Using Complex Impedances in these RLC Circuit Calculations

  • #1
19
0

Homework Statement:

Learning

Relevant Equations:

z=a+bj, angle = tan^-1(b/a) -
R = R, C = 1/jwC, L = jwL
Complex Impedance.PNG
my ans.PNG
 
Last edited by a moderator:

Answers and Replies

  • #2
19
0
Can someone help me with what i do next
 
  • #3
344
69
For part (a), what you have done so far looks correct. You should continue to simplify the expression (get it into a simpler fraction). I'm assuming you were asking about part a - if not, do correct me.

[EDIT]: you may want to 'rationalise' the denominators of the fractions to help reach the required form. If you are taking an AC circuit theory class, I am guessing that you have seen complex numbers before? If not, let me know and I can explain further/ point you towards some resources.
 
  • #4
19
0
Yeah part a
But when I try to simplify it it just goes horrible wrong
 
  • #5
344
69
Yeah part a
But when I try to simplify it it just goes horrible wrong
Okay, why don't you type out your steps here and we can see where the problem is coming from.

Generally, we know that two parallel impedances yield ## Z_{total} = Z_1 // Z_2 = \frac{Z_1 \cdot Z_2}{Z_1 + Z_2} ## which follows directly from the formula you wrote. It might be easier to write out the final impedance that way
 
  • #6
19
0
Yeah part a
But when I try to simplify it it just goes horribly wrong
For part (a), what you have done so far looks correct. You should continue to simplify the expression (get it into a simpler fraction). I'm assuming you were asking about part a - if not, do correct me.

[EDIT]: you may want to 'rationalise' the denominators of the fractions to help reach the required form. If you are taking an AC circuit theory class, I am guessing that you have seen complex numbers before? If not, let me know and I can explain further/ point you towards some resources.
I’ve seen it before. But there’s a lot happening here. And I don’t know where to start
 
  • #7
19
0
Okay, why don't you type out your steps here and we can see where the problem is coming from.

Generally, we know that two parallel impedances yield ## Z_{total} = Z_1 // Z_2 = \frac{Z_1 \cdot Z_2}{Z_1 + Z_2} ## which follows directly from the formula you wrote. It might be easier to write out the final impedance that way
1/Z_{total} = 1/Z_{1} + 1/Z{2} in parallel
 
  • #8
344
69
I’ve seen it before. But there’s a lot happening here. And I don’t know where to start
Lets take ## Z_1 = R + \frac{1}{j\omega C} ## and ## Z_2 = j \omega L ##. The formula then becomes:
$$ Z_{total} = \frac{(R + \frac{1}{j\omega C}) \cdot ( j \omega L)}{( j \omega L) + (R + \frac{1}{j\omega C})} $$

Now you could multiply by, for example, ## j \omega C ## for the numerator and denominator. I don't think I can just give the answer
 
  • Like
Likes etotheipi
  • #9
19
0
1/Z_{total} = 1/Z_{1} + 1/Z{2} in parallel
And in series it’s Z(total) = Z(1) + Z(2)
 
  • #10
344
69
Please see post #8. This question will mainly just involve being careful with algebra.
 
  • #11
etotheipi
Gold Member
2019 Award
2,541
1,479
@NaS4 remember that if you have something of the form ##\frac{a+bi}{c+di}##, you can always multiply it by ##1##,$$\frac{a+bi}{c+di} = \frac{a+bi}{c+di} \frac{c-di}{c-di}$$and then separate into real and imaginary parts.
 
  • Like
Likes Master1022
  • #12
19
0
Yeah i get that. But the formula i got is not the same as the one you are using
 
  • #13
344
69
Yeah i get that. But the formula i got is not the same as the one you are using
It would really help to see some working. To get to the expression I have written, we can multiply the expression you have in the third line of your working by ## (R + 1/j\omega C ) ## and ## j \omega L ## (both numerator and denominator)
 
  • Like
Likes etotheipi
  • #14
19
0
solution1.PNG
 
  • #15
19
0
Is this correct?
 
  • #16
344
69
Is this correct?
In the beginning, ## j \omega L + \frac{1}{j\omega C} \neq j(\omega L + \frac{1}{\omega C}) ## as the j in the denominator should make it negative (i.e. ## j(\omega L - \frac{1}{\omega C}) ##)
 
  • #17
344
69
I think you might also be missing a term where ## j \omega L R ## multiplies the ## R ## in the first line
 
  • #18
19
0
In the beginning, ## j \omega L + \frac{1}{j\omega C} \neq j(\omega L + \frac{1}{\omega C}) ## as the j in the denominator should make it negative (i.e. ## j(\omega L - \frac{1}{\omega C}) ##)
Ahh yes I see it
 
  • #19
19
0
I think you might also be missing a term where ## j \omega L R ## multiplies the ## R ## in the first line
Damn so many mistakes. Thx man. I’ll change it and post my next try
 
  • #21
344
69
@Master1022 How does this solution look?
That looks correct to me! It might be easier to turn the ## (\omega L - \frac{1}{\omega C}) ## into a ## 1 - \omega^2 LC ## term (remembering to multiply through by ## (\omega C)^2 ## ). This looks like it will be easier to work with for the next parts of the question.

It is also useful to think about dimensions when working through these types of problems. Often seeing dimensions that match can be indicative of correct working (e.g. seeing a ## \omega ^2 LC ## or ## \omega ^ 2 RC ##)
 
  • Like
Likes DaveE
  • #22
19
0
That looks correct to me! It might be easier to turn the ## (\omega L - \frac{1}{\omega C}) ## into a ## 1 - \omega^2 LC ## term (remembering to multiply through by ## (\omega C)^2 ## ). This looks like it will be easier to work with for the next parts of the question.

It is also useful to think about dimensions when working through these types of problems. Often seeing dimensions that match can be indicative of correct working (e.g. seeing a ## \omega ^2 LC ## or ## \omega ^ 2 RC ##)
Ok I’m lost now. How did you get (1 - w^2LC)
 
  • #23
344
69
Ok I’m lost now. How did you get (1 - w^2LC)
Apologies, so you can get ## \omega^2 LC - 1 ## and then multiply that term by ## (-1)^2 = 1 ## so your denominator becomes ## (\omega C R)^2 + (\omega^2 LC - 1)^2 = (\omega C R)^2 + (1 - \omega^2 LC )^2 ##. It's just a multiplication by 1 (won't matter because the term is squared).

You can also get there if you multiply the numerator and denominator by ## j ## in either line 3 or 4 of the working.

Hope that makes sense.

So just to clarify:
$$ (\omega^2 LC - 1)^2 = (\omega^2 LC - 1)^2 \cdot 1 = ((\omega^2 LC - 1)(-1))^2 = (1 - \omega^2 LC)^2 $$
 
  • #24
19
0
@Master1022 So i've done most of the tasks now. But I'm abit unsure of task c) ii)
solution3i.PNG
solution3ii.PNG
 
  • #25
19
0
That 11.5k V sounds abit to high 😅
 

Related Threads on Using Complex Impedances in these RLC Circuit Calculations

Replies
7
Views
706
Replies
9
Views
1K
Replies
6
Views
2K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
14
Views
4K
Top