0
\$\begingroup\$

Are there any mathematicians in here? I am currently reading my ECE text's chapter on AC machines and i referred to some material in the chapter that first discussed AC circuits.

In the photo below of the math, I am unsure of how on the right hand side, the subtraction of two phasors became a multiplication of 2 phasors... Specifically, following the red arrow, how did the term furthest to the right go from the form it had I*<150-theta to (1 - 1*<120) ??

enter image description here

I do not recall seeing this anywhere else in the chapter or on the review of complex arithmetic in the appendix and it is sort of bugging me. I am not seeing how they got from that line to the next line in the solution process.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ you should ask this in math.stackexchange.com site, not here.That forum is totally dedicated to mathematical problems. \$\endgroup\$
    – Anklon
    Jun 21, 2016 at 22:40

2 Answers 2

1
\$\begingroup\$

First of all, multiplication of two complex number is multiplication of there magnitude and addition of there angle. like this: $$ I_{\Delta}\angle150 ^{o} =I_{\Delta}\angle120 ^{o}\times 1\angle30 ^{o} $$ Now: $$ (I_{\Delta}\angle30^{o}-\theta ^{o})-(I_{\Delta}\angle150 ^{o}-\theta ^{o}) =(I_{\Delta}\angle30^{o}-\theta ^{o})-(1\angle120 ^{o} \times (I_{\Delta}\angle30 ^{o}-\theta ^{o})) =(I_{\Delta}\angle30^{o}-\theta ^{o})\times(1-1\angle120 ^{o}) $$

\$\endgroup\$
1
\$\begingroup\$

All they are doing is factoring out the common term. The first term is the common term. The second term has a 1 from the first factor and the second part is simply shifting the angle from 150 to 120 degrees so it becomes common with the common term which has an angle of 30 degrees.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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