Using Complex Impedances in these RLC Circuit Calculations

[NaS4]

Homework Statement

Learning

Relevant Equations

z=a+bj, angle = tan^-1(b/a) -
R = R, C = 1/jwC, L = jwL

 

[NaS4]

Can someone help me with what i do next

 

[Master1022]

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.

 

[NaS4]

Yeah part a
But when I try to simplify it it just goes horrible wrong

 

[Master1022]

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

 

[NaS4]

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

 

[NaS4]

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

 

[Master1022]

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

 

[NaS4]

1/Z_{total} = 1/Z_{1} + 1/Z{2} in parallel

And in series it’s Z(total) = Z(1) + Z(2)

 

[Master1022]

Please see post #8. This question will mainly just involve being careful with algebra.

 

[etotheipi]

@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.

 

[NaS4]

Yeah i get that. But the formula i got is not the same as the one you are using

 

[Master1022]

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)

 

[NaS4]

 

[NaS4]

Is this correct?

 

[Master1022]

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}) ##)

 

[Master1022]

I think you might also be missing a term where ## j \omega L R ## multiplies the ## R ## in the first line

 

[NaS4]

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

 

[NaS4]

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

 

[NaS4]

@Master1022 How does this solution look?

 

[Master1022]

@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 ##)

 

[NaS4]

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)

 

[Master1022]

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:
$$ (\omega^2 LC - 1)^2 = (\omega^2 LC - 1)^2 \cdot 1 = ((\omega^2 LC - 1)(-1))^2 = (1 - \omega^2 LC)^2 $$

 

[NaS4]

@Master1022 So I've done most of the tasks now. But I'm abit unsure of task c) ii)

 

[NaS4]

That 11.5k V sounds abit to high

 

[Master1022]

What happened to the denominator in the expressions for ## a ## and ## b ##? I think you have forgotten those?

[EDIT]: I realized that you included them in the working, but not the initial expression. Please ignore the above statement.

That 11.5k V sounds abit to high

That is true and does definitely stand out to me. However, this is what can happen in resonant circuits with very high Q-factors... I just did a quick calculation, and the number I got was slightly larger than your answer (but on the same order of magnitude). Maybe another member might notice an error that I have overlooked

 

[Master1022]

Also, in the expression for ## b##, I think the central term in the numerator should have the ## C ## as ## C^2 ##. However, I am not sure whether that makes much difference given the order of magnitudes. I will go back to check the original expression to ensure that your formula is exactly the same as what I got

 

[NaS4]

Also, in the expression for ## b##, I think the central term in the numerator should have the ## C ## as ## C^2 ##. However, I am not sure whether that makes much difference given the order of magnitudes. I will go back to check the original expression to ensure that your formula is exactly the same as what I got

Master thanks a lot for the help man. You really are a lifesaver

 

[DaveE]

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 ##)

This is an incredibly useful tool when working with this sort of complex algebra. OK, actually, useful in most of physics, really. This allows you to spot errors in equations without having to go through the whole derivation. Maybe you won't know why it's wrong, but you will know it's not correct.

So, to elaborate on @Master1022s method:
R, 1/(ωC), ωL all have units of ohms. If I see a term in your equation like (R + L), I know without much analysis that it's wrong; that should be something like (R + ωL), or (R/ω + L). You just won't see solutions in physics where you have (ohms + volts), (force + energy), (flow + pressure), etc.; these rarely have any useful physical meaning. In the (R + L) example the units are (ohms + ohms*seconds).

BTW, you can't use this to verify correctness completely. It won't spot some errors, like the difference between (R + ωL) and (R + 1/(ωC)), or R vs. 2R.

If you want to know more look up "dimensional analysis".

 

[NaS4]

This is an incredibly useful tool when working with this sort of complex algebra. OK, actually, useful in most of physics, really. This allows you to spot errors in equations without having to go through the whole derivation. Maybe you won't know why it's wrong, but you will know it's not correct.

So, to elaborate on @Master1022s method:
R, 1/(ωC), ωL all have units of ohms. If I see a term in your equation like (R + L), I know without much analysis that it's wrong; that should be something like (R + ωL), or (R/ω + L). You just won't see solutions in physics where you have (ohms + volts), (force + energy), (flow + pressure), etc.; these rarely have any useful physical meaning. In the (R + L) example the units are (ohms + ohms*seconds).

BTW, you can't use this to verify correctness completely. It won't spot some errors, like the difference between (R + ωL) and (R + 1/(ωC)), or R vs. 2R.

If you want to know more look up "dimensional analysis".

Thx man. I’ll look up dimensional analysis

 

[DaveE]

Thx man. I’ll look up dimensional analysis

It can get scary. Look for the useful bits. If they get too abstract (math, philosophy and such) that may a good place to stop, or not, if you like that stuff.

 

[etotheipi]

It can get scary. Look for the useful bits. If they get too abstract (math, philosophy and such) that may a good place to stop, or not, if you like that stuff.

Good place to start is Orodruin's article:
https://www.physicsforums.com/insights/learn-the-basics-of-dimensional-analysis/

 

[Master1022]

Just a quick update, I have checked my expression:
$$ Z = \left( \frac{\omega^4 R L^2 C^2}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) + j \left( \frac{\omega L (1 - \omega^2 LC + (\omega R C)^2)}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) $$
and think it is equal to yours (thus we are either both correct... or both wrong !)

Quick question about the problem statement: are the units of ## \omega ## a typo when it says Hz? Usually we expect ## \omega ## to be in rad/s (a conversion factor of ## 2 \pi ## is needed).

Am doing the calculations in a Python notebook, so will post pictures here to show how my calculations are resulting

 

[NaS4]

Just a quick update, I have checked my expression:
$$ Z = \left( \frac{\omega^4 R L^2 C^2}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) + j \left( \frac{\omega L (1 - \omega^2 LC + (\omega R C)^2)}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) $$
and think it is equal to yours (thus we are either both correct... or both wrong !)

Quick question about the problem statement: are the units of ## \omega ## a typo when it says Hz? Usually we expect ## \omega ## to be in rad/s (a conversion factor of ## 2 \pi ## is needed).

Am doing the calculations in a Python notebook, so will post pictures here to show how my calculations are resulting

Yeah the w = 2(pi)f but since the question tells us that w is 200

 

[Master1022]

Yeah the w = 2(pi)f but since the question tells us that w is 200

Ah okay, that explains the discrepancy I had earlier. After setting ## \omega = 200 ##, I did get the same value of the gain and the final voltage as you (give or take some rounding errors) - the computer output ## 11551.24... ## (V), so this does agree with your result.

Hope that is of some help.