find phasor current

Discussion in 'Electronics Homework Help' started by Ledwardz, Dec 24, 2010.

1. Ledwardz

43
0
Dec 21, 2010
Hi, we have been given exam papers to revise from with no solutions....... yes i know its stupid and means i don't know if what i am doing is right.

ive attached the question and here is my answer

4.46A with a phase angle of 30.38.

but if im doing it wrong i need to know........

my working is as follows:

Vt = 25v to angle of 53 + 33v with no angle
= 58v to angle 53

The magnitude of impedance = 13 (square root of 5 +12 squared)
the angle of impedance = 22.62 degrees. (arctan 5/12)

then i = V/Z

so i get 4.46 < 30.38 A

if any one could confirm that id appreciate it. Thanks, Lee.

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2. barathbushan

223
0
Sep 26, 2009
The whole method of solving the problem is right and the calculation of IMPEDENCE is also right , but the voltage computation is absolutely
wrong !!

i wont re-solve the entire problem , but i will tell you that , you just cant add the voltage magnitudes and angles like you have done , as it is clearly shown here

so learn your complex algebra well , and you will be able to tackle such problems

3. Ledwardz

43
0
Dec 21, 2010
soo

i should get Voltage = (58cos53 + j53sin53)
and impedence = (13cos22.62 + j13sin22.62)

then i got rid of sin n cos

(34.9 + j46.3) / (12 + j5)

then multiply by complex conjugate (12 - j5)

which = 3.85 + j2.25

magnitude = 4.46
angle = 30.3

so answer = 4.46 to angle of 30.3 A ???

4. Ledwardz

43
0
Dec 21, 2010
ahh crap i got same answer...... nm thats obviously wrong ill have another look. AHHHH.... i get it ha ha ha ha yeah okay u mean adding the voltages at the start is wrong h aha ha ha well i look like an idiot now. o well.

Last edited: Dec 25, 2010
5. Ledwardz

43
0
Dec 21, 2010
okay ill try again

voltage magnitude = 52V
and angle = 22.6

I think that should be right?

6. barathbushan

223
0
Sep 26, 2009
no its wrong , let me give you the answer its
52.63<24.15 volts

one more method of solving the addition is converting both the voltages
25<53 and 33<0
to rectangular form , then add the real and imaginary parts
after that re-convert them to the polar form

use the above method is used if you have difficulty with polar form , but it's
not a good way to solve , please go through basic complex algebra

7. Ledwardz

43
0
Dec 21, 2010
i really dont know where i am going wrong as far as i am aware adding in polar goes something like this. i tried using example but exchanged the sin and cos for their actual numbers so when i squared it was easier.

=(25cos53 + j25sin53) + (33cos0)
= 15.04.. + j 19.966.. + 33
= 48.04.. + j19.966..
then magnitude is square root of em squared
n arctan for angle.

have i got it all wrong? It seems near the answer, maybe i have just done something stupid?

8. barathbushan

223
0
Sep 26, 2009
yes that's the answer just convert 48.04 + 19.996j into polar form , then you will get the right answer of 52.63<24.15 volts

magnitude = sqrt (48.04^2 +19.996^2 )=52.02
angle = tan-1(19.996/48.04) =22.56

ie 52.02<22.56 (approx to my answer of 52.63<24.15 )

9. Ledwardz

43
0
Dec 21, 2010
Thanks! i appreciate the help.