When are we going to append the "j" term when analyzing ac circuits. When i say j term, im pertaining to opposition to current/impedance. In the image below, notice that Xl is not written as jXl. When are we going to use the j term and when are we not going to use it?



"notice that Xl is not written as jXl"

You've answered your own question. The problem is that \$X_L\$ is not written as \$j X_L\$. :-)

While it also has the dimension Ω, just like impedance, \$X\$ is the symbol for reactance, not impedance, that would be \$Z\$.

\$ Z_L = R + X_L = R + j \omega L\$

The factor \$j\$ indicates a 90° phase shift with respect to the resistance which does not have a phase shift. For an inductance the phase shift is positive, for a capacitance it's negative, since

\$ Z_C = R + X_C = R + \dfrac{1}{j \omega C} = R - j \dfrac{1}{\omega C}\$

It's because of the 90° angle between resistance and reactance that you can't add them arithmetically, like 5 Ω + 12 Ω = 17 Ω. \$X\$ and \$R\$ create a right angled triangle, so you have to apply Pythagoras to find the magnitude of the impedance:

\$ |Z_L| = \sqrt{|R|^2 + |X_L|^2} \$

so for \$R\$ = 5 Ω and \$X_L\$ = \$j\$ 12 Ω we get a magnitude of 13 Ω instead of 17 Ω.

Right, back to your calculations. The reason it seems to go in the right direction, even without \$j\$ is that you applied Pythagoras implicitely in the first line already; you wrote

\$ V_S^2 = (V_1 + V_R)^2 + V_L^2 \$

and not

\$ V_S^2 = (V_1 + V_R + V_L)^2 \$

or simply

\$ |V_S| = |V_1 + V_R + V_L| \$

That's why you won't see \$j\$ anymore; you got rid of it at the start.

  • \$\begingroup\$ I think your answer is more clear than my one. Please be free to downvote my post if necessary. I'm just a CSE Dip student but I'm sure I'm a pretty good enthusiast , I did spend whole day designing and reviewing and troubleshooting circuits. I need your dowvote if necessary. thank you. \$\endgroup\$ Oct 19 '12 at 11:56
  • 1
    \$\begingroup\$ @sandun - thanks, but I don't downvote an answer because another one would be better. For me downvoting means "this is pretty bad", and yours isn't. "better" or "worse" is in the eye of the beholder. The asker can choose which one answers best his question. \$\endgroup\$
    – stevenvh
    Oct 19 '12 at 13:33
  • \$\begingroup\$ Isn't it \$Z_C = jX_C = -j\frac{1}{\omega C}=\frac{1}{j \omega C}\$? \$\endgroup\$ Oct 19 '12 at 22:23
  • \$\begingroup\$ @Alfred - Yes, of course. (I'm an idiot) \$\endgroup\$
    – stevenvh
    Oct 20 '12 at 8:06

Impedance is a complex number and, like any complex number, can be expressed as the sum of a real number and an imaginary number.

\$Z = R + jX\$

The real part of \$Z\$ is \$R\$ while the imaginary part of \$Z\$ is \$X\$.

The crucial point here is that \$X\$, like \$R\$, is real.

We multiply \$X\$ by \$j\$, the imaginary unit, when we combine the parts into a complex number.

In AC analysis, \$Z\$ is the impedance, \$R\$ is the resistance and \$X\$ is the reactance. For emphasis, note that the reactance is a real number.

To make the distinction even more clear:

\$X\$ is a reactance

\$jX\$ is an impedance, a complex number, with zero real part, \$R = 0\$

Finally, capacitive and inductive reactance are:

\$X_C = -\dfrac{1}{\omega C}\$

\$X_L = \omega L\$

Capacitive and inductive impedance are:

\$Z_C = -j\dfrac{1}{\omega C} = \dfrac{1}{j \omega C} = j X_C\$

\$Z_L = j \omega L = j X_L\$


Please note that X1 is just a magnitude and jX1 is a vector.

This comes from a phasor diagram.There Vl, Vs and V1 means the magnitude only. This equation is come from vector addition equation , where, if a and b are two vectors and | | is conventional notation for vector's magnitude, and when teata is the angle between a and b.

|a+ b|^2 = |a|^2 + |b|^2 - 2 |a||b| cos ( teata )

When the teata is 90deg, cosine(teata) becomes zero and this becomes,

|a+ b|^2 = |a|^2 + |b|^2

sound familiar now? So j or i does not come for magnitudes. For a example |jX1| = |X1| , that's the reason.

Hope you understand.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.