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I'm trying to find the quality factor of this circuit but I'm not sure how to do it. I know how to find it for a series RLC circuit and completely parallel RLC circuit, but I don't know what to do when the resistor is in series as shown here.

Any help is appreciated!

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2 Answers 2

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The voltage source can be assumed to be ideal and therefore have zero internal impedance. So you now have a voltage source in series with 100 KΩ. This can now be replaced with a current source in parallel with 100 KΩ. This is just a Thevenin-Norton transform which many textbooks describe. Now that everything is in parallel, the Q is easy for you to complete for your homework.

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I'd take the bandwidth-centric approach to this:

By analysis of the circuit, find the cutoff frequencies.

Then, follow the formula

$$B = f_{cutoff,upper}-f_{cutoff,lower} = \frac{f_{center}}Q\\ Q = \frac{f_{center}}{f_{cutoff,upper}-f_{cutoff,lower}}$$

Personally, I never got that warm with the component-value based approaches to calculating Q; point is that to me, it's a "sharpness" measurement of a resonant circuit, and the "logical" definition of that is the ratio between center frequency and bandwidth.

Generally, I find that "Q factor" is a nice measure, but far too often abused as a simplistic abstraction. For example, the Q factor of this circuit doesn't tell you anything about how the heat-converted power is distributed across frequency.

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