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If we are given a parallel or series LCR circuit, we know the quality factor of these circuits ( which we can see in many books). But if we are given a LCR circuit with the three components connected in series or parallel as we like, say resistor is connected in series to both inductor and capacitor but the inductor and capacitor are connected in parallel. Now here we have a simple deviation from our traditional series or parallel circuit. so wouldn't this change in configuration also change the quality factor of the circuit. If yes, what factors do we need to consider to find out the quality factor of any such arbitrary LCR circuits?

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so wouldn't this change in configuration also change the quality factor of the circuit. If yes, what factors do we need to consider to find out the quality factor of any such arbitrary LCR circuits?

Any rearrangement of components in a circuit results in different equation that describes system behavior. An LRC circuit has two storages of energy (an inductor and a capacitor) and its behavior can be described via second-order differential equation of the general form

$$\ddot{y} + \frac{\omega_n}{Q} \dot{y} + \omega_n^2 y = f(u) \tag 1$$

where $Q$ is the quality factor, $\omega_n$ is undamped angular frequency of the oscillator (natural frequency), and $u$ is the system input (excitation).

If you want to determine the quality factor of a system, you need to first write differential equation that describes the system behavior and then compare it to the general form given in Eq. (1).


Other well-known form of the second-order differential equation is

$$\ddot{y} + 2 \zeta \omega_n \dot{y} + \omega_n^2 y = f(u) \tag 2$$

where $\zeta$ is the damping ratio. From the damping ratio you can easily determine how system response will look like:

  • undamped or permanent oscillations for $\zeta = 0$,
  • underdamped for $\zeta < 1$,
  • critically damped for $\zeta = 1$, or
  • overdamped for $\zeta > 1$.

The former two have oscillatory response, and the latter two have asymptotic response (i.e., no overshoot) to a stepwise excitation. If you are studying in the field of engineering, I cannot stress enough how important it is to understand behavior of second-order systems.


Example - Series RLC circuit

Here I just give a simple example for series RLC circuit connected to a voltage source with zero internal resistance. We choose input to be voltage source $v_s$ and output to be voltage across the capacitor $v_C$. Note that the choice of input and output variables depends on the particular problem you are trying to solve. The circuit equation is

$$v_s = v_R + v_L + v_C$$

Now we need to write the above equation only in terms of $v_s$ and $v_C$. The three voltages on the right-hand side can be written via current as

$$v_R = i R, \qquad v_L = L \frac{di}{dt}, \qquad v_C = \frac{1}{C} \int i dt$$

From the equation for voltage across the capacitor, the current can be written as $i = C \frac{dv_C}{dt}$ and the original equation now becomes

$$v_s = RC \frac{dv_C}{dt} + LC \frac{d^2 v_C}{dt^2} + v_C$$

By introducing substitutions $u = v_s$ and $y = v_C$ the final circuit equation is

$$\ddot{y} + \frac{R}{L} \dot{y} + \frac{1}{LC} y = \frac{1}{LC} u$$

Comparing this equation to Eqs. (1) and (2) the system parameters are

$$\omega_n = \sqrt{\frac{1}{L C}}, \qquad \zeta = \frac{R}{2} \sqrt{\frac{C}{L}}, \qquad Q = \frac{1}{R} \sqrt{\frac{L}{C}}$$

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  • $\begingroup$ The second order differential equation is found out by Kirchhoff's voltage law, right? So the variable will be the charge. Am I correct? $\endgroup$
    – Igris
    Mar 24, 2022 at 18:19
  • $\begingroup$ @Igris Yes, the equation is found by applying Kirchhoff’s laws. You are free to choose input $u$ and output $y$. The input is usually voltage source, and the output is usually voltage drop on either resistor, capacitor or inductor. But input and output can be whatever you need, it depends on particular problem. $\endgroup$ Mar 24, 2022 at 18:22
  • $\begingroup$ @Igris I added a simple example to show how to get expression for quality factor of a series RLC circuit. $\endgroup$ Mar 25, 2022 at 8:36
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Now here we have a simple deviation from our traditional series or parallel circuit. so wouldn't this change in configuration also change the quality factor of the circuit.

Yes, of course it would. It's not the same circuit anymore. There would be similarities, but you can't use the common equations for series or parallel LCR circuits.

If yes, what factors do we need to consider to find out the quality factor of any such arbitrary LCR circuits?

You'd need to do circuit analysis on the proposed configuration and calculate its bandwidth and attenuation. Any 2nd year EE student could do this (should be able to do this...)

This was touched on already --- see here: $Q$ factor of parallel RLC circuit in series with a capacitor and resistor

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