How do I calculate the equivalent resistance of this type of circuit?

Question

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I get the overall formula, I'm not sure what to do about the 1 mF and the 1000 μH. Do I convert them in some way or ignore them?

What if I'm trying to find total power dissipated or voltage across this circuit from a current?

Answer 1

This circuit looks spooky at first glance but it is actually three paralleled impedances. The below simulation shows the circuit when properly redrawn:

When simulated, it shows an impedance mostly ohmic and of flat magnitude across frequency:

To analyze this network symbolically, you can use the brute-force analysis or the fast analytical circuits techniques abbreviated FACTs as described in my last book. This is a second-order circuit (two poles then) and I can infer from the arrangement that there are also two zeroes. The general form of this transfer function is simply:

\$Z_{in}(s)=R_0\frac{N(s)}{D(s)}\$

The dc resistance is immediate: place all energy-storing elements in dc (short the inductor and open-circuit the capacitor) and determine the resistance by inspection:

This is the first part of the FACTs analysis and I leave the rest to you. The brute-force analysis given below:

Sunday morning edit:

I have spent a few minutes to extract the complete transfer function and it is given below. The complete TF was obtained by inspection only - no equation or complicated manipulations - and tested against the brute-force version that I did not purposely expand. Both answers are rigorously identical.

Answer 2

Unclear. In the simplest case of DC resistance, you substitute inductors by shorts and capacitors by open circuits.

Regarding total power and voltage, again, it depends how the circuit is being excited: DC, harmonic AC, square wave, etc.

Answer 3

This can be easily understood by basic circuit theory concepts. Impedance of inductor with inductance L is given by jwL, where w is angular frequency of source. Similarly, for impedance of capacitance C is given by 1/(jwC). Now replacing these values in circuit, we can easily simplify the circuit as follows.

R2 and inductance L are in series, giving equivalent of W=(10k+j0.001w). Similarly, R5 and R6 are in series, giving X=0.02MOhm. Now R3 and capacitance are in parallel, giving equivalent of Y=j11w/(1+j11w). Now Y is in series with R4 giving Z=Y+9kOhm.

Finally W,X and Z are in parallel, whose equivalent could be easily calculated. Exact value of impedance can be calculated by substituting w.

Thanks