Discussion in 'General Electronics Chat' started by NewGuy01, Aug 10, 2009.

1. NewGuy01 Thread Starter New Member

Aug 9, 2009
5
0
Hello,

I was just wondering if someone could please check my maths and show me where I went wrong...

Im trying to find the total add admittance of this circuit:

http://www.flickr.com/photos/41254456@N05/3807839423/

So far I have:

Y = (1 / 2<0°) + (1 / 2.5<-90°) + (1 / 1.41<45°)

Multiplying across by
(2<0°)(2.5<-90°)(1.41<45°) / (2<0°)(2.5<-90°)(1.41<45°)

Yeilds :

Y = (2.5<-90°)(1.41<45°) + (2<0°)(1.41<45°) + (2<0°)(2.5<-90°)

Y = (3.53<-45°) + (2.82<45°) + (5 < -90°)

Converting to cartesian format =

2.5 - 2.5i + 1.99 +1.99i - 5i

= 4.49 - 0.51i = 4.52 < -6.48° = Y.

This is fine and grand, however if I just get the reciprocals and sum them :

Y = (1 / 2<0°) + (1 / 2.5<-90°) + (1 / 1.41<45°)

Y = (0.5<0°) + (0.4<90°) + (0.71<-45)
= 0.5 + 0.4i + 0.5 -0.5i
Y = 1 - 0.1i = 1 < 5.71°

These two answers are completely different, shouldn't they be the same?
Or have I gone wrong with my math somewhere?
Ive tried this like 3 time and got the same answers each time.

Any help is massively appreciated...

Last edited: Aug 10, 2009

May 11, 2009
5,939
1,222
3. NewGuy01 Thread Starter New Member

Aug 9, 2009
5
0
Thanks mate but I know what admittance is...

My problem is how I ended up with two completely different answers using two (as far as I know) valid methods...

4. t06afre AAC Fanatic!

May 11, 2009
5,939
1,222
The reciprocal theorem is valid also for complex admittance and impedance. So if you find it more convinient to find the complex impedance you can use the 1/x or reciprocal theorem. But as I said the law of complex math will apply...The reciprocal value of a complex number is not 1/x as it is for real numbers. Remember that

5. NewGuy01 Thread Starter New Member

Aug 9, 2009
5
0
Im pretty sure my reciprocals are right though, isn't

1 / (2.5 < -90°) the same as (1 < 0°) / (2.5 < -90°)?

And phasor division is equal to (R1 / R2 < θ1 - θ2)
So 1 / (2.5 < -90°) = (0.4 < 90°)

I still dont see where I went wrong....

6. kkazem Active Member

Jul 23, 2009
160
26
Hi NewGuy01,

With that handle chosen, what are you going to do when you've been a long-time member? Change it to oldguy01?

Anyway, enough humor. I'm a 30+ year experienced EE with my BSEE from UC, Irvine. This is a fairly simple problem, but you did make a couple of math errors that I'll point out to you.

The first thing to do in this problem is to compute the admittance, Y = 1/Z, of the R1 + Y(C) (recall that admittances in parallel add). ALso, recall that when you invert a complex number, the polarity of the imaginary part gets inverted. Therefore, to solve this problem, first, get the admittances of R1 and C seperately, then simply add them in the complex domain as follows:
Y(R1) = (1/2) = 0.5, Y(C) = (-1/j2.5) = +0.4j, Now add them to get:
Y(R1 & C) = (0.5+0.4j). Note the polarity of the capacitance admittance is positive since it's the reciprocal of a negative imaginary term.

Next, well take on R2 & L by noticing that they are in series and since series impedances add, well do that first, then invert the sum to get the admittance, then we can add to the R1&C admittance to get the final answer. Z(R2) = 1, Z(L) = +j1, we add them to get Z(R2+L) = (1+1j), and now, since Y=1/Z, we compute Y(R2&L) = 1/(1+1j) = (0.5 - 0.5j). Again, note that reversal of polarity on the imaginary term of the reciprocal.

For the total Y, we can simply add the 2 previous admittances since they are in parallel as follows: Ytotal = Y(R1 & C) + Y(R2 & L) = (0.5+0.4j) + (0.5 - 0.5j) = (1 - 0.1j). Or in polar coordinates: Ytotal = 1.005<-0.09967) radians or in degrees: Ytotal = (1.005<-5.711) degrees.

That's it. I hope it helps and I hope it cleared-up your math issue, which again is that when a complex number is inverted, it's imaginary part gets a sign inversion.

Best of luck.

Regards,
Kamran Kazem
kkazem