Hi, My question is from Chapter 9.3, phasor addition Example 9.3part a. at the e

Question

Hi, My question is from Chapter 9.3, phasor addition Example 9.3part a. at the end of the first part. there is 43.03+j20.64 =47.72*angle*(25.63o) My question is that, how did they find out to take 47.72common in the rectangular form to give the angle. i.e. value of rin the rectangular form Hi, My question is from Chapter 9.3, phasor addition Example 9.3part a. at the end of the first part. there is 43.03+j20.64 =47.72*angle*(25.63o) My question is that, how did they find out to take 47.72common in the rectangular form to give the angle. i.e. value of rin the rectangular form

Explanation / Answer

whenever you are finding a phasor what you need to do is take themagnitude part out as common and the rest is used for calculatingangle for example if you are given a vector A+jB then to express it inphasor notation you take out common A2+B2 thusyou multipy and divide by it to obtain from A+jB = A2+B2 (A+jB)/A2+B2 whereA/A2+B2 is taken as cos and B/A2+B2 is taken as sin respectively in your question 43.032+20.642 = 47.72 also tan = B / A or = 25.63o

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