A series LCR circuit is required to have a resonant angular...

Given $\omega_0 = 1000 \text{ rad s}^{-1}$, $\Delta \omega = 40 \text{ rad s}^{-1}$, and $R = 20 \ \Omega$.

The bandwidth of a series LCR circuit is given by $\Delta \omega = \dfrac{R}{L}$.

Substituting the given values:

$40 = \dfrac{20}{L} \Rightarrow L = \dfrac{20}{40} = 0.5 \text{ H}$

The resonant angular frequency is given by $\omega_0 = \dfrac{1}{\sqrt{LC}}$.

Squaring both sides:

$\omega_0^2 = \dfrac{1}{LC}$

Substituting the values of $\omega_0$ and $L$:

$(1000)^2 = \dfrac{1}{0.5 \times C}$

$10^6 = \dfrac{2}{C} \Rightarrow C = 2 \times 10^{-6} \text{ F} = 2 \ \mu\text{F}$

Therefore, the inductance is $0.5 \text{ H}$ and the capacitance is $2 \ \mu\text{F}$.

Answer: $0.5 \text{ H}$ and $2 \ \mu\text{F}$