# Extract the angle polar numbers

1. Jan 5, 2012

### commelion

hi

can anybody advise on this, need to find the angle out of the following

2300=2000+(0.5<x).(4<90)

got as far as

300/4<90 = 0.5<x

= 75<90= 0.5<x

is this correct and anybody able to finish this off

regards

2. Jan 5, 2012

### Mentallic

What is 4<90 and 0.5<x ?

3. Jan 5, 2012

### fbs7

I think he means polar notation

a $\angle$ ω = a ( cos ω + j . sin ω )

but the OP formula seems to have a typo of some kind; the only expression I can imagine that makes sense is this

2300 = 2000 . ( 0.5 $\angle$ x ) . ( 4 $\angle$ 90 )
= 2000 . 0.5 . 4 cos ( 90 - x )

x = 35.0996°

May be that, or not...

Last edited: Jan 5, 2012
4. Jan 6, 2012

### commelion

hi yes it is polar notation, got my notation and common sense mixed up (late one last night)

the equation repersents the excitation current angle of a sychronous generator, ( Ia ∠ x ) stands for stator amps at some angle

(2300∠ 15 ) = (1154.7∠ 0) + ( Ia ∠ x ) . ( 4 ∠ 90 )

can some check if

( Ia ∠ x ) = ((2300∠ 15 ) - (1154.7∠ 0) / ( 4 ∠ 90 ))

= (305.4∠-61)

can someone confirm this for me please, if im wrong please correct ?

5. Jan 6, 2012

### fbs7

ah, it's Ia instead of 4... 4 would never get a valid value... that makes sense now; also, if I remember electrics right, you're doing a complex number multiplication, not a vector inner product... so ( a ∠ b ) . ( c ∠ d ) is (ab) ∠ (b+d)... instead of (ab) ∠ (b-d) as I wrote above

so

(2300 < 15) = (1154.7 < 0 ) + ( I < x ) . ( 4 < 90 )
(2221.63 + j 595.28) - ( 1154.7 + j 0 ) = ( I < x ) . ( 4 < 90 )
(1066.93 + j 595.28 ) / ( 4 < 90 ) = ( I < x )
( 1221.76 < 29.16 ) / ( 4 < 90 ) = ( I < x )
( 305.44 < -60.84 ) = ( I < x )

it looks good; confirming result with my trusty Texas calculator:

(2300 < 15) = (1154.7 < 0 ) + ( 305.44 < -60.84 ) . ( 4 < 90 )
(2300 < 15) = (1154.7 < 0 ) + ( 1221.76 < 29.16 )
(2300 < 15) = (2299.99 < 15)

6. Jan 7, 2012

### commelion

thanks for that

regards