Phasor representaion math

1. Jan 21, 2010

nhrock3

http://i47.tinypic.com/2mg2iyf.jpg
i ant to find the current in phasor representation.
i made found the Z total 210+j232.5
the current sourse is 2170
$$I=\frac{2170}{310+j232.5}=1454.56-j1610.4$$

but the calculation igives another result
why?

Last edited: Jan 21, 2010
2. Jan 21, 2010

cepheid

Staff Emeritus
The first problem that comes to mind is that the 210 ohms (real part) should actually be 310 ohms.

Also, do you know how to express a complex number in such a way that the imaginary part appears only in the numerator (i.e. there is no j on the bottom)?

3. Jan 21, 2010

nhrock3

sorry on paper i wrote 310
and i did multiply the numerator and denominator by the "opposite"
of the compex number that is on the denominator

and i got a totaly different result

who is correct?

4. Jan 21, 2010

cepheid

Staff Emeritus
The solution in the jpg image you posted is correct. You have to multiply the expression in your first post (numerator and denominator) by the complex conjugate of the impedance and then simplify.

5. Jan 21, 2010

nhrock3

$$I=\frac{2170(310-j232.5)}{387.5}=1176-j1302$$

i did that as you see
why i still get the wrong result ?

6. Jan 21, 2010

vela

Staff Emeritus
It looks like you took the square root of the bottom. You should have zz*=|z|2 there, not |z|.

7. Jan 21, 2010

thanks:)