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I have just started to learn about AC network analysis and have some questions about "j" (or "i" on my calculator), the imaginary unit. My book doesn't go into a great deal about this, and jumps right into formulas and substitutions (more practical approach, not theoretical). So, what exactly does J represent?

I see that if I draw a complex-plane (y-axis being imaginary, x-axis being real), and draw a unit circle on it, a 90° angle is \$\sqrt{-1}\$, which is "j". I see that I can use this substitution in phasor form when, say, solving for the voltage across a capacitor when the current through it is known:

$$V = \frac{I}{j \omega C}$$

Can someone help me understand this?

To be honest, this question is pretty vague because I'm not even sure how to ask about what J is; it's that foreign to me. I would like a common-sense explanation (big-picture) of it's meaning and purpose in AC circuit analysis. I'm not necessarily looking for a rigorous mathematical explanation (although any necessary mathematical explanation is welcome).

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Algebra is case-sensitive. J and j are different things. –  TRiG Jul 30 '14 at 16:49
You might want to look at the questions under the complex-numbers tag on math.SE: math.stackexchange.com/questions/tagged/… –  The Photon Jul 30 '14 at 19:07
Of course what you find on math.SE will leave open the really interesting question: Why are complex numbers useful in engineering? –  The Photon Jul 30 '14 at 19:09
@The Photon: The answer is on Wikipedia: en.wikipedia.org/wiki/Phasor I can summarize it here, but given the dynamics of voting on SE sites, it would be "wasted bullets". –  Respawned Fluff Jan 13 at 18:03
@RespawnedFluff, Are you responding to what you wanted to respond to? –  The Photon Jan 13 at 18:04

3 Answers 3

up vote 5 down vote accepted

If you put a minus sign in front of the number "5" it becomes "-5".

Try and look at this differently. Try thinking that it rotates the number "5" (tied to the origin by a piece of string of length 5) through 180 degrees to become "-5"

OK so far? Negative signs are the same as rotating thru 180 degrees...

Why not extend this further to produce something you can "stick" in front of a positive number that rotates it thru 90 degrees - in EE this is usually called "j" and it acts to rotate a value (about the origin) thru 90 degrees counter-clock wise i.e. if you did it twice (j*j) you'd get 180 degrees ("-").

From this gem of knowledge you can therefore say j*j = -1 therefore, j = \$\sqrt{-1}\$

Just as a minus sign can rotate any positive value thru 180 degrees it can rotate any vector or phasor thru 180 degrees. The same applies to the j operator - it rotates any vector or phasor thru 90 degrees counter clockwise.

EDIT - forgot part of question: -

substituting j into the impedance of a capacitor. Remember the basic formula for a capacitor is Q=CV and therefore differentiating the variables we get: -

\$ I = \dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$

This tells us that for a sinewave applied voltage across a capacitor, the current will also be a sinewave but differentiated into a cosine like this: -

enter image description here

If you tried to calculate the impedance (V/I) of a capacitor from the V-I relationship you'd get into trouble because when I passes thru zero, V is NOT zero so you get infinities. If on the other hand you apply a "j" to bring current in phase with voltage the math works out fine - current and voltage are aligned and impedance based on instantaneous values of V/I makes sense.

I'm aware that you are just starting out so I've tried to keep this both accurate and simple (maybe too simple for some?).

If you look at the inductor, the "j" can be applied to the voltage to align it with the current hence "j" is in the numerator for inductive reactance and j is in the denominator for capacitive reactance. There are subtleties lying around here that hopefully will make sense as you learn more - it's actually no coincidence that "j" appears to "follow" omega when it comes to impedances - my explanation doesn't cover that and neither does your question!

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I found your answer to be very helpful, especially with your mention of using j to bring the waveforms in phase; this helped me understand its use because I remember that voltage leads current by 90* for pure inductance, and vice-versa for pure cap. Thanks! –  eestack Jul 31 '14 at 5:08

In pure maths we use \$ i \$ to represent the prime square root of \$-1\$.

The other square root of \$-1\$ being \$-i\$.

If you imagine a number line with real numbers placed horizontally. We can now add a second number line going vertically containing the imaginary numbers.

We have now created a system of complex where every point on the plane is represented by a real and imaginary part e.g. \$ 4 + i \ 3 \$ represents a point that is 4 units along the real axis and 3 units up the imaginary axis.

Because a point in two dimensional space can now be represented as a single number, calculations involving 2-dimensional vectors are simplified.

In electronics, when considering systems supplied by a single frequency sine wave, we are taught initially to draw phasor diagrams. Then later to use complex numbers to get to deal with these problems.

We also use \$ j \$ instead of \$ i \$ but the meaning is identical. It’s just to avoid confusion because in electronics \$i\$ is often used for current.

If you would like a little more insight take a look at this question: What are imaginary numbers? from the Mathematics Stack Exchange site.

Or take a look here: A Visual, Intuitive Guide to Imaginary Numbers.

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Thanks for your help and references to some additional reading! –  eestack Jul 31 '14 at 5:09

In mathematics imaginary unit is a very helpful number used to solve equations with higher than 2 order. It was introduced just.... to the test, and it works pretty until today. This provides for obtaining at least one root in every polynomial.

In electronics imaginary unit represents the energy stored in our circuit. So, in capacitor, it is the energy stored in it. It also represents phase shift in circuit, when we are dealing with sinusoidal signals.

I think you should more precise your question, or just write questions which bother you in points.

For example... If your circuit's impedance will be represented only by imaginary unit, not by real, your bill for energy will be... zero :)

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