Adding & dividing phasors: 123∠20°+69∠-10°=?, (35+j50)/(13-j7)=?

[jbjaidee]

Adding & dividing phasors: 123∠20°+69∠-10°=?, (35+j50)/(13-j7)=?

Perform the operations on the phasors shown below: (Answers must be in the same format as the original problem.)

123∠20°+69∠-10°=

(35+j50)/(13-j7)=

Thanks for the help.

 

[Deleted member 4993]

Perform the operations on the phasors shown below: (Answers must be in the same format as the original problem.)

123∠20°+69∠-10°=

(35+j50)/(13-j7)=

Thanks for the help.

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for

 

[Dr.Peterson]

Perform the operations on the phasors shown below: (Answers must be in the same format as the original problem.)

123∠20°+69∠-10°=

(35+j50)/(13-j7)=

Thanks for the help.

The easiest way to add is to convert to rectangular form, add components, and convert back.

The easiest way to multiply or divide is in polar form: add/subtract angles, multiply/divide lengths.

Have you tried? Surely your lessons demonstrate the process, so you have something to work from? We need to see what you can do in order to help. Just giving examples (as a book can do) is not enough.

 

[Deleted member 4993]

Perform the operations on the phasors shown below: (Answers must be in the same format as the original problem.)

123∠20°+69∠-10°=

(35+j50)/(13-j7)=

Thanks for the help.

For problem #2, you can treat the phasors as complex numbers like a + ib.

and then multiply by (a - ib)/(a - ib)

 

[jbjaidee]

Ok so for the phasor here is what I did:

123∠20°+69∠-10°=

123cos(20)=115.58 69cos(-10)=67.95
123sin(20)=42.07 69sin(-10)=-11.98

116+j42 + 68-j12 = 184+j30

Sqrt(184^2+30^2)=186.43
Tan-1(30/184)=9.26

186∠9°=

The second one I had trouble with.

(35+j50) / (13-j7)

Sqrt(35^2+50^2)=61.03 Sqrt(13^2-7^2)=10.95
Tan-1(50/35)=55.01 Tan-1(-7/13)=-28.30


61∠55° / 11∠-28°= 6∠-2°

6cos(-2)=5.99
6sin(-2)=-.21

6-j.2=

I don't think that is the correct answer, if you can shed some light that would be great.

 

[Dr.Peterson]

Ok so for the phasor here is what I did:

123∠20°+69∠-10°=

123cos(20)=115.58 69cos(-10)=67.95
123sin(20)=42.07 69sin(-10)=-11.98

116+j42 + 68-j12 = 184+j30

Sqrt(184^2+30^2)=186.43
Tan-1(30/184)=9.26

186∠9°=

Looks good. I also mentally sketched the vectors, and expected something around 190∠10°, so this looks even better.

The second one I had trouble with.

(35+j50) / (13-j7)

Sqrt(35^2+50^2)=61.03 Sqrt(13^2-7^2)=10.95
Tan-1(50/35)=55.01 Tan-1(-7/13)=-28.30


61∠55° / 11∠-28°= 6∠-2°

6cos(-2)=5.99
6sin(-2)=-.21

6-j.2=

I don't think that is the correct answer, if you can shed some light that would be great.

First, 61/11 = 5.545; you shouldn't round until the end, assuming you were told the answer should be rounded, because early rounding leads to cumulative errors. Even better, 61.03/10.95 = 5.57.

Second, you appear to have divided the angles; lengths are divided, angles are subtracted: 55.01 - (-28.30) = 83.31°.

As Khan mentioned, you can also divide complex numbers as complex numbers, avoiding the need to convert twice in this case (though the work is somewhat harder in itself):

Given (35+j50) / (13-j7), we multiply numerator and denominator by the complement, (13+j7):

(35+j50)(13+j7) = 455 + j245 + j650 + 350j^2 = 455 + j245 + j650 - 350 = 105 + j895
(13-j7)(13+j7) = 169 + 49j^2 = 169 - 49 = 120

So the answer you get the other way should agree with 105/120 + j895/120.

Finish each way, and see if they do agree. (There will probably be some difference due to rounding.)

 

[jbjaidee]

The answer I got isn't even close.

(35+j50) / (13-j7)

Sqrt(35^2+50^2)=61.03 Sqrt(13^2-7^2)=10.95
Tan-1(50/35)=55.01 Tan-1(-7/13)=-28.30

61.03∠55.01° / 10.95∠-28.30°= 5.57∠83.31°

5.57cos(83.31)=.649
5.57sin(83.31)=5.53

.65-j6=

 

[Dr.Peterson]

The answer I got isn't even close.

(35+j50) / (13-j7)

Sqrt(35^2+50^2)=61.03 Sqrt(13^2-7^2)=10.95
Tan-1(50/35)=55.01 Tan-1(-7/13)=-28.30

61.03∠55.01° / 10.95∠-28.30°= 5.57∠83.31°

5.57cos(83.31)=.649
5.57sin(83.31)=5.53

.65-j6=

First, I made a small (but stupid) mistake; my denominator should not be 169 + 49j^2 = 169 - 49 = 120, but 169 - 49j^2 = 169 + 49 = 218. That will change the expected results.

Second, you made essentially the same mistake you made before: the square of -7 is (-7)^2, not -(7^2). So the denominator's length is not sqrt(13^2-7^2)=10.95, but sqrt(13^2+7^2)=14.76. (It's worth noting that this can never be less than the absolute value of either component; your 10.95 was less than 13.)

Try again, and it should work. What you did, apart from the one little error, was fine.

 

[jbjaidee]

Ok I reworked the problem:

Sqrt(35^2+50^2)=61.03 Sqrt(13^2-(-7^2))=14.76
Tan-1(50/35)=55.01 Tan-1(-7/13)=-28.30

61.03∠55.01° / 14.76∠-28.3°= 4.13∠83.31°

4.13cos(83.31)=.48
4.13sin(83.31)=4.10

.48+j4=

 

[Dr.Peterson]

Ok I reworked the problem:

Sqrt(35^2+50^2)=61.03 Sqrt(13^2-(-7^2))=14.76
Tan-1(50/35)=55.01 Tan-1(-7/13)=-28.30

61.03∠55.01° / 14.76∠-28.3°= 4.13∠83.31°

4.13cos(83.31)=.48
4.13sin(83.31)=4.10

.48+j4=

That's what I get, though I round both components to the same precision, 0.48 + j4.10.