Series/Parallel impedance problem

[Blastrix91]

Homework Statement

Problem description:
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Homework Equations

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$$Z = Z_1 + Z_2 + ... +Z_n (Series),$$

$$\frac{1}{Z}= \frac{1}{Z_1} + \frac{1}{Z_2} + ... + \frac{1}{Z_n} (parallel)$$

The Attempt at a Solution

$$\frac{1}{Z}= (\frac{1}{i \omega C} + R)^{-1} + (i \omega L + R)^{-1}$$

$$\frac{1}{Z}= (\frac{R + i \omega C R}{i \omega C R})^{-1} + (i \omega L + R)^{-1}$$

$$\frac{1}{Z}= \frac{i \omega C R}{R + i \omega C R} + \frac{1}{i \omega L + R}$$

$$\frac{1}{Z}= \frac{i \omega C R (i \omega L + R)}{(R + i \omega C R)(i \omega L + R)} + \frac{R + i \omega C R}{(R + i \omega C R)(i \omega L + R)}$$

$$\frac{1}{Z}= \frac{i \omega C R (i \omega L + R) + R + i \omega C R}{(R + i \omega C R)(i \omega L + R)}$$

$$\frac{1}{Z}= \frac{i C L \omega^2 + C R \omega + L \omega - iR}{i C L \omega^2 + C R \omega + C \omega - i}$$

$$\frac{1}{Z}= \frac{i C L R \omega^2 + C R^2 \omega + L \omega R - iR^2}{i C L R \omega^2 + C R^2 \omega + C \omega R - i R}$$

$$R^2 = \frac{L}{C}$$

$$\frac{1}{Z}= \frac{i C L R \omega^2 + L \omega + L \omega R - i \frac{L}{C}}{i C L R \omega^2 + L \omega + C \omega R - i R}$$

.. and here I'm stuck. I need to eliminate those frequencies. (My approach may have been wrong all along?)

 

[gneill]

The Attempt at a Solution

$$\frac{1}{Z}= (\frac{1}{i \omega C} + R)^{-1} + (i \omega L + R)^{-1}$$

$$\frac{1}{Z}= (\frac{R + i \omega C R}{i \omega C R})^{-1} + (i \omega L + R)^{-1}$$

That first term, the rearrangement of the capacitor's admittance, doesn't look right. Rework the algebra.

Regarding your approach, note that if the overall impedance Z remains constant with respect to frequency, then the overall admittance Y = 1/Z is also constant.

What can you say about the derivative of a constant?

 

[Blastrix91]

You are right I screwed up there. I worked out a new solution by the same principle:

$$\frac{1}{Z}= \frac{(i \omega C)}{1+R} + \frac{1}{i \omega L+R}$$

$$\frac{1}{Z}= \frac{(i \omega C)(i \omega L+R) + 1 + R}{(1+R)(i \omega L+R)}$$

$$Z= \frac{(1+R)(i \omega L+R)}{i \omega C(i \omega L+R)+1+R}$$

Still can't see it

I don't get what you're saying in your second sentence though. (derivative of a constant is zero) Is that part of solving the problem?

 

[gneill]

You are right I screwed up there. I worked out a new solution by the same principle:

$$\frac{1}{Z}= \frac{(i \omega C)}{1+R} + \frac{1}{i \omega L+R}$$

$$\frac{1}{Z}= \frac{(i \omega C)(i \omega L+R) + 1 + R}{(1+R)(i \omega L+R)}$$

$$Z= \frac{(1+R)(i \omega L+R)}{i \omega C(i \omega L+R)+1+R}$$

Still can't see it

I don't get what you're saying in your second sentence though. (derivative of a constant is zero) Is that part of solving the problem?

Still a problem with the first term. Start with the series impedance:

$R + \frac{1}{j ω C} = \frac{j ω R C + 1}{j ω C}$

multiplying top and bottom by j:

$= \frac{ω R C - j}{ω C}$

The admittance term is the inverse of this, so: $\frac{ω C}{ω R C - j}$ and altogether:
$$\frac{1}{Z} = Y = \frac{ωC}{ωRC - j} + \frac{1}{R + jωL} $$

If Y is a constant w.r.t. ω then, as you say, its derivative w.r.t. ω must be zero, right? So take the derivative and find out what the value of R² must be that makes it so.

 

[Blastrix91]

Okay so I've taken the derivative with respect to ω

$$\frac{d Y}{d \omega}= - \frac{i C}{(C R \omega -i)^2} - \frac{i L}{(R+ i L \omega)^2} = 0$$

I'm supposed to calculate the parenthesis and isolate R² from there?

 

[gneill]

Okay so I've taken the derivative with respect to ω

$$\frac{d Y}{d \omega}= - \frac{i C}{(C R \omega -i)^2} - \frac{i L}{(R+ i L \omega)^2} = 0$$

I'm supposed to calculate the parenthesis and isolate R² from there?

The idea is to simplify the derivative (giving it a common denominator) and then isolating R². But your derivative does not look right. There should be three terms in the initial expansion of the derivative.

 

[Blastrix91]

My math calculator program says it's correct? What's wrong with it?

Tried finding R² from the derivative i calculated:

$$- \frac{iC(R+iLω)^2}{(CRω−i)^2 (R+iLω)^2} - \frac{i L(CRω−i)^2}{(CRω−i)^2 (R+iLω)^2} = 0$$

$$\frac{- iC(R+iLω)^2 - i L(CRω−i)^2}{(CRω−i)^2 (R+iLω)^2}= 0$$

$$\frac{- iC (R^2+2 i L R ω-L^2 ω^2) - i L(-1-2 i C R ω+C^2 R^2 ω^2)}{(-1-2 i C R ω+C^2 R^2 ω^2) (R^2+2 i L R ω-L^2 ω^2)}= 0$$

$$\frac{-C i R^2-2 C i^2 L R ω+C i L^2 ω^2 + i L+2 C i^2 L R ω-C^2 i L R^2 ω^2}{(-1-2 i C R ω+C^2 R^2 ω^2) (R^2+2 i L R ω-L^2 ω^2)}= 0$$

I plotted all this into a math calculator. The denominator in the last term was incredibly huge.

Btw thanks for the help so far. I at least got the idea behind solving the problem now.

 

[gneill]

My math calculator program says it's correct? What's wrong with it?

My mistake; I failed to recognize that you'd already applied a simplification.

Tried finding R² from the derivative i calculated:

$$- \frac{iC(R+iLω)^2}{(CRω−i)^2 (R+iLω)^2} - \frac{i L(CRω−i)^2}{(CRω−i)^2 (R+iLω)^2} = 0$$

$$\frac{- iC(R+iLω)^2 - i L(CRω−i)^2}{(CRω−i)^2 (R+iLω)^2}= 0$$

$$\frac{- iC (R^2+2 i L R ω-L^2 ω^2) - i L(-1-2 i C R ω+C^2 R^2 ω^2)}{(-1-2 i C R ω+C^2 R^2 ω^2) (R^2+2 i L R ω-L^2 ω^2)}= 0$$

$$\frac{-C i R^2-2 C i^2 L R ω+C i L^2 ω^2 + i L+2 C i^2 L R ω-C^2 i L R^2 ω^2}{(-1-2 i C R ω+C^2 R^2 ω^2) (R^2+2 i L R ω-L^2 ω^2)}= 0$$

I plotted all this into a math calculator. The denominator in the last term was incredibly huge.

Btw thanks for the help so far. I at least got the idea behind solving the problem now.

Since the derivative is being set to zero, you can ignore the denominator. You only need to expand the numerator and collect the R² terms.