# Quick question, phasors and impedance

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1. Nov 8, 2014

### clurt

On line S=0.5*V*I why is the minus sign removed from in front of the "I" angle. 25.38 specifically. Could is be a typo?

http://imgur.com/eUmEovo

2. Nov 8, 2014

### ZetaOfThree

It's not a typo. $I$ is complex conjugated, hence the star $I^*$. When you complex conjugate a phasor, you change the sign of the angle.

3. Nov 8, 2014

### clurt

Do you always flip the minus sign when when using S=VI? And what about the half in front, all the time too?

4. Nov 9, 2014

### zoki85

Half in the front when you use peak values of V and I

5. Nov 11, 2014

### Formagella

if the formula is $$S = \frac{1}{2} \cdot V_s \cdot I^*_s$$ so you take $I_s^*$ which is the complex conjugate of $I_s$. Since it's written in a modulus and phase format, you just have to invert the sign of the phase of $I$ to obtain it (draw both numbers in the complex plane and it's evident that the modulus stays the same but the phase changes).
I don't think you should memorize rules of flipping that sign or not, just read the formula (it's S = VI*, not S = VI). If you weren't able to see why the minus sign changed given the formula, then you either don't know the notation used here for the complex conjugate (in mathematics classes I've usually seen it written as $\overline{I}$, but it's written the other way in engineering) or you need to study complex numbers.