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I was reading a textbook and came across the following

$$ \\I_1 = Ae^{j0} = A \\I_2 = Be^{-j\frac{\pi}{2}} \\\bf{I} = I_1 + I_2 \\If\ A=10\ and\ B=20 \\|I| = \sqrt{A^2 + B^2} \\\angle I = -tan^{-1}(B/A) \\I = 27.98e^{j30.4^{\circ}} $$

From my knowledge, of complex numbers, shouldn't \$\angle\bf{I} \$ be \$\angle \bf{I} = tan^{-1}(B/A)\$ since both B and A are in the first quadrant?

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  • \$\begingroup\$ You can't just use the values of A and B since they are magnitudes of the phasor. They will always be positive since they are magnitudes. \$\endgroup\$
    – DKNguyen
    Commented Feb 1, 2021 at 6:25

2 Answers 2

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I = A - jB is a complex number and it is in the fourth quadrant. So argument of I must be negative. Mind that I_2 has angle -pi/2.

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Positive x and negative y means Quadrant IV.

enter image description here

\$\vec {I_1} = 10\ ∡ 0^\circ\ A\$

\$\vec {I_2} = 20\ ∡ - \frac {\pi} {2}\ A = 20\ ∡ - 90^\circ\ A \$

\$\vec {I_1} + \vec {I_2} = 10 - j20\ A = 22.36\ ∡ -63.43^\circ\ A \$

This does not agree with your \$ 27.98\ e^{j30.4^\circ} \$, but that is the math.

y is larger than x, so angle has to be greater than \$45^\circ\$.

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