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Sorry, will provide more information now.

I am working through this question from an all about circuits webpage (See attached webpage link) and have having trouble seeing how they calculate an impedance value.

Here is the example circuit:

enter image description here

Their method uses the reciprocal method for finding the parallel impedance of a capacitor C2 and inductor L combined in series, and then in parallel with a resistor R. I have worked out the capacitor and inductor in series ok. The LC series impedance is -j1.525k ohm.

I cannot get the same answer they have in the table that I have circled below in my calculator using the circled parallel forumula: enter image description here

What I am asking is could someone show me what numbers go in that formula to get their answer I have circled in the screenshot?

This is the webpage of the question I am following if you want to see it:

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-5/series-parallel-r-l-and-c/

Thanks David

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    \$\begingroup\$ What results are you getting? When i calculate it i get the same as stated in the table. \$\endgroup\$
    – Linkyyy
    Commented Oct 9, 2018 at 18:29
  • \$\begingroup\$ Oh, I think I have to remove j from the denominator to get this: Z = (R^2 * Xl--c^2) / (R^2 + Xl--c^2) + j(R^2 * Xl--c) / (R^2 + Xl--c^2). I am now getting the correct answer. Is this the easiest method for parallel impedance calculations or are there other methods? Did you use this formula? \$\endgroup\$
    – David777
    Commented Oct 9, 2018 at 18:50
  • \$\begingroup\$ Im not sure about the formula you wrote, but essentially just the basic parallel impedance formula Z1*Z2/(Z1 + Z2) \$\endgroup\$
    – Linkyyy
    Commented Oct 9, 2018 at 22:11
  • \$\begingroup\$ I get a magnitude of 449.1 too. \$\endgroup\$
    – Andy aka
    Commented Oct 10, 2018 at 12:02
  • \$\begingroup\$ Hi everyone, yeah I am now getting the correct answer. I was making a mathematical mistake with the complex numbers. Thanks everyone! \$\endgroup\$
    – David777
    Commented Oct 11, 2018 at 9:22

2 Answers 2

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What I am asking is could someone show me what numbers go in that formula to get their answer I have circled in the screenshot?

The numbers are: \$Z_R=470\$ and \$Z_{L--C_2}=-j1523.3\$

They result in: \$Z_{R//L--C_2}=429.15 - j132.41\$

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Well, the input impedance is equal to:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{\text{j}\omega\text{C}_1}+\frac{1}{\frac{1}{\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}_2}}+\frac{1}{\text{R}}}\tag1$$

With:

$$\omega=2\pi\cdot\text{f}\tag2$$

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