Obtain the resonant frequency

Inductance, L = 2.0 H

Capacitance, C = 32 μF = 32 × 10−6 F

Resistance, R = 10 Ω

Resonant frequency is given by the relation,

$\omega_{\mathrm{r}}=\frac{1}{\sqrt{L C}}$

$=\frac{1}{\sqrt{2 \times 32 \times 10^{-6}}}=\frac{1}{8 \times 10^{-3}}=125 \mathrm{~s}^{-1}$

Now, Q-value of the circuit is given as:

$Q=\frac{1}{R} \sqrt{\frac{L}{C}}$

$=\frac{1}{10} \sqrt{\frac{2}{32 \times 10^{-6}}}=\frac{1}{10} \times \frac{1}{4 \times 10^{-3}}=25$

Hence, the Q-Value of this circuit is 25.