# RLC ac network

1. Feb 11, 2013

### freshbox

My Working for part (i); Determine the total impedance Zt:

Z1=70+j0 + 0+j50
=70+j50
= 86.023<35.537°Ω

1/Z2=1/-j80 + 1/86.023<35.537°
=j0.0125+0.0116<-35.537°

My question is do 35.537° need to be divided by 1 like 86.023? If not why?

Thanks.

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Last edited: Feb 11, 2013
2. Feb 11, 2013

### freshbox

My Full working:

Z1=70+j0 + 0+j50
=70+j50
= 86.023<35.537°Ω

1/Z2=1/-j80 + 1/86.023<35.537°
=j0.0125+0.0116<-35.537°
=j0.0125+0.0094-j0.00674
=0.0094-j0.00549
=0.0108<-30.28°
Z2=92.59<-30.28°

Zt= 100+92.59<-30.28°
=192.59<-30.28°

Ans is: 183.34<-14.82°

3. Feb 11, 2013

### Staff: Mentor

You've lost some accuracy due to rounding and truncation of intermediate values. Keep more decimal places for intermediate values and round final values for presentation. Z2's magnitude will turn out to be slightly smaller than you've found, and its angle's magnitude a bit larger.
Oops. You can't add the polar value to a Cartesian value like that; you need to deal with the angle.

Last edited: Feb 11, 2013
4. Feb 11, 2013

### freshbox

Oops, do you mean I have to change 192.59<-30.28° to rect form since I'm doing addtion?

Changing to rect form:
R1=100+j0
Z2=79.95-j46.68

Zt=100+j0+79.95-j46.68
Zt=179.95-46.68

Ans is still wrong, kinda far :(

5. Feb 11, 2013

### Staff: Mentor

Fix your accuracy issue for determining Z2.

6. Feb 11, 2013

### freshbox

1/Z2=1/-j80 + 1/86.0232<35.5376°
=j0.0125+0.01162477<-35.537°
=j0.0125+0.009459473-j0.006756747
=0.009459473+j0.005743253
=0.011066462<31.26373489°
=90.36311575<-31.26373489°
=77.24125762-j46.89

Zt=100+j0+77.24125762-j46.89

7. Feb 11, 2013

### Staff: Mentor

8. Feb 11, 2013

### freshbox

-_- what am i'm thinking.. Thanks alot gneill!!!

1/Z2=1/-j80 + 1/86.023<35.537°
=j0.0125+0.0116<-35.537°

How come the degree cannot/don't have to divide by 1?

9. Feb 11, 2013

### Staff: Mentor

You mean why does the angle of some complex number Z just get negated when you perform 1/Z? The angles a geometrical aspect of the number, denoting the direction of the phasor. Angles get added and subtracted, but not divided or multiplied.

You can see how the result occurs if you work out the reciprocal in Cartesian form:

$$Y = \frac{1 + 0j}{A + Bj}$$

Compare the polar angles of (A + Bj) with the angle of the result of the division. Also, you'll find that in the general case (Z1<θ1) / (Z2<θ2) = (Z1/Z2)<(θ1 - θ2) .

10. Feb 11, 2013

### freshbox

Thank you sir for your time and help :)