RLC ac network

  • Thread starter freshbox
  • Start date
  • #1
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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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Answers and Replies

  • #2
290
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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
gneill
Mentor
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2,877
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°
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.
Zt= 100+92.59<-30.28°
=192.59<-30.28°
Oops. You can't add the polar value to a Cartesian value like that; you need to deal with the angle.
 
Last edited:
  • #4
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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
gneill
Mentor
20,936
2,877
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 :(

Fix your accuracy issue for determining Z2.
 
  • #6
290
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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
Zt=177.24-j46.89 (Wrong Answer)

Book answer is 183.34<-14.82°
 
  • #7
gneill
Mentor
20,936
2,877
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
Zt=177.24-j46.89 (Wrong Answer) <--- Convert to polar

Book answer is 183.34<-14.82°

Book answer is in polar form. Convert your Zt to polar.
 
  • #8
290
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-_- what am i'm thinking.. Thanks alot gneill!!!

Oh I want to ask
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
gneill
Mentor
20,936
2,877
-_- what am i'm thinking.. Thanks alot gneill!!!

Oh I want to ask
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?

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
290
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Thank you sir for your time and help :)
 

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