An AC circuit has $$R = 100\,\Omega$$, $$C = 2\,\mu F$$ and $$L = 80\,\text{mH}$$, connected in series. The quality factor of the circuit is:

We have a series $$RLC$$ circuit whose resistance, capacitance and inductance are given as $$R = 100\,\Omega,\; C = 2\,\mu\text{F} = 2 \times 10^{-6}\,\text{F},\; L = 80\,\text{mH} = 0.08\,\text{H}.$$

The quality factor $$Q$$ of a series resonant circuit is defined by the standard relation

$$Q \;=\; \dfrac{\omega_0 L}{R},$$

where $$\omega_0$$ is the angular resonant frequency. First, we find $$\omega_0$$. For a series $$RLC$$ circuit, the resonance condition gives

$$\omega_0 = \dfrac{1}{\sqrt{LC}}.$$

Substituting the given inductance and capacitance, we get

$$\omega_0 = \dfrac{1}{\sqrt{\,L\,C\,}} = \dfrac{1}{\sqrt{\,0.08 \times 2 \times 10^{-6}\,}} = \dfrac{1}{\sqrt{1.6 \times 10^{-7}}}.$$

Now, $$1.6 \times 10^{-7} = 16 \times 10^{-8} = (4^2) \times 10^{-8},$$ so

$$\sqrt{1.6 \times 10^{-7}} = 4.0 \times 10^{-4}\; \text{(approximately)}.$$

Hence,

$$\omega_0 = \dfrac{1}{4.0 \times 10^{-4}} = 2.5 \times 10^{3}\,\text{rad s}^{-1}.$$

Now we return to the quality-factor formula:

$$Q = \dfrac{\omega_0 L}{R} = \dfrac{\left(2.5 \times 10^{3}\right)\left(0.08\right)}{100}.$$

Simplifying numerator first,

$$\left(2.5 \times 10^{3}\right)\left(0.08\right) = 2.5 \times 0.08 \times 10^{3} = 0.20 \times 10^{3} = 200.$$

Now dividing by the resistance:

$$Q = \dfrac{200}{100} = 2.$$

So, the quality factor of the given series $$RLC$$ circuit is $$2$$.

Hence, the correct answer is Option A.