voltage law equation going around that loop.
Once again, adding up the voltage drops.
So, a voltage increase will be a negative,
that would be a negative 10 E to the J 30 plus going across this capacitor,
the voltage drop across that capacitor will
be the current I times the impedance.
So, that will be I times the impedance which is
a negative J 100 plus the voltage drop across the resistor which is going
to be I times 50 plus the voltage drop across that inductor is going to be
I times the impedance of that inductor which is
J 75 and the sum of those terms then equals zero.
We want to solve for I the current in this circuit.
So we'll factor out the I from these common terms
here and bring this term here on over other side and we get then
that I times a negative J 100 plus 50
plus J 75 is equal to a positive 10 E to the J 30.
We have a negative J 100 plus J 75 that gives us a negative J 25 plus 50.
Dividing both sides by that term then gives us that I is equal
to 10 E to the J 30 divided by,
writing the real part first 50 minus J 25.
You get out your calculator.
First of all, let's convert just this denominator here to polar form.
It's a little bit easier to see what's going on here if we do so.
In polar form, that's equal to 55.9 E
to negative J 26.57.
Now, we can see there the division,
it's 10 E to the J 30 divided by 55.9 E to the minus J 26.57.
So 10 divided by 55.9 gives us the magnitude.
The magnitude of the numerator divided by the magnitude of
the denominator gives us 0.179.
Now, the phase E,
it will be E to the J 30 in the numerator,
E to the minus J 26.57 in the denominator.
We've already demonstrated that when dividing numbers involving exponents,
it's the exponent of the numerator minus the exponent in the denominator.
So it's J 30 minus and minus it becomes plus a 26.57 that
gives us a positive J 56.57 degrees.
This entire thing here is the phasor representation of the current.
We can now take this back into the time domain
by recognizing that the amplitude of the current
is going to be the magnitude of this phasor or let's be explicit now.
I of T then is equal to amplitude of 0.179.
It will be a cosine term.
We came in as a cosine will come back to the time domain as a cosine term.
Cosine, the frequency doesn't change that's still
1,000 T. But now the phase of the current is equal to 56.57 degrees.
So plus 56.57 degrees.
That then is the time domain expression for the current.
The voltage, the source driving this circuit is oscillating at
1,000 radians per second and the current is oscillating 1,000 times per second.
It's going back and forth changing at
the same frequency that the source is going into it.
The source has an amplitude of 10 volts.
The current flowing has an amplitude of 0.179 Amps.
So, the amplitude of the current is smaller than the amplitude of
the voltage and finally let's look at the phase.
The phase angle of the voltage source is 30 degrees, positive 30 degrees.
The angle of the current is 56.57 degrees.
So, we'd say then that the current has been shifted ahead
of the source by the 26.57 degrees.
Here are plots of those,
the voltage in the red,
V of T is equal to 10 cosine of 1,000 T plus 30 degrees.
I of T we have now found to be equal to 0.179,
cosine of 1,000 T plus 56.57 degrees.
So here's the voltage. It's got an amplitude of
10 volts and it has been shifted to the left by 30 degrees.
The current has an amplitude of
0.179 and it has been shifted to the left 56.57 degrees.
So the difference, I can draw this very well.
I don't know if I can or not.
That distance right there is
a phase shift of the current ahead of the voltage.
We say that the current is leading the voltage,
the current peaks out before the voltage peaks.
The current crosses the zero before the voltage crosses the zero.
The current is leading the voltage by that
much which corresponds to again that 26.57 degrees.
Just one note about what appears to be an inconsistency of units.
Not only does it appear to be an inconsistency of units,
it isn't inconsistency of units.
What are the units of omega, the radial frequency?
That's 1,000 radians per second.
Yet we're specifying the phase shift of 30 degrees.
We're specifying it in degrees.
To be consistent and frankly,
we're going to leave it in this form I think simply because it's
a lot easier for us to visualize.
We have a better intuitive feel for
phase shifts in terms of degrees than we do radians but if wherever we're
doing calculations we'll need to convert E to the this into
degrees per second or change this to radians.
Each course the change of radius is going to just be simply
the 30 degrees times Pi divided by 180.
So that phase shift there,
the 30 degree phase shift represents
Pi sixth phase there
and somebody could do the same for the phase shift and the current.
But the point is let's just drive this home.
This circuit here, the circuit that we just
analyzed did not change the frequency.
The frequency of the current flowing in
here is the same as the frequency of the voltage.
What did change was the amplitude and the phase.
Using phasor analysis we were able to determine the phase
on the amplitude and phase of the current.