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I am currently working on a problem of determing the steady-state value of v using nodal analysis. I understand all the steps up until the step of converting my j6 / -1-j to the phasor value that I circled within the red rectangle below.
Could someone give me some insight on what was achieved here and why?
Well, it is just converting the numerator and denominator from complex numbers to magnitude-phase (phasor) notation, which is really short-hand of imaginary exponential.
Numerator
j6 is a vector with no real component (only has imaginary component). So in phasor terms, the magnitude is 6, and the angle is 90° (if it had been negative (-j6), the angle would have been -90°). The angle is measured positive in the anti-clockwise direction, with respect to the positive real axis.
$$ j6 = 6\angle{90°} $$
Denominator
-1-j has real and imaginary components equal to -1. So from the origin it looks like an arrow pointing to the bottom-left in the complex plane. So the angle is -135°, and the magnitude is \$(1^2+1^2)^{1/2} = \sqrt{2} \$
$$ -1-j = \sqrt{2} \angle{-135°} $$
To answer the question in the comments of why \$-j=\frac{1}{j} \$:
When you divide phasors, the resulting magnitude is the quotient of the magnitudes, and the resulting angle is the difference between the angles.
$$ -j = 1\angle{-90} = \frac{1\angle{0}}{1\angle{90}} = \frac{1}{j} $$
-j = 1 / j. Would you mind explaining this as well?
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– theGreenCabbage
Jun 8 '14 at 21:16
So from the origin it looks like an arrow pointing to the bottom-left in the complex plane. So the angle is -135° How did you determine this precise angle?
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– theGreenCabbage
Jun 8 '14 at 21:26
e^01.439*i.
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– theGreenCabbage
Jun 8 '14 at 21:37
