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### Course: Electrical engineering > Unit 2

Lesson 5: AC circuit analysis- AC analysis intro 1
- AC analysis intro 2
- Trigonometry review
- Sine and cosine come from circles
- Sine of time
- Sine and cosine from rotating vector
- Lead Lag
- Complex numbers
- Multiplying by j is rotation
- Complex rotation
- Euler's formula
- Complex exponential magnitude
- Complex exponentials spin
- Euler's sine wave
- Euler's cosine wave
- Negative frequency
- AC analysis superposition
- Impedance
- Impedance vs frequency
- ELI the ICE man
- Impedance of simple networks
- KVL in the frequency domain

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# ELI the ICE man

Remember the relationship between current and voltage in an inductor and capacitor with the help of ELI the ICE man. Created by Willy McAllister.

## Want to join the conversation?

- ELI the ICE man is awesome.(14 votes)
- Hello,

In developing impedance for inductor or capacitor you based on the fact that the input signal is e^(jwt) but when you are developing relationship in this video the input is cos(wt) and when deriving the relationship for V/I on inductor for example with are current of the form cos(wt) we get V = -Lw*sin(wt) --> V/I = -Lw*tan(wt) which is not jwL can you please explain why you are still using jwL - what am i missing?(6 votes)- You have taken current waveform as I = 1*cos(wt), assuming amplitude (or peak) of current is equal to 1, and it's not shifted ( since phi = 0). Now calculating V = - L*di/dt , we get V = (-L)
**w**(-sin(wt)) = wL*sin(wt).

It may be obvious that sin(wt) is nothing but cos(wt) shifted by 90 degs to the right (you can refer the graphs of both). Therefore sin(wt) = cos(wt - 90).

Thus, finally we get V = wL*cos(wt - 90).

Now to keep things simple, note that multiplying by complex factor 'j' also shifts the waveform or phasor (to be exact) right by 90 degs. Hence, cos(wt - 90) = j*cos(wt).

From here you can get, V/I = jwL.

P.S. I have not involved complex phasor algebra which anyone explaining this would have possibly done, only to keep things straight. You may refer to that topic if you want.

Hope that helps :)(5 votes)

- So, inductors and capacitors are actually wave filters... isn't it?(5 votes)
- Yes. Inductors and capacitors act as frequency filters.(6 votes)

- I don't understand how the multiplication of j with the cosine function causes it to shift by 90 degrees. In earlier videos we saw that complex numbers would rotate 90 degrees on the complex plane. I'm finding a lot of trouble reconciling that with this situation, because this is a normal plane.

In the complex plane, the real and imaginary parts of the complex number would keep alternating and switching signs because of multiplying j and that made sense, but I'm finding it difficult to get how the cosine function on its own is affected in any way by the multiplication of j.(7 votes) - I get most of this, but I don't quite understand why 1 / j = - j (at7:30in the video). Why doesn't it work like a normal reciprocal?(3 votes)
- I'm a bit lost about the concept of a phasor. Is there more on this, in Khan Academy, or can you speak to what a phasor is exactly? Is the magnitude of the phasor in the real "axis" and the angle a rotation towards the imaginary "axis" in a complex plane? Or is there a better way to think of it?(3 votes)
- Hello Michael,

A phasor is a shortcut!

It is a shorthand way of describing a sinusoid. Instead of writing:

v(t) = 170COS(2*pi*60t + 30°)

we write:

170 ∠ 30° V_peak

or

120 ∠ 30° V_RMS

Know that phasors are very handy to show the relationships between voltage and current. This is especially important when the circuit has reactive components such as inductors or capacitors. For example, a motor will likely have an inductive component (coil of wire and such). The current phasor will lag the voltage phasor.

Please leave a comment below if you have any questions.

Regards,

APD

P.S. The phasor notation makes the math simple! For example, to calculate the power consumed by a motor you can multiply the RMS voltage and the RMS complex conjugate of the current phasor. The real part of this complex number in the real power consumed by the motor. No worries, this is much easier to do than it sounds...

