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I understand the concept that the voltage lags behind the current in a capacitor, which is why we have to use imaginary numbers to find the impedance of a simple Series RC-circuit like this:

RC Series circuit

Which gives the following vector diagram:

vector Diagram of RC circuit

What i dont understand, however, is why you have to multiply by -j and not j when calculating the capacative reactance. Based of my limited understanding of the complex plane and rotation, multiplying by j would rotate the reactance 90 degrees counter-clockwise, bringing it in phase with the regular resistance. How come this is not the case?

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  • \$\begingroup\$ Rotation in the standard mathematical circle is counter-clockwise, not clockwise. The reason is arbitrary, but probably because rotating a +x by 90 degrees to a +y made more sense than the other choice. Because of the way that euler's works, \$e^{i\:\theta}=\operatorname{cos}\left(\theta\right)+i\:\operatorname{sin}\left(\theta\right)\$, the rest just falls out for you. See Use of Complex Impedance. Also, (1,0) becomes (cos θ, sin θ), when rotating counter-clockwise. Convenient. \$\endgroup\$ – jonk Apr 12 '18 at 17:00
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    \$\begingroup\$ Hi, Jeppe. I sincerely think you should probably read "Theory and Calculation of Alternating Current Phenomena," 3rd edition, by Charles Proteus Steinmetz (and Ernst J. Berg), 1900. Chapter IV discusses the graphical (and more intuitive) approach. It then transitions into WHY the algebraic method, the one you are asking about and which is covered in Chapter V, was selected to replace the graphical methods. The graphical details get "quite busy" on the paper, very quickly. Yet if you learn the algebraic approach, you are much less likely to make mistakes. But graphics is much more intuitive. \$\endgroup\$ – jonk Apr 13 '18 at 0:04
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Based of my limited understanding of the complex plane and rotation, multiplying by j would rotate the reactance 90 degrees clockwise

This is incorrect. When measuring angles a positive angle is counterclockwise, not clockwise. This is the case in basic trigonometry/geometry as well.

In an RL circuit you would multiply by j.

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  • \$\begingroup\$ Apologies, that was a typo. I have edited the post with the correction. With this correction though, why does multiplying by j not rotate the reactance counter-clockwise, bringing it in phase with the resistance? If you multiplied by -j, doesn't that just bring the reactance 180 degrees out of phase? \$\endgroup\$ – Jeppe Apr 12 '18 at 17:17
  • \$\begingroup\$ The resistance does not have a complex (reactive) component. No amount of multiplying a real number by j will give you a real number. The reactive component will ALWAYS be on the "j" axis \$\endgroup\$ – DerStrom8 Apr 12 '18 at 17:28
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If you multiply a real variable with j it would be in the plane on the imaginary axis. so multiplying with j turns it counterclockwise, -j turns it clockwise.

Why it is -j by the capacitor is because $$Z_C=\frac{1}{j \omega C}$$ multiplying with $$\frac{j}{j}$$ generates a -j in the numerator. Z is the so called impedanze (ac-resistance) of a component the others are $$Z_L=j \omega L$$ and $$Z_R=R$$. so your total impedance is $$Z=R-\frac{j}{\omega C}$$.

You can calculate the impedance with the laplace transformation, like i made it in my picture.

Hope i could help

enter image description here

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