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Homework Statement:
 Learning
Relevant Equations:

z=a+bj, angle = tan^1(b/a) 
R = R, C = 1/jwC, L = jwL
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Okay, why don't you type out your steps here and we can see where the problem is coming from.Yeah part a
But when I try to simplify it it just goes horrible wrong
I’ve seen it before. But there’s a lot happening here. And I don’t know where to startFor 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.
1/Z_{total} = 1/Z_{1} + 1/Z{2} in parallelOkay, 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
Lets take ## Z_1 = R + \frac{1}{j\omega C} ## and ## Z_2 = j \omega L ##. The formula then becomes:I’ve seen it before. But there’s a lot happening here. And I don’t know where to start
And in series it’s Z(total) = Z(1) + Z(2)1/Z_{total} = 1/Z_{1} + 1/Z{2} in parallel
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)Yeah i get that. But the formula i got is not the same as the one you are using
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}) ##)Is this correct?
Ahh yes I see itIn 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}) ##)
Damn so many mistakes. Thx man. I’ll change it and post my next tryI think you might also be missing a term where ## j \omega L R ## multiplies the ## R ## in the first line
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.@Master1022 How does this solution look?
Ok I’m lost now. How did you get (1  w^2LC)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 ##)
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).Ok I’m lost now. How did you get (1  w^2LC)