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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. • 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? • 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 :)
• 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. • I get most of this, but I don't quite understand why 1 / j = - j (at in the video). Why doesn't it work like a normal reciprocal? • So, inductors and capacitors are actually wave filters... isn't it? • 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? • 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.

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...

• I have a friend called Eli who was in the same building as me while i was watching this video. 🤣🤣🤣 • 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 you should explain about that part at . 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)

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.

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

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 ;
(1 vote) 