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I've been given the above circuit and asked to find the equivalent impedance, however after drawing it half a dozen different ways I cannot find a way to simplify it. I can simulate it, and I get (what I think is the right answer) of -0.934-j1.467 ohms, but I'd like to know if there's a way to do it manually, and if there's some useful trick that I'm missing out on here. Or maybe its something simple!

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  • \$\begingroup\$ en.wikipedia.org/wiki/Y-%CE%94_transform \$\endgroup\$ – John D Feb 2 '18 at 21:44
  • \$\begingroup\$ I believe you're overthinking this because it is easier than you think :) If you're going to place an imaginary number as your reactance of the capacitor or inductor, then treat it just like any other (complex) number. Imagine if you replace all of those components as blocks of impedance with the same values. Reduce the circuit. Just remember, when you have a reactance and resistance in series, you may have to reduce it to impedence \$Z=\sqrt{R^2+X^2}\$ \$\endgroup\$ – KingDuken Feb 2 '18 at 21:44
  • \$\begingroup\$ Please use the schematic editor when you post a question like this. Sure, you have to spend some time learning how to use it. Sure, you have to spend some time putting the schematic in. It appears you took the path offering you the least possible effort. That isn't fair to anyone else. It's your question and you should apply yourself so that we may spend the least amount of time trying to help you. It should not be handled the other way around. Had you used the schematic editor, we'd have labels and that would make communication easier and faster for us. \$\endgroup\$ – jonk Feb 2 '18 at 22:31
  • \$\begingroup\$ By the way, that's a simple bridge in parallel to a cap. Should be easy to manually process. For bridge solve: (1) \$\frac{V_X}{Z_1}+\frac{V_X}{Z_2}+\frac{V_X}{Z_5}=\frac{V}{Z_1}+\frac{V_Y}{Z_5}\$; (2) \$\frac{V_Y}{Z_3}+\frac{V_Y}{Z_4}+\frac{V_Y}{Z_5}=\frac{V}{Z_3}+\frac{V_X}{Z_5}\$; (3) \$Z=\frac{V}{\frac{V_X}{Z_2}+\frac{V_Y}{Z_4}}\$. (Note: V should cancel out if you did it right.) Then apply your remaining cap in parallel to Z. Done. Since you didn't label anything, I'll leave it to you to work out how I labeled things as an illustration of the value you'd add had you done so. \$\endgroup\$ – jonk Feb 2 '18 at 22:46
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As John D points out: you can do a Delta to Y conversion. Or Y to Delta. That is where you take a set of three resistors (or impedances) and convert them to three other values.

His link points to the wikipedia where they use all resistors. But instead of R you can use a Z and it will still work.

schematic

simulate this circuit – Schematic created using CircuitLab

In the schematic above you e.g. take Z2, Z3 and Z5 (That is a Y-formation or T-formation) and make them into a Delta Za, Zb Zc.

schematic

simulate this circuit

I assume you can take it from there.

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Looks like you have a three branch parallel circuit with a bridge.

enter image description here

Google how to solve the resistance of a bridge circuit and apply it to your circuit. Chances are it won't be using caps and inductors but you seem to already know about how to calculate the impedance of a basic circuit so its just more of the same thing.

I found this link that walks you thru a simple one but I'm sure you can find others that might be better.

Solving a resistor bridge circuit

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  • \$\begingroup\$ "Google how to solve the resistance of a bridge circuit and apply it to your circuit.", lol. \$\endgroup\$ – Harry Svensson Feb 3 '18 at 0:14

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