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?
Electric Circuits – Understanding Quality Factor of LCR Circuit
capacitanceelectric-circuitselectrical-resistanceinductanceresonance
Related Solutions
I mean, can I replace this configuration by one capacitor with one resistor in series such that this resistor is equivalent to the other two?
The answer is actually no.
For a single resistor and capacitor in series, the real part of the impedance is independent of frequency, i.e., the real part acts like a resistor.
$Z_s = R_s + \frac{1}{j \omega C}$
However, for the circuit you describe, the real part varies with frequency. This is easily seen by noting that at zero frequency, the impedance is real and equal to the value of the parallel resistance. At "infinite" frequency, the impedance is real and equal to the parallel combination of the two resistances.
The equivalent impedance of the circuit you describe is:
$Z_{eq} = R_p || (R_s + \frac{1}{j \omega C})$
This can be expressed as the sum of a real part and an imaginary part but the real part involves the radian frequency $\omega$.
So, although we can write the above as the sum of a real (resistive) part and an imaginary (reactive) part, the real part acts like a frequency dependent resistor.
If your circuit were to be operated at a single frequency, then, in that limited context, the answer is yes, one can replace the two resistors with an equivalent resistor for that particular frequency.
There's a really awesome trick for problems like this. This is going to be a long post but the method presented makes problems like this really easy.
The idea is to turn the series branch $C_2$, $R_2$ into an effective parallel $R$ and $C$. See the diagram. The effective parallel values are denoted $C_{2,p}$ and $R_{2,p}$. Parallel capacitances just add, so the total capacitance is now $C+C_{2,p}$. Parallel resistances add in parallel so the total resistance is now $R||R_{2,p} = \left( 1/R + 1/R_{2,p} \right)^{-1}$. Since we now have a purely parallel circuit, you can stick these values into your formula for $Q$ (which was wrong in the OP, by the way, but I edited it).
Of course, to actually do any of this we have to understand how to solve for $R_{2,p}$ and $C_{2,p}$.
Before we do that I want to simplify some notation. It is extremely useful to define $Z_{LC} = \sqrt{L/C}$. This is the "characteristic impedance" of a resonant mode, and it will show up all over the place. With this definition, the equation for the $Q$ of a parallel $RLC$ resonator is $$Q = R/Z_{LC}$$ which is really easy to remember: if $R\rightarrow \infty$ then no current flows through the resistor so there's no energy loss, and as we can see $Q\rightarrow \infty$.
Series/parallel equivalence
Suppose we have a series resistance $R_s$ and reactance $X_s$. The total series impedance is $$Z_s = R_s + i X_s .$$ We want to find the equivalent parallel circuit. The impedance of a parallel resistance $R_p$ and reactance $X_p$ is $$Z_p = \frac{iR_pX_p}{R_p + iX_p} = \frac{R_pX_p^2 + iR_p^2X_p}{R_p^2 + X_p^2}$$ Now define a new symbol $Q_p \equiv R_p/X_p$. Using this we can rewrite $Z_p$ as $$Z_p = \frac{R_p}{1+Q_p^2} + iX_p \frac{Q_p^2}{1+Q_p^2}.$$ Since this is now just a sum of a real number and an imaginary number, it's obvious what the equivalent series values are: $$R_s = R_p \frac{1}{1+Q_p^2} \qquad X_s = X_p\frac{Q_p^2}{1+Q_p^2} . \qquad (*)$$ Unfortunately we have solved the problem in the wrong direction: we found series values in terms of parallel ones instead of the other way around. To solve this problem, define $Q_s \equiv X_s/R_s$, and divide the two equations in $(*)$ by one another to find $$Q_s = Q_p .$$ Now we can easily invert $(*)$ to find $$R_p = R_s(1 + Q^2) \qquad X_p = \frac{1 + Q^2}{Q^2} X_s$$ where we now write $Q$ instead of $Q_s$ or $Q_p$ because we just showed that they are equal. We now have the parallel values in terms of the series values. The best part is that almost always when you have a circuit like the one in the original post, you have $Q \gg 1$, which simplifies the transformation equations considerably to $$R_p \approx R_s Q^2 \qquad X_p \approx X_s . $$ The take-home message is that the equivalent parallel resistance is transformed to a much larger value, and the equivalent reactance is basically the same as the series value.
Solve the original problem
In the original problem we have $$ \begin{align} R_s &= R_2 \\ X_s &= \frac{1}{\omega C_2} \\ Q_e &= \frac{X_s}{R_s} = \frac{1}{\omega C_2 R_2}. \end{align} $$ where I've written $Q_e$ to indicate that this is the $Q$ of the "external" circuit. The equivalent parallel values are $$ \begin{align} R_{2,p} &\approx R_2 Q_e^2 \\ X_p &\approx X_s \rightarrow C_{2,p} \approx C_2 \end{align} $$ We now have a new fully parallel circuit with $$ \begin{align} \text{resistance} &= R||R_{2,p} \\ \text{capacitance} &= C + C_{2,p} \approx C \qquad \text{assuming }C \gg C_2 \\ \text{inductance} &= L \end{align} $$ The $Q$ of the circuit is $$ \begin{align} Q &= \text{resistance} / Z_{LC} \\ &= \left(\frac{1}{R} + \frac{1}{R_2 Q_e^2} \right) ^{-1} / Z_{LC} \\ \frac{1}{Q} &= \frac{Z_{LC}}{R} + \frac{Z_{LC}}{Q_e^2 R_2} \\ &= \frac{1}{Q_i} + \frac{1}{Q_c} \end{align}$$ where we've defined $Q_i \equiv R / Z_{LC}$ which is the internal $Q$ of the circuit without the external series branch, and $Q_c \equiv Q_e^2 R_2 / Z_{LC}$ is the extra $Q$ induced by the coupling. In other words, when you add the series branch, the total $Q$ of the resonance winds up being a parallel combination of two components:
$Q_i$: The $Q$ you would have without the coupling to the series branch.
$Q_c$: The $Q$ you would have if $R$ were absent. This part comes from the coupling to the series branch.
This is, of course, just a result of the fact that $R$ and the effective parallel resistance of the series branch $R_{2,p}$ add in parallel.
We now write down a useful expression for $Q_c$. First write $$ Q_e = X_s / R_s = \frac{1}{\omega R_2 C_2} .$$ Since we're talking about properties near resonance, we take $\omega \approx 1/\sqrt{LC}$ giving $$Q_e = \frac{\sqrt{LC}}{R_2 C_2}.$$ Then for $Q_c$ we get $$ \begin{align} Q_c &= \frac{Q_e^2 R_2}{Z_{LC}} \\ &= \frac{R_2 L C \sqrt{C}}{R_2^2 C_2^2 \sqrt{L}} \\ &= \frac{Z_{LC}}{R_2}\left( \frac{C}{C_2} \right)^2 . \end{align} $$ The constitutes a full solution to the problem.
Summary
$$\frac{1}{Q} = \frac{1}{Q_i} + \frac{1}{Q_c}$$ where $$Q_i = \frac{R}{Z_{LC}}$$ and $$Q_c = \frac{Z_{LC}}{R_2}\left( \frac{C}{C_2}\right)^2.$$ The approximations made here are that $C_2 \ll C$ and $Q_s \gg 1$. The approximation $Q_s \gg 1$ is pretty good for $Q_s>3$.
Best Answer
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:
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
$$\omega_n = \sqrt{\frac{1}{L C}}, \qquad \zeta = \frac{R}{2} \sqrt{\frac{C}{L}}, \qquad Q = \frac{1}{R} \sqrt{\frac{L}{C}}$$