L7 3 3 Example 1 with numbers

>> 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.