Ref: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-ac-analysis/v/ee-eulers-sine-wave(5 votes)

- I have a friend called Eli who was in the same building as me while i was watching this video. 🤣🤣🤣(5 votes)
- i think you should explain about that part at4:45. about adding 90° to phasors .

it's so confusing

i spent lot of time and i think it's :

instead of assuming Acos(wt+phi) to be current let's assume e^jwt to be current so jwL*e^jwt = Voltage = e(notation used in video)

So basically when we multiply j*e^jwt , it's equivalent to :

[jsin(wt)+ cos(wt)]* j

which is equal to -sin(wt)+ j(cos)

Now take like a paper and plot a complex plane of e^jwt

and j*e^jwt .

if we plot they would have rotated 90°

Now basically in EE they want to assume current is cos()

So cos = Re(e^jwt)

Re() is taking the real part of imaginary number

Anyways take the real part of j*e^jwt and you'll get -sin(wt)

guess what -sin(wt) = cos(wt+90°)

go on google and type this:

-sin(x) and cos(x+1.5708)

1.5708 rad is 90°

you'll see that

so we can say if current is e^jwt then cos(x+90°)= Re(j*e^jwt)

You get what i mean , kinda lazy

so instead of directly plugging cos(x) as current they plug e^jwt and take the real part after multiplying by j.

it took me days to understand.

also about willy putting :

jwLCos() instead of jwL*e^jwt

bcs think about . jwL is the impedance only for imaginary current went through somehow .

you can't have a real jwL impedance

so why did he mix a real e^jwt with cos()

ie : jwL*e^jwt ✅

---> jwL*Acos(wt)❓❗

this is good enough

but about that phasor part;i

basically a phasor is a way to represent a sinusoid function . it doesn't differentiate between cos() and sin() and tan(). it can also be used for other purposes like vectors

but it's the interpretation of the context that help know what they mean in that context

2 ways to represent a phasor; to draw it or it's notation.

the notation only gives you the magnitude and angle . but from that you deduce the length and widht

the phasors are generally in complex plane .

if i want to draw Acos(x+5°) = A∠5°.

again they don't really mean anything

but if you draw them then it's in a x-y plane ;there should be an arrow that has length A and angle 5° from the x line.

just type on gooogle phasor arrow

now traditionally they consider the y axis to mean the ***j*** and x to mean the real axis

it doesn't have any real meaning but nvm

if you multiply that phasor that somehow represent a normal cos() and also a complex number .

it would mean to multiply the complex number by j: A∠5°.*j

so remember the arrow that we drew, it's gonna rotate 90° and now the angle of the arrow from x axis is 90°+5°

and this arrow had the interpretation to mean a Acos(angle)

meaning the angle of the arrow is the ''angle'' in cos.

cos(x) = phasor0° but jcos(x) = j*phasor+0° = phasor+90° = cos(x+90°)

jcos(x) = cos(x+90°) ?? <-- this doesn't make much sense

but again cos(x) was supposed to be e^jwt and take the real part

hope it helped try to graph in 3d real axis , imaginary axis , and time ;(2 votes) - Hi everyone, I have a question: in order to get rid of the j in e=jwLI0cos(wt+symbol) we multiply j and shift the degree to +90 degree. Wouldn't that becomes je instead of just e now? I always have trouble understanding complex number and I would look it up more but how do we explain it in this case? Why multiplying j has no effect to the left side of the expression e=jwLI0cos(wt+symbol)?(1 vote)
- I think we didn't multiply the "j", since the imaginary unit "j" denotes as a 90 degree phase shift we can just simplify it to "e=ωLIo∠(ϕ+90)".(1 vote)

