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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.

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  • spunky sam orange style avatar for user Yamandu
    ELI the ICE man is awesome.
    (10 votes)
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  • blobby green style avatar for user Guy .
    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)
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    • leaf green style avatar for user Anurag Singh
      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)
  • piceratops seed style avatar for user gowtham
    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)
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  • purple pi teal style avatar for user Michelle Grove
    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?
    (3 votes)
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  • piceratops ultimate style avatar for user fernando.lopez.ele304
    So, inductors and capacitors are actually wave filters... isn't it?
    (4 votes)
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  • orange juice squid orange style avatar for user Michael Country
    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)
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    • purple pi purple style avatar for user APDahlen
      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)
  • leafers ultimate style avatar for user J05hu4 6r33n
    I have a friend called Eli who was in the same building as me while i was watching this video. 🤣🤣🤣
    (3 votes)
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  • blobby green style avatar for user LawrencefSu
    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)
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  • leaf orange style avatar for user 🌿
    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)

    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 ;
    (1 vote)
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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.