## Video transcript

- [Voiceover] Okay, it's
time to introduce you to a new friend, ELI the ICE man. ELI the ICE man is a friend
of every electrical engineer and what we've been taking
about is A/C analysis, and in A/C analysis we
limit ourselves to one type of signal and that's a sinusoid, and the sinusoid we like is called cosine. We say cosine of omega t plus phi. Omega represents the radiant
frequency of the cosine, here it's shown in blue. That radiant frequency is omega, and phi is the phased
layer, the phased shift. And if we look here we see this
isn't really a cosine wave, because the peak is a little
before zero, time equals zero, so this distance right here
is the lead, the phase leed, and that's phi. So when phi is a positive
number, this whole cosine wave is shifted a little bit to the left. That's what we mean by phase shift. So in these kind of signals, our input into our favorite
components, we're gonna get a relationship between the
voltage and the current in those components, and it's
related by the impedance. We define the idea of
impedance as the ratio of voltage to current, we
gave that the symbol z. Now in this video, instead of using v as my variable for
voltage, I'm gonna use a different letter, I'm gonna use e. E is short for EMF or electromotive force, and it's really commonly
used, almost as often as v, for representing voltage,
and I'll show you why I wanna use e in a little bit. And another way I can
write this just as easily, e equals z times i. And this looks a lot like
Ohm's law, and what we're gonna find out here is we can
apply this in addition to applying it to resisters, we can apply it to capacitors and inductors. So first off we're gonna look
at our friend, the inductor. And we're gonna look at the equation e equals z i for an inductor. I'm going to assign i to be
a sinusoid, so i is gonna be equal to some magnitude,
we'll call it i naught, cosine omega t plus phi. So I'm gonna say my
current is a cosine wave of this magnitude with this
phase delay, and that's shown in blue here, so this here is i. And now let's write e
in terms of this i here, so I can write e equals z,
now what is z for an inductor? The impedance of an inductor
is j omega l, and what is i? i is sitting right here,
and I'm gonna represent i like this, I'm gonna
represent i as a phasor, or a phasor representation. And we said that that
could be represented as i, the magnitude of the current
indicated at the angle of phi, so these are equivalent
representations of i, this is the time domain representation, and this is the phasor representation. Now what we have out here in
front of i is a scaling factor, there's this complex j that
we'll take care of in a second, and there's omega l. So omega is the frequency and
l is the size of the inductor. Now for the purposes of
this video, when I plot out the voltage over here in
orange, we're gonna assume that this scaling factor omega l
is one, just so that we can focus on the timing relationships between the current and the voltage. When we talked about complex
numbers, multiplying by j, multiplying something by j represents a rotation of plus 90 degrees. And so I can write this as e equals, let's put the scaling factor
out there, and we'll have i naught, which is the original
magnitude of the current, and phi, it's changed here,
phi changes, phi becomes phi and this multiplication
by j here corresponds to adding 90 degrees to phi. So multiplying by j corresponds
to a 90 degree phase shift, and if I draw here, this is
now e, and the phase shift, we decided this distance right here, this distance right here
is phi, and this distance right here is a phase leed of 90 degrees. And you'll notice I key off
the peaks of these wave forms, because that's the easiest
place to see the leed. So when I move to the left,
that corresponds to a lead of plus 90 degrees. So in an inductor, in an inductor, we say that e leeds i by 90 degrees. Alright, now let's do
it for our capacitor. And we'll do the same kind
of thing here for capacitor we'll assign the same
current, we'll say i equals some current, i naught, times
cosine of omega t plus phi, and now let's work out the
voltage across the capacitor. So the voltage across the
capacitor, e, is the same thing we have here, e equals zi or I can write e in the capacitor equals z. Now what is impedance of a capacitor? It's one over j omega c. That's z, and i we represent
the same way as we did before, i naught at an angle of phi. So now let's carefully
do this multiplication. e equals one over j times one
over omega c times i naught at an angle of phi. So here's this one over j term, now I can rewrite one over j as minus j. Now we're multiplying
something by minus j, and multiplying by minus j
corresponds to a rotation of minus 90 degrees, so I
can write e one more time like this, e equals one over omega c, here's the scale factor,
here's the original current magnitude, and
I get the angle of phi, this time minus 90 degrees. So this minus sign here
corresponds to a lag, a phase lag, so here's our original current
here, let me label that, here's i, and now we have our
voltage, e looks like this, here's e, and what we
see, let me go out here and measure it here,
here we have a phase lag, where it's pointing to
the right of 90 degrees. And that we call a lag. We can summarize that, we
can say in a capacitor, we say e lags i. And an equivalent way to say this is we could say that i leeds e. I leeds voltage. So I can actually put boxes
around these two results, here and here. Now there's a lot of sign
flipping going on here, and there's actually an
easy way to remember this, and I wanna introduce you
to someone who can help you remember this, and his
name is ELI the ICE man. So what can Eli tell us? ELI tells us that in an
inductor, an l, voltage, leeds current, and over
here in a capacitor, c, current leeds voltage. That's the message from ELI the ICE man, he helps us remember the
order that voltage and current change in inductors and capacitors. He's gonna be your friend for a long time.